Feeds:
Posts

Amazing Things we Find in Textbooks – The Real Second Law of Thermodynamics

Probably many, most or all of my readers wonder why I continue with this theme when it’s so completely obvious..

Well, most people haven’t studied thermodynamics and so an erroneous idea can easily be accepted as true.

All I want to present here is the simple proof that thermodynamics textbooks don’t teach the false ideas circulating the internet about the second law of thermodynamics.

So for those prepared to think and question – it should be reasonably easy, even if discomforting, to realize that an idea they have accepted is just not true. For those committed to their cause, well, even if Clausius were to rise from the dead and explain it..

On another blog someone said:

Provide your reference that he said heat can spontaneously flow from cold to hot. And not from a climate ‘science’ text.

I had cited the diagram from Fundamentals of Heat and Mass Transfer by Incropera and DeWitt (2007). It’s not a climate science book as the title indicates.

However, despite my pressing (you can read the long painful exchange that follows) I didn’t find out what the blog owner actually thought that the writers of this book were saying. Perhaps the blog owner never grasped the key element of the difference between the real law and the imaginary one.

So I should explain again the difference between the real and imaginary second law of thermodynamics once again. I’m relying on the various proponents of the imaginary law because I can’t find it in any textbooks. Feel free to correct me if you understand this law in detail.

The Real Second Law of Thermodynamics

1a. Net heat flows from the hotter to the colder

1b. Entropy of a closed system can never reduce

1c. In a radiative exchange, both hotter and colder bodies emit radiation

1d. In a radiative exchange, the colder body absorbs the energy from the hotter body

1e. In a radiative exchange, the hotter body absorbs the energy from the colder body

1f. This energy from the colder body increases the temperature compared with the case where the energy was not absorbed

1g. Due to the higher energy radiated from a hotter body, the consequence is that net heat flows from the hotter to the colder (see note 1)

The Imaginary Second Law of Thermodynamics

2a. – as 1a

2b.  – as 1b

2c.  – as 1c

2d.  – as 1d

2e. In a radiative exchange, the hotter body does not absorb the energy from the colder body as this would be a violation of the second law of thermodynamics

Hopefully everyone can clearly see the difference between the two “points of view”. Everyone agrees that net heat flows from hotter to colder. There is no dispute about that.

What the Equations Look Like for Both Cases

Now, let’s take a look at the radiative exchange that would take place under the two cases and compare them with a textbook. Even if you find maths a little difficult to follow, the concept will be as simple as “two oranges minus one orange” vs “two oranges” so stay with me..

Here is the example we will consider:

We will keep it very simple for those not so familiar with maths. In typical examples, we have to consider the view factor – this is a result of geometry – the ratio of energy radiated from body 1 that reaches body 2, and the reverse. In our example, we can ignore that by considering two very long plates close together.

E1 is the energy radiated from body 1 (per unit area) and we consider the case when all of it reaches body 2, E2 is the energy radiated from body 2 (per unit area) and we consider that all of it reaches body 1.

We define Enet1 as the change in energy experienced by body 1 (per unit area). And Enet2 as the change in energy experienced by body 2 (per unit area).

Radiation Exchange under The Real Second Law

E1 = εσT14; E2= εσT24 (Stefan-Boltzmann law)

Enet1 = E2 – E1 = εσT24 – εσT14

Enet2 = E1 – E2 = εσT14 – εσT24

Therefore, Enet1 = -Enet2

Under The Imaginary Second Law

Enet1 = – E1 = -εσT14

Enet2 = E1 – E2 = εσT144 – εσT24

Therefore, Enet1 ≠-Enet2 ; note that ≠ means “not equal to”

This should be uncontroversial. All I have done is written down mathematically what the two sides are saying. If we took into account view factors and areas then the formulae would like slightly more cluttered with terms like A1F12.

In the case of the real second law, the net energy absorbed by body 2 is the net energy lost by body 1.

In the case of the imaginary second law, there is some energy floating around. No advocates have so far explained what happens to it. Probably it floats off into space where it can eventually be absorbed by a colder body.

Alert readers will be able to see the tiny problem with this scenario..

What the Textbooks Say

First of all, what they don’t say is:

When energy is transferred by radiation from a colder body to a hotter body, it is important to understand that this incident radiation cannot be absorbed – otherwise it would be a clear violation of the second law of thermodynamics

I could leave it there really. Why don’t the books say this?

Engineering Calculations in Radiative Heat Transfer, by Gray and Müller (1974)

Note that if the imaginary second law advocates were correct, then the text would have to restrict the conditions under which equation 2.1 and 2.2 were correct – i.e., that they were only correct for the energy gain for the colder body and NOT correct for the energy loss of the hotter body.

Heat and Mass Transfer, by Eckert and Drake (1959)

Note the highlighted area.

Basic Heat Transfer, M. Necati Özisik (1977)

Note the circled equations – matching the equations for the “real second law” and not matching the equations for the “imaginary second law”. Note the highlighted area.

Heat Transfer, by Max Jakob (1957)

Note the highlighted section, same comment as for the first book.

Principles of Heat Transfer, Kreith (1965)

Note the highlighted sections. The second highlight once again confirms the equation shown at the start, that under “the real second law” conditions, Enet1 = – Enet2. Under the “imaginary second law” conditions this equation doesn’t hold.

Fundamentals of Heat and Mass Transfer, Incropera and DeWitt (2007)

Note the circled section. This is false, according to the advocates of the imaginary second law of thermodynamics.

And the very familiar diagram shown many times before:

From "Fundamentals of Heat and Mass Transfer, 6th edition", Incropera and DeWitt (2007)

Conclusion

There are some obvious explanations:

1. Professors in the field of heat transfer write rubbish that is easily refuted by checking the second law – heat cannot flow from a colder to a hotter body.

2. Climate science advocates have crept into libraries around the world, and undiscovered until now, have doctored all of the heat transfer text books.

3. (My personal favorite) Science of Doom is refuted because these writers all agree that net heat flows from the hotter to the colder.

4. Look, a raven.

Relevant articles – The Real Second Law of Thermodynamics

Notes

Note 1 – Strictly speaking a hotter body might radiate less than a colder body – in the case where the emissivity of the hotter body was much lower than the emissivity of the colder body. But under those conditions, the hotter body would also absorb much less of the irradiation from the colder body (because absorptivity = emissivity). And so net heat flow would still be from the hotter to the colder.

To keep explanations to a minimum in the body of the article in 1e and 1f I also didn’t state that the proportion of energy absorbed by each would depend on the absorptivity of each body.

323 Responses

1. […] Amazing Things we Find in Textbooks – The Real Second Law of Thermodynamics […]

2. SoD

You have obviously tried to cherry pick passages(which perhaps could have been more carefully written) to back up your claim that radiation from a colder body can increase the temperature of an object at a higher temperature.
This for the nth time is the standard definition of heat.

Heat is the process of transferring energy from a higher temperature body to a lower temperature body.

Heat has the thermodynamic capacity to do work.

If you wish I can supply a long list of quotes to back up the correct physics definition.

Unless you can give us examples where a colder bodies radiation DOES WORK between the lower temperature and higher temperature sinks, I for one will stick with the Physics definition of HEAT.

• I’ve been trying to get SoD to wrap himself in a cold, wet blanket to see if his body temp will rise. LOL!

3. Bryan:

Do you agree that the radiation from a colder body will be absorbed by a hotter body?

That is the question at issue.

In the past you have said this is not possible and does not happen.

The people who write heat transfer textbooks and go by the titles of professors of mechanical engineering say it does happen.

Look, a raven!

4. scienceofdoom

You are opening new posts at a record rate.
Hopefully this is not done so as to avoid answering awkward questions like this one unanswered from

Absorption of Radiation from Different Temperature Sources……………………………………..

To make matters crystal clear please, once again, for the cameras

One BB source produces 200J/s landing on a surface per second with total absorption.
The spectrum is centred around 2um
The first source is then removed and replaced by the second BB source producing 200J/s but centred around 50um and is totally absorbed.

If you are correct then the two sources will produce identical effects every second.

I think the 200J of 2um centred radiation will have far more capacity to do WORK than the 200J of 50um centred radiation.
What do you think?

5. I’ll take a look at your question over on that article.

6. scienceofdoom

…..”I’ll take a look at your question over on that article.

It really is important to get these fundamental definitions right, however boring or dull it might seem to you.
Otherwise you could be like the speck builder who builds a fancy house without laying proper foundations.
You know the rest.

7. Oh, no! Thermodynamics and engineering have been coopted to spread the climate lies too!

(just kidding)

8. on October 7, 2010 at 2:22 pm | Reply Leonard Weinstein

Science of Doom,
Bryan and others are referring to Heat rather than energy transfer. Your occasional loose us of the phrase cold body also transferring Heat to the hot body (which you have pointed out that you were actually referring to energy transfer) caused that confusion to Bryan. I was not confused once you clearly stated what you meant, but it would stop the confusion to others if you stayed with energy flux or transfer, or photon transfer, or radiation energy transfer. If with the clear distinction Bryan does not still get it, then you are wasting your time trying to correct someone that does not understand basic thermodynamics.

9. Leonard Weinstein

I’m afraid that SoD has gone a bit further than having a loose definition of heat.

He/she thinks that 200J of radiation centred around 2um is exactly thermodynamically equivalent to 200J centred around 50um.

• I’m a bit confused. Why is the wavelength of the radiation important if all the energy is absorbed. Shouldn’t the both 200J be equivalent as long as they are both totally absorbed by the surface (in each case the surface will have 200 J more energy).

• Yes, it has been pointed out before that 200J = 200J, but Bryan is steadfastly ignoring that.

I’m not sure or not if Bryan has given up on the idea that individual photons are not centered around any sort of spectrum; they are one specific wavelength.

There is also an equation for determining the energy in a photon. It takes more photons at a longer wavelength to equal the same amount of energy as a group of photons at a shorter wavelength, but 200J is still 200J.

• I think Bryan’s confusion here comes from optics considerations. It’s true that 200 W/m2 of sunlight can be efficiently converted to useful work, while 200W/m2 of the thermal IR in downwelling IR, say, can’t. But it isn’t a frequency issue, but a beam geometry one. Sunlight emanates from a small source, and can be focussed to achieve high temperatures. IR from the sky cannot be focussed, so no high temperature can be achieved. And generally, a given power density at low frequency will have been emitted by a larger area, which will inhibit focussing.

On the other hand, a 15 micron CO2 laser provides a parallel beam, and can be used to cut steel. Its energy could be highly efficiently converted to mechanical work

• Nick Stokes

…..”But it isn’t a frequency issue, but a beam geometry one.”…….

The Sun also transmits in the Infra Red and a filtered parallel beam should not be too hard to obtain.
The longer wavelength 200J/s (15um) would not have the quality of 200J/s (2um).
If equal areas collected each radiation then undoubtedly the higher frequency radiation would be capable of conversion into other energy forms with a much higher efficency.

10. Bryan and SOD: Much of the confusion about the second law of thermodynamics arises because energy is transferred between molecules over UNOBSERVABLE, sub-nanometer distances by COLLISIONS, but over OBSERVABLE distances (kilometers in the atmosphere) when transferred by RADIATION. The laws of thermodynamics were described before we understood what happens on a molecular scale. However, all of the laws of thermodynamics have been derived from statistical mechanics – the application of the simple laws of physics to large numbers of colliding molecules. At a molecular level, entropy arises from disorder of individual molecules and has nothing to do with the direction of heat (kinetic energy) transfer between molecules!

According to the scientific consensus cited above by SOD, radiation transfers energy from cooler objects to hotter objects, but more radiation flows from hotter objects to cooler ones – making the net flow from hot to cold. Critics say this violates the Second Law. If we examine conventional heat transfer at a molecular level, is kinetic energy (heat) ever transfered from a slower moving molecule to a faster moving molecule during a collision? The answer is yes, it happens all the time. The clearest example occurs when a slower moving (cooler) argon atom hits a faster moving (hotter) argon atom directly from the side. Due to the geometry of the collision, no energy is transferred from the hotter argon atom to the cooler one and some of the kinetic energy of the cooler argon is transferred to the hotter one. Is this a violation of the Second Law? No, the Second Law refers to the net result of a large number of collisions. On a molecular level, energy is sometimes transferred from colder molecules to hotter molecules by BOTH COLLISIONS AND PHOTONS. When radiation from the warm surface transfers energy to the colder atmosphere, this is analogous to molecular collisions that transfer kinetic energy from faster molecules to slower ones. When radiation transfers energy from the colder atmosphere to the warmer surface, this is analogous to molecular collisions that transfer energy from slower molecules to faster ones. Neither process violates the second law – which refers to the net transfer of energy.

If energy were never transferred from colder molecules to hotter molecules, we would not observe a Boltzmann distribution of thermal energy with a few molecules possessing much more energy than average. The hotter molecules are only hot because they received kinetic energy from cooler molecules during collisions.

Critics may object individual molecules have kinetic energy, but not a defined temperature – making it is incorrect to call them hotter or colder. They may say that temperature and heat refer to macroscopic phenomena and the second law only applies to net transfer of energy on a macroscopic scale. And they would be right. However, the individual molecule in the atmosphere that emits a photon towards the earth does not have a defined temperature and neither does the molecule on the surface that absorbs the photon. We don’t apply the second law to individual molecules exchanging energy by collisions and we shouldn’t try to apply it to individual molecules exchanging energy by means of a photon.

In conventional heat transfer, everything happens on an unobservable scale (less than 1 nanometer). With radiation, energy transfer takes place over kilometers in the atmosphere and 1.5*10^8 kilometers from the sun to the earth. To some extent, applying traditional thermodynamic principles about heat flow to the earth, atmosphere and sun is like observing our solar system from Alpha Centauri. Intelligent Alpha Centaurians – Bryan is thinking like an Alpha Centaurian – can analyze photons from our solar system, but they can’t observe of the details of the energy transfer between the earth, its atmosphere and the sun. All the Alpha Centaurians see is confirmation of the Second Law of Thermodynamics – that energy only flows from the hotter regions to cold intra-stellar space.

If Bryan wanted to observe transfer of thermal energy between molecules by collisions in a manner analogous to the way SOD describes energy transfer between the surface and the atmosphere by radiation, he would need to shrink himself down to the size of an electron. At that size, he would observe some collisions that transfer energy from cold to hot, but more transfer energy from hot to cold. Upon returning to normal size, the phenomena SOD describes would not seem so foreign.

• This is an important point. In 1850, when Clausius’ book was published, there was not much instrunentation for observing heat fluxes. They were always inferred from changes in heat content, so they were always nett fluxes.

With radiation, but not conduction, it is now possible to break a flux down into observed directional components. This makes possible the sort of arguments that we hear. With closely observed turbulent convection, you can also discriminate directional components, if you want to, and could probably devise many new second laws.

But the real law comes down to quantifying a state variable – entropy. And what counts is its nett amount and movement – not the pathways by which it got there.

• “The answer is yes, it happens all the time. The clearest example occurs when a slower moving (cooler) argon atom hits a faster moving (hotter) argon atom directly from the side. Due to the geometry of the collision, no energy is transferred from the hotter argon atom to the cooler one and some of the kinetic energy of the cooler argon is transferred to the hotter one. Is this a violation of the Second Law? No,”

The answer is no. However, your description of the molecules as being hot and cold is misleading. The molecule moving the fastest on the x axis, has more kinetic energy on that axis. However, on the y axis it might be slower than the molecule that collides with it on that axis. The slower x-mover is a faster y-mover and can transfer energy along the y axis.

11. Frank

…”According to the scientific consensus cited above by SOD, radiation transfers energy from cooler objects to hotter objects, but more radiation flows from hotter objects to cooler ones – making the net flow from hot to cold.”…

I agree with the above and the difference is what we call Heat.

But also the spectra from the hot object will illustrate energy of a higher quality than that of the lower temperature object.

What about Sods other idea that 200J of 2um centred photons is exactly thermodynamically equivalent to 200J of 50um centred photons?

• None of this has anything to do with my post, which carefully tried to explain why simple ideas about heat only flowing from hot to cold are derived from a macroscopic view of the world are as inappropriate for individual photons as they are for individual collisions. I think I have bought new (although too lengthy) perspective to the psuedo-dilemma of energy flow from hot to cold and would be interested in a constructive scientific discussion of why my rational may be right or wrong.

The “higher quality” of 200 Joules of energy with a 25-fold difference in wavelength (and therefore blackbody source temperature) has nothing to do with entropy OR the atmospheric greenhouse effect. There isn’t even a 2X difference between the atmosphere and the surface that receives radiation from it.

12. on October 7, 2010 at 6:19 pm | Reply Nullius in Verba

Not that I disagree with you – I’m just posting this for fun.

I note that when dealing with non-black bodies, all your sources assume that they are grey. What if they are not?

Hot body emits at frequency band 1 at which body 1 is black and body 2 is transparent or reflective. Cold body emits at frequency band 2 at which both cold and hot body are black.

What happens?

(Rather amusingly, a good example would be sunlight shining on greenhouse glass.)

.

PS. You need to be careful to word your second law to allow for refrigerators.

13. Frank

….”None of this has anything to do with my post”….

The top paragraph comes directly from your post.

I found nothing specific to disagree with in the rest.

There will as you say on occasion be an example of conductive energy transfer from a slow moving particle speeding up a faster one
With as you say statistically the vast majority going the other way.

The idea of shrinking folk down down to a molecular level to observe individual photon atom interactions was in fact one I had thought of with SoD in mind.

SoD often asks what happens when a photon from the colder surface strikes the hotter surface.
However I considered it too polemical a request.

The wide difference in wavelengths in the example was to crystallize a flaw in what seems to be SoDs view of energy and radiation.
You still seem to be sitting on the fence on this issue.

14. I’m not quite sure I understand your basic positions, Bryan and SOD. Give me a basic position on the following scenario: we have two separate totally vacuum chambers (A and B), each a 10 foot cube. Each contains a steel ball of roughly identical size at 300K. Chamber B, however, also contains an aluminum ball at 100K. Will the steel ball in chamber B cool less quickly than the steel ball in chamber A (the environments exterior to the chambers being equal)?

And if radiation from the steel ball in chamber B heats up the aluminum ball, will the aluminum ball stop absorbing the radiation from the steel ball when the balls’ temperatures equate?

15. Frank on October 7, 2010 at 3:28 pm:

Excellent explanation.

Statistical thermodynamics came long after the formulation of the basic thermodynamics laws.

One of the question they tried to answer was why does net heat flow from hotter to colder? Why not the reverse?

Statistical thermodynamics proves that with two bodies in a closed system, the overwhelming probability is that the hotter one cools down and the cooler one warms up until both end up in equilibrium.

Not “inevitable”, just “overwhelmingly probable”.

“Of all the gin joints in all the towns in all the world, she walks into mine.” – kind of the opposite of that but when the number of gin joints in the world is 10^25 higher. You find that yes, she always walks into this one. Perhaps a subject for another article at some stage.

• Thanks for the kind words. Comparing energy transfer by molecular collision with energy transfer by photons provides a solid rational why the Second Law is compatible with the greenhouse effect. During individual collisions, the molecule with the greatest kinetic energy was gain or (more often) lose energy. When photons pass between the surface and the atmosphere, the molecules with the greater kinetic energy can gain or (more often) lose energy. The concepts of temperature and entropy don’t become important until we discuss what happens when large numbers of molecules exchange energy by collisions or by photons and then the NET flow of energy is “always” from hot to cold. The only difference between energy transfer by collision and by photons is the distance over which energy is transferred. (Why couldn’t I write it this succinctly the first time?)

16. Bryan:

I seemed to have missed it, but do you think that the writers of these textbooks believe that energy from colder bodies is absorbed by hotter bodies?

17. Sod:

Statement 1a is a mis-statement: the physical property under consideration is “heat”, pure and simple; “net” is a counting term.

Statement 1c is incomplete: it’s missing the qualification “in a vacuum”. The condition is tacit in your illustration and in the text book extracts you cite. In a material environment, the direction of heat flow will depend on the temperature of the environment relative to those of the two bodies. If it exceeds those of both bodies, heat will flow from the environment to both bodies, as required by the entropy law; and vice versa. If it lies between the temperature of the two bodies, heat will flow from the hot body to the environment and from the environment to the cold body, again as mandated by the entropy law.

The maths assumes equal emissivities and, as you note, ignores absorptivity for the sake of simplicity. However, the bodies could have different emissivities and absorptivities, in which case the net energy loss from the hot body is unlikely to be equal to the net energy gain by the cold one. A material environment could make up the difference as required by the 1st law of thermodynamics; but how would that law be obeyed in the case of a vacuum?

Taking up Frank’s suggestion of eye-balling electrons to get a better handle on the issue, I see great confusion. In the atoms of the hot body, which is simultaneously losing energy and (allegedly) gaining it from the cold one, the electrons are simultaneously transitioning from higher to lower, and from lower to higher, energy orbits. Echoing DSL’s question, would this continue even at equilibrium?

If statistical mechanics has rendered classical thermodynamics obsolete, somebody should tell the punters to stop paying makers of refrigerators and air conditioners for something they could have for free from nature.

http://www.ce.utexas.edu/prof/hodges/site 2006/documents/thermodynamics.pdf

• John: Classical thermodynamics was invented to describe the behavior of large numbers of molecules in terms of work and “state functions” such as entropy, enthalpy and “free energy”. With the later development of statistical mechanics, the empirical state functions of classical thermodynamics were derived by applying Newtonian and quantum mechanics to large numbers of molecules. The situation is somewhat analogous to deriving PV=nRT (an empirical formula first created to explain observations made by Boyle and Charles) using the kinetic theory of gases. (The kinetic theory of gases postulates that the pressure of a gas is caused by collisions between molecules and the walls of their container and that the temperature of a gas is proportional to the total kinetic energy of all of the molecules.) Neither thermodynamics nor the ideal gas law are obsolete; instead these empirically derived laws can be understood in terms of the fundamental physics of molecules.

You wrote: “In the atoms of the hot body, which is simultaneously losing energy and (ALLEGEDLY) gaining it from the cold one…” If collisions of gas molecules only transfered energy from faster moving molecules soon to slower moving ones, all of the molecules in a gas soon would be traveling at the same speed. If you accept that the speed of all of the molecules in a gas is not the same (and instead follows the Boltzmann distribution), you logically should believe that some collisions transfer energy from atoms with less kinetic energy to ones with more. This doesn’t violate the second law, because thermodynamics only describes heat flow between large collections of a molecules.

Thus I suggested that we would be easier to understand the actual behavior of colliding molecules if we could watch colliding molecules on molecular scale. In fact, you can do so at: http://intro.chem.okstate.edu/1314f00/laboratory/glp.htm

When we are discussing energy transfer by individual photons, the situation is the same as with collisions. A photon may travel in either direction between a molecule in a colder location and a molecule in a hotter location, but the net flow of energy between a large number of molecules MUST be from hot to cold. The only difference is that photons can transfer energy over kilometers while collisions transfer energy over nanometers.

We can avoid some confusion by remembering that a single molecule has kinetic energy, but not a “temperature”. Temperature comes into existence when a collection of molecules collide to produce local temperature equilibrium. So a single collision OR PHOTON can transfer energy from a molecule with more kinetic energy to a molecule with less kinetic energy, but technically neither process actually transfers energy from hot to cold. Hot, cold, and temperature are terms that technically apply only to large collections of molecules – collections which are large enough that the NET flux of energy is always from hot to cold. When a photon is transferred in either direction between a CO2 molecule 10 kilometers above the ocean and a water molecule in the ocean, we don’t know the temperature or the exact kinetic energy of the CO2 or water molecules involved. (Temperature is undefined for a single molecule and the probability of a molecule having a given kinetic energy when the bulk gas around it has a given temperature is described by Boltzmann’s distribution!) We do know that the atmosphere 10 kilometers up is colder than the ocean and that the net flux of photons between these location is in the upward direction.

• Frank:

I appreciate your lucid explanation. However, I have difficulty visualising molecular collisions in air (apart from not being able to activate the demonstration you pointed to). Air is mainly space, as the production of liquid air, which reduces volume by 98%, demonstrates. The range of molecules is measured in nanometres. Doesn’t this combination imply that the molecular content of air is characterised by distances between them much larger than their individual ranges? How do they collide? Wouldn’t “collisions” more likely be between molecules and photons (which presumably occupy the entire volume of the atmosphere)?

18. SOD, you are really tireless. I am following with interest your series on the imaginary second law, “yes, Planck was right” and so on. In case you need some “help”. You can use an idea (Section 1.6) from Bohren and Clothiaux, “Fundamentals of atmospheric radiation”. They give an interesting point of view on why does the atmosphere “heat” the surface (more or less the same you have already explained 20+ times). Still, I think some readers will not get the idea. I really appreciate that you continue insisting. Good luck

19. Nick Stokes,

Nitpick: While it may be possible to construct a 15 micrometer CO2 laser, the usual wavelength is 10.6 micrometers. Other lines in the 9-11 micrometer range are also used.

http://www.rp-photonics.com/co2_lasers.html

20. Frank,

Finally a voice of reason with respect to the inappropriateness of a black body model applied to the atmosphere/surface radiative heat exchange even with the caveat that we ignore all the conduction and convection going on in the real world. However, you lost me at the point where you say “When radiation transfers energy from the colder atmosphere to the warmer surface, this is analogous to molecular collisions that transfer energy from slower molecules to faster ones.” I can’t do the statistical calculations, but I doubt that any photons from the top of the atmosphere make it all the way to the surface without interception by other molecules along the way. Three paragraphs later, you essentially claim that the case for radiation as opposed to conduction. I’m grasping for air here, but I don’t think the distances are km scale.

The surface will warm up from the atmosphere only when the surface cools enough that the net heat can move in that direction according to the laws of thermodynamics. During the day heat transfer moves the other way. Happens every day/night.

• When I illustrated my post with the example of a photon traveling between a CO2 molecule 10 kilometers above the surface of the ocean and a water molecule in the ocean, I chose 10 kilometers simply because I wanted a location in the atmosphere that unambiguously would be colder than the water. I wanted to avoid any meaningless debates the direction of net heat flow, but stupidly created a distraction about the probability of a photon traveling this far.

For those who are unwilling to accept what SOD tells them physics books actually say, I wanted to clearly illustrate why the principles of thermodynamics were never meant to apply to individual photons or individual collisions. The concept of temperature (hot or cold) doesn’t apply to individual molecules. Due to Boltzmann’s distribution, no one can know the difference in total energy of the molecules emitting and absorbing the photons (even though we know the temperature at these locations). Seen from this perspective, the entire debate is absurd! The Second Law only applies to the NET energy flux from a large number of collisions or photon transfers.

21. Chic Bowdrie,

You are correct that Frank’s example of a photon emitted by a CO2 molecule 10 km above the surface reaching the surface directly is extremely unlikely. Extremely unlikely, however, is not impossible. But that’s not the point. Most of the energy radiatively exchanged between the atmosphere and the surface in the 15 micrometer CO2 band is done within a few meters or less of the surface so the temperature is essentially identical and there is no net flow in either direction. But there is a net flow of energy coming in from the sun which has to be lost to space for the average temperature to remain more or less constant. If there is less flux to space because of ghg’s in the atmosphere, then the surface temperature must go up until the average fluxes in and out balance.

22. John Millet from October 8, 2010 at 12:28 pm:

..If statistical mechanics has rendered classical thermodynamics obsolete..

Frank has already answered this point, but for the sake of completeness, since I raised it..

No, that wasn’t what I said.

Statistical mechanics has helped to explain why classical thermodynamics is true in terms of the statistics of particles.

Classical thermodynamics showed that the second law of thermodynamics was true. It wasn’t clear why without statistical mechanics.

..Statement 1c is incomplete: it’s missing the qualification “in a vacuum”. The condition is tacit in your illustration and in the text book extracts you cite..

The examples from the text books are the simplest examples. Many pages are filled with more advanced problems.

This post is about simply demonstrating that experts in the field of heat transfer believe that radiation from colder bodies is absorbed by warmer bodies. The best way to demonstrate that point is with the simplest cases.

• Sod:

Your post wants us to focus on particular words in the various text books to the effect that the warmer body absorbs radiation from the cooler one. However, the words are incidental to, and need to be read in the context of, the authors’ purpose. This is to demonstrate that the quantity of heat flowing between two bodies (which see each other) is proportional to the difference between the fourth powers of their temperatures:

q = k (T1^4 – T2^4)

To illustrate how the formula is derived, they expand the equation – in effect re-positioning the bodies so they don’t see each other, when the heat flowing from each is proportional to the fourth power of its temperature.

In this state of mutual blindness, let the temperatures become the same and then bring the bodies back into eye contact. The heat flowing between them is zero. This would be the result if neither body either emitted or absorbed radiation. Do you think that the authors, had they been demonstrating this case, would have used words to that effect; or to the opposite effect, that both bodies simultaneously emit and absorb equal quantities of radiation?

The former option would imply an unexcited atomic state; the latter, maximum excitation, which seems anomalous given zero energy change. This can be illustrated by bringing the bodies, say a couple of plates of equal dimensions, into contact so that heat transfer, if any, is by conduction, not radiation. At temperature equilibrium there is no atomic excitation. What calmed them? Why would parting the plates excite the atoms on both surfaces?

Incidentally, I feel better knowing that statistical mechanics, far from rendering classical thermodynamics obsolete, has affirmed its validity – the refrigerator business is not a scam and it’s quite rational to acquire one and pay for electricity to keep my beer cold!

23. I like this website. Kudos to SoD for creating a space for inquiring minds to jump in. Having only recently tuned in, I have a couple more questions. I beg your pardons if this is too painfully covered many times already.

How certain are we of the typical values reported for incoming and outgoing radiation? Can we say unequivocally that the planet is out of balance? If so, are elevated surface and seawater temperatures rising as a direct result?

But is the converse necessarily true? Couldn’t a radiation balance still result in elevated atmospheric temperatures due to heat transfer from deepwater or the earth’s crust?

24. Chic Bowdrie:

How certain are we of the typical values reported for incoming and outgoing radiation?

At top of atmosphere the measurements are by satellite and as a result they cover the planet very well. Therefore, we are very certain of the typical values reported for incoming and outgoing radiation.

At the surface it is quite different. Pyranometers to measure solar radiation and pyrgeometers to measure the atmospheric radiation are expensive. Some measurements of the downward atmospheric radiation (DLR) are in the posts The Amazing Case of Back Radiation, along with the subsequent two posts.

The upward surface measurements are much more certain because we know temperature, and therefore from the Stefan-Boltzmann equation we can know the upward surface radiation very precisely.

Can we say unequivocally that the planet is out of balance?

No. For this we need to consider the top of atmosphere balance and the uncertainty in the satellite measurements is around 5W/m^2.

If so, are elevated surface and seawater temperatures rising as a direct result?

We can see that the heat stored in the planet has increased over the last few decades. The only way this could have happened is that the energy in has exceeded the energy out.

But is the converse necessarily true? Couldn’t a radiation balance still result in elevated atmospheric temperatures due to heat transfer from deepwater or the earth’s crust?

Theoretically yes, but every estimate of this heat flow from the interior produces values that are tiny by comparison.

25. I am not a wealthy person. On the other hand, the government has more money than god. I pay taxes — send money to the government. The government provides me with certain goods and services — food stamps, roads, public safety, national defense — greater in value than the taxes I pay.

1. There is a flow of money from the impecunius (me) to the flush (the federal government).
2. The net flow of money, goods and services is from the federal government to me.

For illustration only. Ignore for the moment the size of the national debt.

26. Reply John Millett 10/09/10 10 am.

There are probably dozens of demos on the Internet. Do a Google search for “applet” and Boltzmann distribution. Applets are small interactive applications used for demos that usually run in Java. Different browsers run different versions of Java. (Or update your browser.)

Most liquids expand about a 1000-fold when they evaporate at atmospheric pressure. For example, one mole of gas occupies 24.4 L (room temp) and one mole of liquid water fills 18 mL. In the liquid phase, surface (electron probability cloud) of the molecules are touching. When they evaporate, they move about 10 molecular diameters apart on the average in each of three directions.

Visualizing the atmosphere: If you condensed the atmosphere, it would make a layer of liquid roughly 10 m thick on the surface. (The 400 ppm of CO2 would be a layer 4 mm thick, a little thinner than glass in a car window). If you imagine that layer evaporating, molecules are touching on each side and the only direction the molecules can move apart is up. Therefore a 10 m layer of liquid air evaporates to a 10 km layer of atmosphere. About 5 kilometers up, the pressure from the weight of the above atmosphere is half the pressure at the surface, so those molecules move twice as far apart and rise a little higher. 3/4 the way through the atmosphere, they are 4X further apart. (Data shows that half the atmosphere lies below 5.5 km and 90% below 16 km.)

Photons don’t have a mass so it could be misleading to think about them colliding. Instead they have a probability (determined by quantum mechanics) of being absorbed when they pass through (or near?) a molecule.

• Frank, thanks:

“…they move about 10 molecular diameters apart on the average in each of three directions”.

By comparison how far do the molecules range?

• The data below is from the following web address. The site contains a lecture deriving these quantities from more fundamental principles of physics. http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/MolecularCollisions.htm

Here are some numbers for O2, N2: the molecular diameter is about 0.3 nm, the mean free path between collisions is about 60 nm, the speed of the molecules at room temperature (v) is approximately 500 meters per sec. (1100 mph), so the molecule has of order 10^10 collisions per second.

27. DSL, This example is actually a good analogy for the greenhouse effect, as the real question is (or should be) not whether the net flow of heat is from the hot body to the cold body (it is) but whether the presence of the cold body can slow down the heat loss of the warm body (it can).

The traditional peoplewhodontagreewithus-ists [moderator’s note, please check the etiquette ) understanding of the 2nd law can be tested by applying it to a more familiar situation — a blanket in a cold room. The blanket is initially the temperature of the room, it gets warmer when you wrap it around yourself but it will never reach body temperature because it is still losing heat to the room. So the 2nd law tells us there will be a net flow of heat from you to the blanket. Doesn’t mean you’d be warmer without it.

Bryan seems to have a different idea about how the 2nd law of thermodynamics disproves global warming, but he hasn’t explained it very well and hasn’t reconciled it with the 1st law of thermodynamics. It is true that sunlight has lower entropy than the Earth’s radiation and is in some sense higher quality energy. You can build a solar panel based on sunlight, do photosynthesis with it, but these can’t work with long wave radiation from the Earth. But IR radiation quite easily becomes heat because that is high entropy energy. It’s harder to do *useful* work with longwave radiation, but it does work all the same, 200 J of absorbed energy must do 200 J of work no matter what kind of energy it is, otherwise we are in violation of the 1st law of thermodynamics.

• 200 J of absorbed energy must do 200 J of work no matter what kind of energy it is, otherwise we are in violation of the 1st law of thermodynamics.

Work and energy in the form of heat content are not interchangeable. Energy can flow from hot to cold objects in physical contact without doing any work at all. You can’t do work if you have only one heat reservoir no matter it’s temperature. If you have two reservoirs at different temperatures and a heat engine, you can only extract part of the energy flow through the engine as work.

You can do work by heating a gas in a vertical cylinder with a free piston with a weight on top. If you heat the gas, the pressure increases and the piston goes up, lifting the weight against the force of gravity and doing work. But that’s only part of a cycle.

• Okay, work isn’t really the right word, I just mean energy transfered. If the ground absorbs 200 J of visible light, it now has an extra 200 J of heat, and likewise if it absorbs 200 J of infrared. If anything the “quality” of the energy argues in the opposite direction of what Bryan thinks it does — energy from solar radiation can be used by plants or photovoltaic cells to do work, energy from infrared can only be turned into heat.

28. The best example of heat transfer from a colder, to a hotter body, is the case of the fire bricks of a blast furnace. You will note that the blast furnace never comes to an equilibrium, there is always a temperature gradient from the pig iron/slag/coke and the outer brick. The vast majority of the heat is focused with in the furnace, the cold walls do increase the internal temperature.

However, the surface of the blast furnace is much larger than the inside, the amount of heat generated within the furnace is exactly matched by heat leakage, from the walls and from the flue.

The same is true for a sphere made out of a radioactive element, like plutonium or uranium. The core is hotter than the outside, heat is transferred from the cooler outside to hotter interior, which is why a sphere is the ‘hottest’ shape.

Now one can do a nice experiment. Build a thin-walled and a thick-walled furnace. Heat them up to the same temperature and then stop loading. The rate at which the larger furnace cools will be much slower than the thin-walled one; it will also take longer to come to its steady-state operating temperature.

So this gives us a simple test for the role of CO2 as an insulator or green house gas. The RATE at which the two poles heat up in the polar summer and cool in the polar winter should be changed by an increase in CO2.

The US and USSR/Russia have the temperature records of the Antarctic interior going back to the 50’s. During that time atmospheric CO2 has increased from 310 to 385 ppm.
These records should be like our blast furnace, high levels of CO2 should increase the rate at which the South Poles warms up and the rate at which it cools should be slowed.

Should CO2 drive AGW be true, one would be able to read it from the difference in the rate of heating and cooling at the poles.

29. DocMartyn,

Should CO2 drive AGW be true, one would be able to read it from the difference in the rate of heating and cooling at the poles.

That might be true if the atmosphere were static. It isn’t. There are substantial heat flows from the equator to the poles through the atmosphere. Also, greenhouse forcing is lowest at the poles because the height of the tropopause is lower at high latitudes resulting in a lower temperature difference between the surface and the tropopause. Antarctica has even lower CO2 forcing than the Arctic because the average altitude of the surface is something like 3 km, making the temperature difference even less than for the Arctic.

Then there’s the distribution of land between the NH and the SH. The higher fraction of land in the NH means it’s likely to warm faster than the SH. So the temperature time series of the Antarctic interior is probably not a very good measure of CO2 forcing.

30. Science of Doom,

On reading your article I had thought you had overdone the copying of so many textbooks and copying of whole pages. Was it really necessary?

Then I see the accusations of cherry picking and loose definition. So yes, repetition and whole pages were necessary.

It would appear the concept of a net flow is beyond some people.

31. This is the first time I have visited this website, and with this shoddy writing and logic, it shall be the last.

Cheers, Charles the Dr.P.H., University of Illinois

• CRS, Dr.P.H.: I’m sure that you will be much happier cheering the usually one-sided and sometimes half-baked science at WUWT. After all, everything you read that supports your preconceptions about AGW MUST be right and everything that contradicts your preconceptions must be wrong. On the other hand, if you think science is a never-ending search for the truth, read one of SOD’s earliest posts:

https://scienceofdoom.com/2009/12/13/understanding-the-flaw/

32. John Millet on October 10, 2010 at 5:42 am:

You just need to follow the maths. It’s pretty simple.

First, confirm that you agree with my comparison of the two ideas in the equations I posted in the article.

I said:

.. Enet1 = -Enet2

whereas you are saying (people who believe that the colder body’s radiation does not get absorbed by the hotter body are saying):

.. Enet1 ≠-Enet2

Do you see the difference? It is easy to separate the two ideas mathematically, and it is not a challenging problem. You don’t need to understand how to perform double integrals, just to understand the “=” sign.

Second, confirm what you think the writers of all these textbooks believe.

I say that they also believe: Enet1 = -Enet2.

This would prove your hypothesis wrong.

Or at least it would prove that the writers of these textbooks disagree with you.

• We define Enet1 as the change in energy experienced by body 1 (per unit area). And Enet2 as the change in energy experienced by body 2 (per unit area).

Radiation Exchange under The Real Second Law

Enet1 = -k(T1^4 – T2^4) (T1>T2)

Enet2 = k(T1^4 – T2^4)

Enet2 = -Enet1:

The exchange: What the hot body loses the other gains.

When T1 = T2, there is no exchange.

There is no Imaginary Second Law.

33. I’m actually very impressed by the discussion here, it is very educational and civil.
Keep it up, many people will benefit from these pages.

34. DeWitt Payne to DocMartyn,

DM:-Should CO2 drive AGW be true, one would be able to read it from the difference in the rate of heating and cooling at the poles.

DW-P: That might be true if the atmosphere were static. It isn’t. There are substantial heat flows from the equator to the poles through the atmosphere. Also, greenhouse forcing is lowest at the poles because the height of the tropopause is lower at high latitudes resulting in a lower temperature difference between the surface and the tropopause. Antarctica has even lower CO2 forcing than the Arctic because the average altitude of the surface is something like 3 km, making the temperature difference even less than for the Arctic.”

“Antarctica has even lower CO2”

Which matters not one damn to the CO2-AGW postulate.
Either CO2 recycles outgoing IR radiation or it doesn’t.
If the CO2 level changes from 300 to 380 ppm or from 150 to 190 ppm one should be able to see the effect.
Should the CO2-AGW postulate be correct, then the heating/cooling curves of the South Pole from 1956-1966 will be statistically different from 1999-2009. If the postulate is correct, the later decade will show a faster rate of heating and a slow rate of cooling.
This is not difficult, physics should work everywhere, you can’t claim that CO2 driven heating works everywhere except the South Pole.

• You seem to have ignored everything DeWitt Payne has said.
But in any case, the atmosphere and interactions with land and oceans are not as simple as a comparison with a blast furnace.

Ozone changes have had an influence.

The problem is that you assume only one variable has changed in the time you have stated. As you say, it’s not difficult, but then again, you have to include all the things that may have had an influence.

35. John Millet on October 10, 2010 at 11:08 am:

By agreeing with the equations I identified under the heading “Radiation Exchange under The Real Second Law” you apparently accept that the warmer body absorbs radiation from the colder body.

This is what I think, along with the writers of all these textbooks.

It seems as if you don’t think this, so perhaps you don’t understand how the equation is derived. Or perhaps you agree with me.

Just to confirm.. The hotter body (body 1) radiates energy and absorbs energy. In your rewritten equation:

Enet1 = -k(T1^4 – T2^4)

– The first term is the emission of radiation, the second term is the absorption of radiation that was emitted by the colder body.

If you believe the hotter body (body 1) does not absorb energy from the colder body (body 2) then your equation will be:

Enet1 = -k.T1^4

Which is it?

• Sod:

“If you believe the hotter body (body 1) does not absorb energy from the colder body (body 2) then your equation will be:

Enet1 = -k.T1^4”

Not so.

Believing the hotter body does not absorb energy from the colder body is entirely consistent with believing, as I think the authors do, that the energy exchange between them is proportional to the difference in the fourth powers of their temperatures. The energy exchange comprises energy changes in each body equal to:

+/- k(T1^4 – T2^4), plus for the colder (T2) body, minus for the hotter one.

That is, Enet1 = -k(T1^4 – T2^4)

Sod:

“Enet1 = -k(T1^4 – T2^4)

– The first term is the emission of radiation, the second term is the absorption of radiation that was emitted by the colder body”.

Not so.

The brackets contain the symbolic equivalent of the text “the difference between the fourth powers of their temperatures”, to which the energy exchange is proportional, as taught by the textbook authors.

True, the expression can be seen as the superposition of two bodies in isolation, each radiating into a zero temperature environment, when the temperature difference is the same as the temperature level. But that doesn’t mean that when the bodies “see each other”, each continues to radiate at its temperature level, as if the other wasn’t there. Rather, each becomes the environment of the other, and the radiation becomes proportional to the difference between the fourth powers of their temperatures. And it is all in one direction – from hot to cold; for there is no radiation from the cold body because the temperature of its environment, the hot body, exceeds its own temperature. There being no cold body radiation, the issue of absorption by the hot body doesn’t arise; nor conflict with the second law of thermodynamics.

• Ah, yes, the case of intelligent molecules.
They can apparently know whether they are on a cooler body and when they realise this, they decide to only absorb and stop emitting!

Neat little trick.

• John, I don’t get it. Are you saying that radiation zooming across space (let’s say across 100 parsecs) from a 1000K body will suddenly disappear or make a U-turn when it realizes that it’s in the presence of a 2000K body? Or will it bounce off the 2000K body without interacting whatsoever with that body? Or does radiation “die out” after a certain distance or time?

It would seem that you’re saying that no other body in the universe could radiate, except for the hottest body.

36. DocMartyn,

Your comparison assumes that over the decades 1956-1966 and 1999-2009 that every other feature of the weather in central Antarctica will average to the same value for the two decades with high precision so that any difference observed in the phase and magnitude of the annual temperature series between the decades can only be caused by CO2. If no significant difference is found then AGW is falsified.

My contention is that the effect in central Antarctica from the increase in CO2 is small and that it will not be possible to separate it from the noise caused by the much larger, IMO, variation in, to list a few possibilities, heat flows from lower latitudes, variations in the stratospheric ozone and possible systematic errors in the temperature measurements (automated weather stations being covered with snow, e.g.).

• I’d like to second your signal-to-noise explanation … which is why it’s time for SOD to discuss the location on the planet which is predicted to be most sensitive to warming by GHGs – “non-hotspot” in the upper tropical troposphere.

37. OK, this is fun. Take two bodies one at Th and the other at Tc, facing each other across a vacuum in a perfectly insolating chamber. Assume that the thermal radiation from the colder one cannot be absorbed by the hotter body. Where does the radiation go?

38. Eli Rabett:

They call this the “infra-red catastrophe”.

Max Planck appeared to solve the late 19th century version of the problem in his own way and got his Nobel prize. At the time many people thought his solution was a bit crazy.

In time, many people will know the names of our intrepid commenters who have defended the honor of the second law of thermodynamics.

Right now it looks as they have done this by driving a big crusher over every part of 20th century physics without even realizing.

But in time, a solution will be found which brings harmony..

39. John Millett on October 11, 2010 at 9:25 am:

You can make up whatever ideas you like to support your belief in the imaginary second law of thermodynamics.

I invite new readers to read what the textbooks writers actually say.

40. Eli/SoD

Yet another strawman repeated endlessly for no purpose and no educational value.

Most things on Earth are not in a vacuum. The textbooks illustrate only the simple case of a vacuum. John Millet is patiently telling you in various imaginative but apparently doomed ways that you are still not yet grasping that in real life, with air present, these textbook simplifications don’t actually work and the real calculation is iterative, including convection and using the correct difference form of the Stefan-Boltzmann equation for radiative heatflow which includes the environmental temperature, which is affected hugely by convection. And our experience of multitudinous real life tests of real life objects versus 3D radiative/convective models tells us that radiative heat flow from a low temperature body is insignificant compared to convective transfer. High temperature objects of course are a different story.

Yes the greenhouse effect exists, yes there is a net flow upwards, yes the 2nd law is only a statistical law (as even Planck was loathe to finally accept), yes the net upflow is small. But people are trying to tell you that the oversimplification of an unreal situation that you so prefer is misleading you into thinking you can ignore one thing in an interactive system then calculate the others separately. While sometimes that is possible it isn’t in 3D heat transfer – you just end up with the wrong numbers. You need to stop thinking about distant radiating objects and start thinking about transfer media and fluid flows if you want to get serious about realistic heat transfer.

In fact though this is the classical mistake of anyone who has never bothered to do realistic 3D heat transfer modeling.

And btw the 2nd law likely isn’t even applicable to the atmosphere.

• “And btw the 2nd law likely isn’t even applicable to the atmosphere.”

Care to elaborate on this? I think you will find the second law applies to all energy transfer.

As far as adding gases etc, this is kinda exactly what this discussion is about, but its been simplified to explain the basic mechanics of energy transfer… Specifically radiative transfer… End o the day, convection and conductive flows of energy, are also dependent on a differential between T1 and T2. This is what limits the flow from the higher temperature body through the lower.

• Have you read much of this blog? SoD has been doing many posts involving much more complicated calculations that account for things like convection. However, he often runs into (in comments) the idea that this can’t possibly be correct because his calculations show the Earth absorbing radiation from a cooler layer in the atmosphere, which violates the 2nd law. Hence this post. This is not a strawman argument just because you wouldn’t make it; that the greenhouse effect violates the 2nd law is a common ——-ist [moderator’s note, please read the etiquette] trope. Roy Spencer got a lot of mail from his readers trying to inform him that the 2nd law disproves the greenhouse effect when he explained that the effect is real. So it was of no educational value to you; that’s fine, don’t take it personally that SoD felt this needed to be said.

41. Warmcast and the intelligent molecule

More likely the intelligent atom – based, as usual, on sensing the environment. Sensing a warmer environment, electrons transition to higher energy orbits in order to accomodate entering photons, or energy absorption; and vice versa. One or the other: absorption or emission. But…..simultaneous absorption and emission with implied attendant electronic chaos which is at its height at temperature equilibrium?

DSL and the vastness of the universe

Radiation does “die out”, to use your term, with distance – we feel the sun’s radiation less keenly than did Icarus. Though the radiation from either of the two stellar bodies starts at effective temperatures higher than the 2.7K of space, it eventually equilibrates to that temperature. Both bodies exist in this low-level energy field; both emit into it; neither absorbs from it. However, depending on the temperature difference and the distance between them, the colder body can find itself in the energy field of the hotter body. It would then cease emitting into the low-level energy space environment and start absorbing from its new neighbour’s high-level one. But would gravity wring even bigger changes in the cold body’s behaviour?

Eli Rabett in search of the missing radiation

Radiation from Tc goes nowhere because it doesn’t start. It is prevented from doing so by its hotter environment, being the photon gas emitted by Th which fills the enclosed space – a material vacuum, not an energy one.

Scienceofdoom

Adding to others’ gratuitous suggestions about future articles, I think an exploration at the atomic level would be useful.

• That’s a rather peculiar approach to radiative interaction. However, how do you propose the mechanism for the “sensing” of the environment works. In order to know about what kind of objects are around it, one must interact with them somehow?

PS. I think it should be pointed out that a single atom can’t simultaneosly emit and absorb radiation (at least by most theories), but as there are quite a lot of them atom thingies about in a body generally, it really isn’t much of a problem, as some of them can absorb radiation while others emit and vice versa.

• Mait:
It’s the same mechanism that makes natural processes follow the entropy law and move systems to states of increasing disorder or entropy, the temperature potential between the body and its environment. What mechanism will tell which atoms to absorb and which to emit?

• John Millet: “It would then cease emitting into the low-level energy space environment and start absorbing from its new neighbour’s high-level one.”
Simple question John.

You have two ‘bodies’ one is hotter or brighter than the other.
Are you seriously suggesting that the least bright of the two stops emitting towards the other body, just because it is less energetic?

If you stood between the two, do you expect the least bright one to go black whilst it absorbs photons from the brighter one?
Then when they reach the same temperature the once cooler body suddenly becomes the same temperature?

This seems to be the universe you are painting, unless I am interpreting your strange models incorrectly.

• Warmcast:

A model universe that obeys the entropy law shouldn’t seem strange. The two bodies are immersed in a medium, if not a material medium then an energy one, or photon gas. (Your question overlooked this condition). The temperature potential between a body and the medium determines its radiative properties – strength and direction. The entropy law requires that the direction of radiation in a system moves the system to states of increasing disorder, or that entropy of the system may not reduce. This requires heat to flow from hot to cold.

If the less bright body of your question is less energetic (colder) than the medium it would cease emitting into the medium (incidentally, towards the other body). It would instead absorb from the medium (incidentally, from the other body).

42. You are of course precisely correct. One day someone will find a way to beat the facts of radiative heat transfer into the adiabatic skulls of the AGW crowd.

The science they perform would get them failing grades in any junior level Chemical Engineering program. They are very good at presenting their information in such a way that untangling what they are saying takes some patience, but it is worth that effort to see the truth.

John Kehr
The Inconvenient Skeptic

43. JamesG:

Yet another strawman repeated endlessly for no purpose and no educational value.

Most things on Earth are not in a vacuum. The textbooks illustrate only the simple case of a vacuum. John Millet is patiently telling you in various imaginative but apparently doomed ways that you are still not yet grasping that in real life, with air present, these textbook simplifications don’t actually work and the real calculation is iterative, including convection and using the correct difference form of the Stefan-Boltzmann equation for radiative heatflow which includes the environmental temperature, which is affected hugely by convection. And our experience of multitudinous real life tests of real life objects versus 3D radiative/convective models tells us that radiative heat flow from a low temperature body is insignificant compared to convective transfer.

Actually, no John Millet isn’t saying that.

The strawman as you put it is my attempt to reconstruct what the various brave yet scientifically inaccurate people put forward in incomplete and contradictory terms.

Let’s take commenter Bryan, our bravest proponent of what I humorously like to call “the imaginary second law of thermodynamics”.

Bryan has subscribed to a number of views throughout the long campaign I have waged to find out what he actually thinks. But one thing has been clear, until quite recently – and paraphrasing now from 100 or 1000 exchanges – “a hotter body cannot absorb energy from a cooler body”.

He has explained in absolutely certain terms that the energy from the colder body “reaches” the hotter body.

But Bryan has never explained where the energy goes after is “reaches” the hotter body. “Mostly reflected” was a view one time explained but never followed up with evidence or explanations when evidence to the contrary was presented. And he never explained how in physics terms the energy could be not absorbed, given that this appeared to contradict basic physics.

And also Bryan has not said “in a vacuum the hotter body does absorb energy from the colder body”.

Let’s take commenter John Millett now beginning to articulate his views on the subject.

He has also not said “in a vacuum the hotter body does absorb energy from the colder body”.

His view appears to be that the colder body somehow doesn’t radiate towards the hotter body.

This is totally contrary to Bryan. And equally wrong.

Let’s take commenter Gord who flourished for a while. His view was that because EM radiation was a vector, therefore, radiation in opposite directions would cancel out. He was wildly entertaining in his own way.

There have been others, whose names are preserved for posterity on this blog.

But these three and many more have precisely and certainly articulated the view that energy from a colder body CAN NEVER be absorbed by a hotter body. No provisos. No “perhaps in a vacuum it’s all good”. No “well it does but convection is much more important”.

So, that’s why I write these articles. Because many people believe a falsehood.

Amazing that you can call it a strawman.

Onto “textbook simplifications” – I will just comment that it is easy to throw accusations around to show how very sophisticated your ideas are.

All I want to present here is the simple proof that thermodynamics textbooks don’t teach the false ideas circulating the internet about the second law of thermodynamics.

I haven’t claimed that this in any way proves anything about the “greenhouse” effect on earth, or the effectiveness of convection, or baroclinic instabilities, or the polar vortex, or AGW. If I have, you can refute it.

If you think something specific is in error in another article – perhaps one of the few on convection, go ahead and point that out.

Many people are very happy to wave their arms and say “this is wrong, that is wrong”.

Awesome.

Make sure it’s poetic, because unless you make specific detailed claims backed up with analysis and maths, it is at best, just poetry.

44. John Kehr

You are of course precisely correct. One day someone will find a way to beat the facts of radiative heat transfer into the adiabatic skulls of the AGW crowd.

The science they perform would get them failing grades in any junior level Chemical Engineering program.

Assuming that you are cheering on JamesG or John Millet..

So it should be very easy for you to point out the specific flaws. After all, “junior level” sounds pretty simple.

Or perhaps you will join the ranks of the many people who make claims they can’t back up.

45. It isn’t a function of complexity but one of application.

The statement of the 2nd law of thermodynamics is not precisely stated in the above article. All bodies radiate energy. That is a given. Net flow of energy cannot go from a colder body to a warmer body. For example. You hand is always radiating energy at th ~37 °C. When you are cold and you put your hand near a fire to warm it, the net flow is from the fire to your hand. You hand still radiates, but gains net energy from the fire.

Much like the atmosphere is on average cooler than the surface of the Earth so the net energy flow is from the surface to the atmosphere. The cooler atmosphere cannot give more energy to the surface than it receives. That would be the correct statement of the 2nd law of thermodynamics in regards to the surface and the atmosphere.

Net energy is what matters in almost all cases so the atmosphere is always receiving more energy than it gives to the surface. By the same token space is always colder than the atmosphere so the atmosphere is always radiating energy into space.

Most of the disagreements stated by both sides on this issue are more about differing terms and definitions. All bodies radiate heat, this includes the atmosphere. Net heat transfer is from warmer to colder bodies. Radiative heat flows both ways, but the colder body receives more than it gives.

Getting those issues confuses would most assuredly impact an engineers grade.

• “You of course are precisely correct”

No, John K, I think John M is going to have to disagree with you. For him, there is no such thing as “net flow.” Molecules “sense” a warmer body and stop radiating toward that body.

• The result being that without greenhouse gases the temperature gradient would be greater and the net loss from the surface would be greater.
Hence with greenhouse gases present the net loss will be lower and the surface stays warmer than would be the case without greenhouse gases.
Ok so the main input is the Sun, never the less the surface warms because of the presence of greenhouse gases.

Or am I missing some more magic?

• The greenhouse effect reduces the rate that the surface loses heat. In effect it prevents the surface from cooling faster.

You “could” look at it like the atmosphere “downwells” energy to the surface, but that is not the proper definition of what is happening.

The greenhouse slows the rate of surface cooling, it does not warm the surface, except in the case where the air is warmer than the surface. That does happen at times and in that case there is warming, but the thermal mass difference means that the air cools much more than the surface warms.

46. Technically the surface is always cooling, but that is a play of words and language, it doesn’t mean that the surface temperature will not get warmer with greenhouse gases present. Increase greenhouse gases and the ‘net’ impact is a warmer surface (excluding other obvious variables that we could argue about) that is still cooling.

I don’t think we disagree do we?

47. If the insulation effect of the atmosphere significantly increased the rate of cooling would decrease and then the surface would warm up. So we do agree in the concept.

The question is what impact does higher CO2 levels have on the insulation effect?

My analysis is that CO2 has a limited effect on the insulation effect of the atmosphere. I am putting that detailed analysis together now.

If I had found that CO2 did in fact cause significant increase in the insulation effect, I would strongly be arguing the other side.

48. Warmcast on October 12, 2010 at 5:26 pm:

The result being that without greenhouse gases the temperature gradient would be greater and the net loss from the surface would be greater.
Hence with greenhouse gases present the net loss will be lower and the surface stays warmer than would be the case without greenhouse gases.
Ok so the main input is the Sun, never the less the surface warms because of the presence of greenhouse gases.

The right way to consider the effect of more “greenhouse” gases, all other things being equal, is to consider how the whole climate system cools to space.

See The Earth’s Energy Budget – Part Three for some more detail on this subject.

Of course, all other things aren’t equal, and so the problem is very challenging to solve. But in the first instance if we can understand the problem without feedbacks then at least we have understood the basics.

49. Gentlemen,

I don’t want to throw cold water on this heated discussion, because like several other commentators, I find the discussion fascinating. However, much of the argument seems to me more semantic than scientific. Here are my philosophical two bits.

The black body models are not too useful to describe the atmosphere as Frank and others have pointed out earlier. But they to help in understanding radiative exchange vs. classical thermodynamics as has been demonstrated here. The energy transfers between black bodies in a vacuum don’t happen in discrete events as if we could replay each exchange and measure the energy transfers that are described by the textbook equations. Scienceofdoom’s 1f statement above in his version of the real 2nd Law helps understand what happens, but that never actually happens. The colder body’s radiation will be absorbed by the hotter body, but never enough to raise the temperature of the hotter body. OK, so much for semantics.

Now here is my virtual experiment. Consider an example of a system of a small hot black body in the proximity of a larger and colder one. Both lose heat to their surroundings. They are identical except for size and temperature, made of material which equilibrates internally instantaneously. The temperatures of the two bodies are plotted with respect to time. At no time does either body warm. The hotter body’s temperature decreases faster by virtue of its smaller mass until such a time that both bodies are the same temperature. Subsequently, the large (previously colder) body is now warmer than the small (previously hotter) body. Both bodies continue to radiate and lose heat. At all times, whichever body was hotter absorbed radiation from the colder one (and vice-versa), but a colder body never warmed a hotter one.

50. Chic Bowdrie:

Now here is my virtual experiment..

You can find this along with some other experiments in The First Law of Thermodynamics Meets the Imaginary Second Law. That is example 2.

When you consider a more complex example with the sun heating two bodies, you find that – from the 1st law of thermodynamics – the hotter body is a little hotter than the case without the second body.

Well, take a look.

51. ScienceofDoom, I will take a look.

I apologize to all female commentators I discounted by addressing my last comment to the guys.

52. Cool to have studied radiation in such detail.
Exchange of energy by radiation in a closed, dry container now seems well covered if not trivialized, horse-flogged and had his straw strewn by the flying monkeys.

I’m naturally anticipating the examination of the other mechanisms of heat transfer that are visibly responsible for the weather we see and experience on a daily basis.

This blog has become a biographical entity recording personal growth – particularly the process of building knowledge around a scotoma.

P.S. did you ever notice that the sun is white? It was never yellow… this still amazes me that I would have picked the yellow crayon to color the sun despite having seen it over a million times…
Isn’t it shocking how one can see what he thinks he’s supposed to see regardless of the truth in his face?

53. There is one point that seems to have been overlooked in the whole AGW debate. The cause of warming of any sort is, obviously, heating. Not the effect of the insulator/blanket/CO2 or any any other passive mechanism. That will merely slow down the rate of loss of heat.

So, AGW can be said to be initially the result of Anthropogenically Generated Warmth — which coincidentally generates the same acronym.

Now, we are talking about the INCREASE in average temperature over that resulting from, say, our pre-industrial baseline. In other words, we (Mankind) are generating more heat. Therefore things get warmer (than before.) What’s the problem?

Now, if you are saying that the extra insulating properties of CO2 or other “greenhouse” gas are causing the temperature to remain hotter than it would be without them, I would say “show me.”

Doubling or tripling the amount of CO2 in a real greenhouse seems to result in no observed temperature increase over that observed without the additional greenhouse gases.

You might as well replace 600 parts per million of your woollen blanket fibres with some fibre proved to have superior insulating properties (although the insulation is more due to preventing conduction), and expect to notice any difference in performance.

If all fossil fuel combustion produced energy was replaced with none greenhouse gas producing energy, would the Earth’s temperature drop? I don’t know, and I’ll guess neither does anybody else. Nothing magic, and if you can produce different facts, I will change my mind.

• This is going way off topic.

The atmosphere in the greenhouse isn’t very thick (as Bryson pointed out) and you would expect a scaled down response to sunlight because it isn’t so thick (in total the environment would be cooler than would be the case for the Earths atmosphere, despite doubling or tripling the CO2). You would I guess expect a lapse rate where by the greenhouse temperature at your feet would be a bit warmer, but because the thickness of the greenhouse atmosphere is thinner than the Earth, you would get more rapid cooling than is possible in a thicker atmosphere. You wouldn’t expect the temperature to increase uniformly, especially if the air is relatively still.

I doubt if you have any experimental evidence or any expert analysis to refer to, that details what is going on.
You have a lot of problems with such a set up, the glass, plants, ventilation etc. that change the way CO2 would play a part.

Plants cool the surrounding environment, it is a way to counter the heat island effect of cities.

Another point is that a farmer would never let the temperature increase and would open up ventilation to maintain a temperature acceptable for the crop.
Maybe you should be asking farmers with failed ventilation systems, whether the greenhouse got to warm?

Basically, unless a very strict experimental environment is maintained by some trained scientists, your ideas have no authority or credibility.

54. Mr. Flynn:

Causes are things that make a difference. If you’re comparing a world like Earth with one CO2 level and another world that’s just the same except for a higher CO2 level, the higher CO2 is the only difference there to discuss– and it leads to higher temperatures in a causal way, by causing the planet to retain more of the heat energy flowing into the earth/atmosphere system. It then warms up slowly until it gets warm enough at the (now higher) TOA (from which infrared can actually escape the atmosphere directly) to radiate back as much energy (in infrared) as it’s receiving from the sun (mostly visible light).

For the rest, either do the math and see what you get, or walk away.* Fussing about ‘real’ greenhouses is not exactly relevant– they warm by reducing losses to convection, and CO2 at that concentration and in that small a region really does make little difference: perhaps you have noticed that the atmosphere is rather deeper than the thin layer of air contained in a greenhouse.

As for the blanket and its fibres- another absurd and unjustified simile- well, I suppose it depends on the properties of those fibres. If they actually cut off heat losses substantially (as that level of CO2 in the atmosphere would indeed do– again, check the maths), then it would make a difference.

Further, showing you is not an option if you can’t be bothered to actually look at the physics beyond blustering about silly comparisons. So what are you asking for? A God’s eye, eternal view so you can see it happening? Or are you proposing the rest of us wait around 50 years for you to realize, whoa! It is getting warmer!

To answer your closing question, my bet would be that the Earth’s temperature would stabilize over a few decades and then, as CO2 fell slowly towards its previous state (rather under 300 ppm), it would return to something like the pre-anthropogenic climate (which was, I gather, on the whole, and very slowly, cooling). But no guarantees– if we’ve actually triggered enough CO2 and methane feedbacks, it could already be beyond our control (on any human timescale). The odds of that go up the higher we push the level of GHGs.

*Note: it’s already been done; see various texts on atmospheric physics and the greenhouse effect– and, for a nice, clear introduction, the treatment of the radiation exchange etc. provided in the many posts to be found here.

55. Long ago I used to wonder about the amount of “warmth” generated by mankind. It is significant. The energy generated is roughly 15 TWhr’s a year. If you assume it was generated at 33.3% efficiency then the warmth produced is 45 TWhr’s worth of heat per year (that also assumes that in the end electricity ends up converting to heat through usage, trying to crank up the heating effect).

That seems like a lot. It really does… Until you realize that the Earth’s surface absorbs that much energy from the sun every 0.08 seconds. Suddenly, I stopped worrying about the warmth that humankind generates. The Earth also sends energy out faster than the surface absorbs it.

John Kehr
The Inconvenient Skeptic

56. An aside for a moment.

What is the claimed total forcing of CO2 in the energy balance. I have been able to find countless estimates for the change in forcing, but nothing for the absolute forcing.

A reasonable range would suffice, especially if there is a reference. I am guessing that the proposed value is 50-100 W/m2, but I have been unable to find anything to verify this.

I am very curious about this though. If you have such an estimate I would greatly appreciate if you let me know what it is.

Thanks,
John

57. Mike Flynn:

There is one point that seems to have been overlooked in the whole AGW debate. The cause of warming of any sort is, obviously, heating. Not the effect of the insulator/blanket/CO2 or any any other passive mechanism. That will merely slow down the rate of loss of heat..

..Now, if you are saying that the extra insulating properties of CO2 or other “greenhouse” gas are causing the temperature to remain hotter than it would be without them, I would say “show me.”

I used insulation as an example in Do Trenberth and Kiehl understand the First Law of Thermodynamics? to explain in very simple terms how the surface radiation can be higher than the top of atmosphere radiation. And how that doesn’t violate any laws of thermodynamics.

Many other people use simple terms to convey an idea about how various trace gases in our atmosphere can increase the surface temperature.

These simple terms aren’t “the theory”.

Radiative physics is a little harder to understand than conduction because it is not in people’s everyday experience. That’s why simple parallels are often drawn.

The whole climate system will heat if less energy is radiated to space than is absorbed from the sun.

But the various radiatively-active gases in our atmosphere are not acting as “insulators” (well, actually they are very poor conductors of heat but that is not what we are talking about).

It is important to understand the actual theory before being able to say whether it has flaws.

58. @Bryson Brown.

Your first paragraph demonstrates that you do not understand what type and quantity of radiation from the Sun reaches the Earth’s surface, or what happens to it thereafter.

If, your direction that I “do the math” involves repeating the same incorrect assumptions about the shape, orbit, albedo, amount and type of the Sun’s radiation, I must respectfully decline.

The Earth is an asymmetrical oblate spheroid. Its albedo varies from day to day, and night to night. It follows a somewhat irregular elliptical orbit, with roughly a 2% variation in distance from the Sun, depending on its relation to its closeness to perihelion and aphelion. Add the slowing in rotational speed, axial precession, axial inclination from the plane of the ecliptic,
other irregular orbital and axial perturbations, and I find myself reluctant to perform any calculations which do not take into account the above, let alone make the observably incorrect initial assumption that the Earth behaves as a “black body.”

Having performed radiative calculations for the “average temperature” of the Earth, you will then get the same wrong answer as everyone else who makes similarly incorrect assumptions. GIGO.

You obviously have no knowledge about real physical greenhouses as opposed to your magical mathematical greenhouse, which “traps” and “stores” heat. This obviously uses New Physics.

Now a few facts, that even you can see.

The Earth has cooled considerably since its formation. The surface temperature is considerably less than some thousands of degrees C.

The atmosphere cools the Earth during the day. Otherwise, the surface would reach temperatures similar to those on the Moon, (after all plant and other life had vanished due to the lack of atmosphere).

Likewise, night temperatures would be colder, if no atmosphere existed.

A final point. CO2 creates no heat. It “traps and stores” heat only until it is warmer than the surrounding atmosphere, at which point the second law comes into operation. Every other gas molecule in the atmosphere is doing the same thing, to a greater or a lesser degree, depending on the physical characteristics of the gas in question.

You have precisely and absolutely no credible evidence that CO2 has any magical powers to create energy — in the form of heat or anything else. If you are convinced otherwise, I suggest you participate in the many forum discussions which take place regarding perpetual motion of various types. They would welcome your energy creation theory with open arms.

@John Kehr.

I agree totally. The problem is that the fanatics that believe that any temperature rise is man made rather than natural, and then proceed to claim that a by-product of fossil fuel oxidation causes warming greater than the amount of heat which led to the formation of the CO2 in the first place.

Hence my comment, as it is obvious that the atmosphere acts as an imperfect insulator at most, and could not account for any rise in average temperature as it contributes no additional energy.

Like you, I cannot find any reason to believe that any warming or cooling that we observe is any more or less than the Universe at work.

I’m off. Life to live. Cheers. Stay warm.

59. […] and Amazing Things we Find in Textbooks – The Real Second Law of Thermodynamics […]

60. I thought these comments had stabilised into some sensible resolution and understanding, even with agreements between pro and skeptical AGW folk (like John Kehr).
But now bizarrely we have Mike Flyn agreeing with John Kehr!??!
Does Mike Flyn actually understand what has been written here?
Has anyone here actually said CO2 creates heat?
From what I can make out Flyn is the only person that is saying that, in which case it is a strawman in the context of this blog.

John Millet… photon gas…bodies that switch off and on…

61. John Kerr,

MODTRAN
http://geoflop.uchicago.edu/forecast/docs/Projects/modtran.orig.html
can give you an estimate for the total forcing from CO2. Using the 1976 standard atmosphere and looking down from 100 km with no clouds, the total emission from 100-1500 cm-1 with 375 ppmv CO2 and a surface temperature of 288.2 K (emissivity 0.98) is 258.799 W/m2. Reduce the CO2 to 0 and the emission is 286.242 W/m2. The difference or forcing at the TOA is 27.443 W/m2. Removing as much water vapor as the interface will allow increases the emission to 336.608 W/m2 or a net of 50.366 W/m2. Remove methane and emission increases to 339.12 W/m2 or a net of 2.512 W/m2. Remove stratospheric and tropospheric ozone and emission would be 346.97 W/m2 for a net of 7.85 W/m2. If the atmosphere were perfectly transparent then the emission would be 360.30 W/m2 for a net of 13.33 W/m2 from nitrous oxide, stratospheric water vapor and other trace ghg’s. So the total clear sky forcing is 101.501 W/m2.

62. From ground level looking up, the numbers are somewhat different.

Standard conditions as above except 0 km looking up: 258.673 W/m2

0 CO2: 230.005 W/m2, net 28.668

0 Water Vapor Scale: 43.7716 W/m2, net 186.2334

0 methane: 40.6944 W/m2, net 3.0772 W/m2

0 ozone: 37.68 W/m2, net 3.0144 W/m2

leaving 37.68 W/m2 for residual stratospheric water vapor, nitrous oxide and other trace ghg’s. Stratospheric water vapor would be responsible for the majority of this emission.

63. SoD

You say that this is my position

……..”and paraphrasing now from 100 or 1000 exchanges – “a hotter body cannot absorb energy from a cooler body”………

Please supply some of the 1000 exchanges to back up your paraphrase.

Perhaps you would like to reread and answer the question I posed

“To make matters crystal clear please, once again, for the cameras

One BB source produces 200J/s landing on a surface per second with total absorption.
The spectrum is centred around 2um
The first source is then removed and replaced by the second BB source producing 200J/s but centred around 50um and is totally absorbed.
The object being radiated on is at 100K

If you are correct then the two sources will produce identical effects every second.

I think the 200J of 2um centred radiation will have far more capacity to do WORK than the 200J of 50um centred radiation.
What do you think?”

You can cover up my answers below to see if you are making any progress!
………………………………………………

From the first law of thermodynamics

200J of 2um centred radiation is equal to 200J of 50um centred radiation.

However according to the second law of thermodynamics

200J of 2um centred radiation is not equal to 200J of 50um centred radiation.

…………………………………………………..

Feel free to quote me for the following real statements.

Things that never happen!

1. Water will not of its own accord flow uphill.

2. An exploding grenade will not reassemble and re pin.

3. Heat will never of its own accord flow from a colder object to an object at a higher temperature.

• I covered this in more detail above, but I’ll put it this way:

If a plant absorbs 200J of 2um centered radiation, it will use some of it to do work — it will be stored in the bonds of sugars and not become heat until later when that sugar is burned.

If a plant absorbs 200J of 50um centered radiation, it will not be able to do work with it, it will simply turn into heat — 200J of it to be exact.

If you do not believe the 200J of absorbed energy becomes heat, what do you believe it becomes?

• Eric L.

I agree with a lot of your comment but you are using the word “heat” here inappropriately.

Heat is defined as the transfer of thermal energy from a high temperature to a lower temperature.

Slack use of the word “heat” has been a feature of SoDs posts.

The word “heat” as used by the “man in the street” is encouraged by SoD.
Its difficult to find a reason for this as he has been corrected many times on this by otherwise prominent supporters of his.

Why does he /she occupy this lonely position?

Perhaps SoD feels that without the atmosphere directly “heating” the Earth surface its difficult to sustain the IPCC position that 33C increase in Earth Surface temperature by the “Greenhouse Effect”

I don’t know of anyone else who continues to use “heat” in the way SoD does.
Can you come up with an answer?

I have already ruled out downright fraud.

64. Mike–

So you have a different view of Earth’s albedo, orbital geometry and the radiation received from the sun as a consequence, than is found in the texts? And you have a different view about CO2 and its properties and their influence on the transmission of IR through the atmosphere, too?

Who here has claimed the Earth is a black body? Who has suggested we set aside seasonal variations in insolation? These are all part of every serious effort to model the climate. And who has suggested making calculations based on an average surface or atmospheric temperature?

“The earth has cooled since its formation” Duh. And irrelevant, given that (unlike some 19th century types like Jules Verne) no one today thinks the heat given off by the earth is an important part of the climate today.

Finally, of course CO2 ‘creates’ no heat– who here has said it does? It does change how heat moves through the atmosphere and where (at what altitude) it escapes to space, removing (when we finally get back to equilibrium) the same amount of energy that the earth absorbs. This is a long way from magic– it’s basic atmospheric physics.

So, if you’re really sure you’re right, you should write it up: new thermodynamics, new atmospheric physics, all with massive implications for the agreed facts on how these things work– it should be an easy Nobel prize!

Best of luck with that.

• 1. View does not count. However, often the assumptions are wrong, with a view to “simplifying” the calculations. Which texts do you mean? In what context?

2. Anybody using equations such as those based on the Stefan-Boltzmann Law. Maybe you can find a definition which doesn’t contain the words “black” and “body”. I can’t.

I agree that every “serious effort” should encompass the variations I mentioned and many more besides. Using that definition, you may be able to tell me where I might find the code for a model that takes reality into account, as you have suggested. The code that I have downloaded and perused appears to be based on assumptions that in some cases have no factual support or justification.

3. I like people who can write Duh. Or puh-leaze, or stupid, ignorant, and all the other similar terms. I didn’t say the observed fact that the Earth has cooled since it was a glowing ball was anything other than a fact. It was merely to point out that averages of temperature may be irrelevant depending on time scale.

4. If you think that heat radiated from the surface of the Earth cannot escape other than by some magic process necessarily involving “greenhouse gases”, good for you. I don’t believe the facts support your contention, but you may be able to provide me with experimental data that shows that the atmosphere acts like anything more than a fairly poor insulator.

I don’t need anything new. It is all there.

As for a Nobel prize, I believe Marshall and Warren received one in 2005. Marshall in particular, believed he was right, and the medical fraternity around the world were wrong. It turned out he was right, they were all wrong, and he got the Nobel, not the “experts.”

Interesting to note that some doctors were fined and threatened with de-registration for telling patients that certain peptic ulcers were caused by bacteria rather than stress. The field of science is littered with similar instances of “fact by consensus”.

Even Al Gore received a Nobel prize. I am not sure what he “wrote up”. Why do you assume I want one?

Regards.

• Mike, you haven’t given any reference for an earlier comment where you claimed doubling or tripling CO2 in a greenhouse does not raise the temperature significantly.
You used it as evidence that the greenhouse effect was not significant.

Where is the data/experimental evidence to back your claim?

65. Bryan,

A BB source with a 50 um peak intensity has a temperature of 58 K. So to get 200 J/sec you would need a surface area of 3117 m2. At 2 um, OTOH, the temperature would be 1449 K and the surface area would be 0.008 m2 for 200 J/s. Your problem is severely underspecified. If all the incident radiation from the 58 K blackbody must be absorbed by the 100 K body, then it must have the same area as the the 58 K body as a parallel plane a small distance away. But then what do you do for all the excess area when the 58 K body is replaced by the much smaller 1448 K body? Where will all the radiation from the 100 K body that can’t physically be absorbed go? Because there is way too little area for it all to be absorbed by 1448 K body.

A better defined experiment would be narrow band radiation at the two wavelengths. If we take a bandwidth of 1 um and a flux of 200 W/m2 then we can calculate the effective source temperature from the Planck equation. For 2 um (1.5-2.5 um), the effective source temperature is 762.75 K. For the same bandwidth at 50 um (49.5-50.5 um) the source temperature would be 4366694 K.

But we still haven’t set up an actual experiment. Suppose we put our light source in the center of a thin spherical shell with a surface area of 1 m2 in deep space far from any star that already contains a power source that produces 5.67 W. The surface of the sphere will be 100 K. Now we turn on our light source that produces 200 W at either 50 um or 2 um. The surface temperature of the spherical shell will then be 245.4 K for either source. 200 J is 200 J and no work is being done.

• DeWitt Payne

“Your problem is severely underspecified. ”

This may be true but it is a response to falsify a physically impossible situation that SoD finds possible.

He apparently thinks that magnitude of radiation (number of Joules) is all that is required to describe its properties.
That 200J of 2um radiation is exactly equivalent to 200J of 50um radiation.

However it should not be too difficult to arrange say 200J/s of parallel radiation of both types to fall successively on say a 5cm square surface and compare the results.

I find it hard to believe you would thing that identical effects would be produced.

If still unconvinced then scale up to Earth like situation.

Exactly (or so we are told by the IPCC) equal amounts of Solar radiation enter the Earths atmosphere as leave the atmosphere in the form of low frequency EM radiation

In this case the Ist law of TD is satisfied.

But to have the reverse i.e. low frequency EM radiation enter the Earths atmosphere, be upconverted, and leave as the same magnitude but higher quality solar radiation is explicitly forbidden by the 2nd Law of TD.

66. Bryan

You say that this is my position:

……..”and paraphrasing now from 100 or 1000 exchanges – “a hotter body cannot absorb energy from a cooler body”………

Please supply some of the 1000 exchanges to back up your paraphrase.

Your central comment used in an article. You had many chances to correct any misconceptions, in the long discussion in Intelligent Materials and the Imaginary Second Law of Thermodynamics.

The refusal to answer a direct question, and the follow up question and your claims that I have cherry picked the sections of these books which state that the hotter body does absorb energy from the colder body – all strongly support your earlier claims.

Well, perhaps your position has changed. That would be wonderful. It’s hard to tell.

Let’s take the case of a body in thermal equilibrium with its surroundings, at a temperature T, which has 200J added from a source with 2um as its peak wavelength. What happens to the temperature?

The equation that relates heat capacity to temperature change:

[corrected]
ΔT = ΔQ/mc, where m = mass of the body, c = specific heat capacity, ΔQ = heat added, ΔT = change in temperature

ΔT = 200/mc – so if we know mass and heat capacity we can find the temperature change.

Let’s take the case of a body in thermal equilibrium with its surroundings, at a temperature T, which has 200J added from a source with 50um as its peak wavelength. What happens to the temperature?

ΔT = 200/mc

Interesting. The equation for temperature change of a body with a certain amount of heat added is a very simple one. When we apply it to the two cases Bryan has supplied we find the answers are identical. In both cases, the 200J leads to exactly the same temperature change.

If we took a different case where perhaps this body was losing heat at a particular rate then we would find that the slow down in temperature change was the same in both cases.

You can’t make up new laws of physics just to support a point of view that you have. Instead, you need to apply the well-known laws of physics to your argument to find out if you are correct.

If the equation relating temperature change to energy added was of this form:

ΔT = Q.f(vs)/mc, where f(vs) is a function of source frequency, vs – perhaps you would have material to work with.

67. DeWitt,

I know what the models of the atmosphere say, but I am trying to find a reference source for such statements. There are also two components to the forcing of CO2, I am looking for a source of energy absorbed by CO2 and a source for the downwelling attributed to CO2.

As far as I can tell such hard statements have not been made.
Thanks,
John

• The source of energy for emission is collisional excitation. For a given temperature a fraction of CO2 molecules calculated using the Boltzmann distribution will be in the excited state. The rate of emission is then determined by the Einstein A21 coefficient for that transition.

See here:

CO2 molecules absorb radiation at the characteristic wavelengths from whatever source, the sun, the ground, other CO2 molecules, and transfer the energy to the rest of the atmosphere by inelastic collision. For Kirchhoff’s Law, absorptivity = emissivity, to be valid, collisional transfer has to be much more probable than radiative transfer.

68. John Kehr

I know what the models of the atmosphere say, but I am trying to find a reference source for such statements.

Radiation and Climate, I.M. Vardavas and F.W.Taylor, Oxford University Press (2007)

Atmospheric Radiation: Theoretical Basis RM Goody & YL Yung, Oxford University Press (1989, 2nd ed)

• SoD,

Do either of those state the total amount of surface radiation converted to warming of the atmosphere by CO2?

Thanks

69. Bryan on October 18, 2010 at 8:31 am:

This may be true but it is a response to falsify a physically impossible situation that SoD finds possible.

He apparently thinks that magnitude of radiation (number of Joules) is all that is required to describe its properties.
That 200J of 2um radiation is exactly equivalent to 200J of 50um radiation.

However it should not be too difficult to arrange say 200J/s of parallel radiation of both types to fall successively on say a 5cm square surface and compare the results.

I find it hard to believe you would thing that identical effects would be produced.

Bryan is finding it difficult to come to terms with a very simple subject.

1. The equation for temperature change of a body doesn’t specify the source temperature of the energy received.

There is a simple equation which you can find in any basic thermodynamics textbook and the temperature change from receiving 200J doesn’t depend on whether that 200J came from a 10000000K body or a 1000K body or a 100K body.

Bryan can’t dispute it, which causes confusion for Bryan, so he describes it as a problem for me..

2. The reason why heat flows from hotter to colder is very simple.

– In the case of radiation – hot bodies radiate much more energy than cold bodies (proportional to the 4th power of absolute temperature)

– In the case of conduction – the heat transfer is proportional to the temperature difference between the two bodies.

Let’s stay with radiation, because it is the subject that causes the most confusion.

So if we have two bodies radiating towards each other, then the 10,000K body will (in general) radiate 100x times as much energy as the 1,000K body.

That is why the colder body heats up and the hotter body cools down. Because the colder body gains energy (Joules or J) and the hotter body loses energy. Not because of some mythical “quality” of the energy.

3. If by some particular experimental basis you arrange for two identical bodies under identical conditions to absorb 200J – one body absorbs the 200J from a 10,000K source, while the other body absorbs the 200J from a 1,000K source – guess what?

The temperature change is identical in both cases.

This is basic physics. Confusing for Bryan and many others, yet still true.

How can you know it is true?

Just look up the equation for temperature change and you will find the equation depends on the mass and heat capacity of the body, and the energy absorbed. Not on anything else.

Simple.

So, if you don’t like a result you have three choices:

1. pretend you don’t have to address the result
3. prove current physics wrong and present a new theory

• SoD:
” – In the case of radiation – hot bodies radiate much more energy than cold bodies (proportional to the 4th power of absolute temperature)

– In the case of conduction – the heat transfer is proportional to the temperature difference between the two bodies.”

Pretty obvious really, even to a ‘person on the street’!
So why is it, some people can’t get their heads around the difference between radiation, conduction and convection?

Do some engineers and public figures struggle with radiation but can cope with conduction?

• My last comment was in the context of the 2nd law.
eg. seems like some people struggle with radiation, but can easily understand the 2nd law when it comes to conduction.

• SoD

Your posts continue to imply whatever satisfies the first law should also satisfy the second law.
The simple equation you provide is from first law.

If still unconvinced then scale up to Earth like situation.

Exactly (or so we are told by the IPCC) equal amounts of Solar radiation enter the Earths atmosphere as leave the atmosphere in the form of low frequency EM radiation

In this case the Ist law of TD is satisfied.

But to have the reverse i.e. low frequency EM radiation enter the Earths atmosphere, be upconverted, and leave as the same magnitude but higher quality solar radiation is explicitly forbidden by the 2nd Law of TD.

• Your hypothetical violation of the 2nd Law is not something that anyone has ever suggested could happen. There is no up conversion involved in the atmospheric greenhouse effect. Net heat flow is always from warmer to colder.

It’s also not possible to conduct your 5 cm2 experiment for the reasons I stated above, but I’ll restate it in terms of time rather than area. An area of 0.0005 m2 at 58 K radiates 3.21E-4 W. So it would take 7.2 days to transfer 200 J. At 1448 K, the same 0.0005 m2 radiates 124.6 W so it would only take 1.6 seconds to transfer 200J.

70. DeWitt Payne

You seem to be implying that it is impossible to provide any source capable of producing 200J/s 50um radiation incident on an area of 5cm2, am I correct?

The reason I am using an example that examines absorption of radiation is that is where the major difficulty that preoccupies Sod and is the source of our major disagreement.

71. DeWitt Payne

….”Your hypothetical violation of the 2nd Law is not something that anyone has ever suggested could happen.”…

In fact I think that the implication of SoDs equivalence of effects for short wavelength and long wavelength radiation of equal magnitude leads directly to that conclusion.

Perhaps SoD could tell us why it does not!

72. Bryan:

In fact I think that the implication of SoDs equivalence of effects for short wavelength and long wavelength radiation of equal magnitude leads directly to that conclusion.

Perhaps SoD could tell us why it does not!

It’s not a difficult subject.

I don’t believe I can explain this simple subject in any simpler terms.

And if you aren’t having a laugh, and really can’t understand it, that’s ok too.

• SoD

I think its you who’s having a laugh.

You must have searched through hundreds of textbooks to find the few above who have been a bit careless with their thermodynamic definitions.

I think the authors would be appalled at your use of their lack of clarity namely to prove that heat flows from a colder object to an object at a higher temperature.

What happened to the 99.999% of textbooks which give correct definitions that you must have discarded to pick your chosen sample.

We notice that there are no Physics Textbooks in your sample.

Why do you think that is?

I don’t know what you hope to gain by misleading people who might not know enough to see through a bogus scam.

Do you think you have to distort the laws of physics to bolster your belief in the so called “greenhouse effect” if so I feel sorry for you, your holding a busted flush.

73. Sod to Bryan:

“If by some particular experimental basis you arrange for two identical bodies under identical conditions to absorb 200J – one body absorbs the 200J from a 10,000K source, while the other body absorbs the 200J from a 1,000K source – guess what?

The temperature change is identical in both cases”.

But the times to effect the identical temperature change wouldn’t be identical. The time difference would reflect the difference in qualities of radiation from the two sources, if I understand Bryan correctly.

SoD to Bryan:
” – In the case of radiation – hot bodies radiate much more energy than cold bodies (proportional to the 4th power of absolute temperature)

– In the case of conduction – the heat transfer is proportional to the temperature difference between the two bodies.”

Not so.

Both conductive and radiative heat transfers rely on a temperature difference between the body and its environment/surroundings. They are proportional to the differences in the first and fourth powers of those temperatures, respectively. That’s what Sod’s textbook extracts show for radiative heat transfer, if I understand correctly.

74. John: The time to effect a temperature change is only a matter of the time over which that amount of energy is absorbed, which is not affected by the frequency of photons absorbed. The rate of energy flow to the body being warmed is all that matters, and this can be equal for a 1,000 K source that delivers a total of 200J over some interval and a 10,000 K source delivering the same. For example, the emitting surface of the 1,000K source could be made large enough to compensate for the difference in energy emitted per unit area by the two sources. In that case the total warming and the rate of warming would be exactly the same.

Bryan: So you think Fundamentals of Heat and Mass Transfer, Heat and Mass Transfer, Principles of Heat Transfer etc. are not physics texts? Pretty silly. Even if these are engineering texts (and indeed one of SOD’s books is labelled as such), engineers actually do have to get the physics right– otherwise, either their designs won’t work or the physics is wrong! The burden of proof is on you at this point: SOD has a good looking collection of sources that are perfectly clear in supporting his (her?) view. Can you find a physics text that tells the story your way? Somehow I doubt it.

• Bryson Brown

….”Can you find a physics text that tells the story your way? Somehow I doubt it.”…….

From University Physics by Harris Benson

page 382

Modern definition of Heat.

Heat is energy transferred between two bodies as a consequence of a difference in temperature between them.

Page 435

…..it is impossible to transfer heat continusly from a cold body to a hot body without the input of work”…….

University Physics Young and Freedman

page 470

Energy transfer that takes place solely because of a temperature difference is called heat flow or heat flow transfer and energy transferred in this way is called heat

pg 559

Heat always flows from a hot body to a cooler body never the reverse.

Hope that helps. It really is important to get the basic definitions right or you will get very confused as the discussion progresses because SoD doesn’t seem to read Physics books either!

• Bryson, does it really work that way? True, by scaling the surface area of the cold body (by 10,000:1 in this case) you can achieve two emitters with equal power. But the power density remains unchanged, being related to temperature and 10,000 times as much for the 10,000K emitter as for the 1000K one. These are the energy fluxes that independently irradiate the identical absorbing bodies and determine the relative times required to absorb identical quantities of energy.

75. Bryan: Still being silly, I see. SOD has been very clear about the distinction between net flow and the one-way flows in radiative exchange (and the similar distinction that applies to random exchanges between particles as net heat is conducted from a hotter body to a colder: this is due to the statistical balance, while occasional small amounts of heat energy are transferred from molecules in the colder material to molecules in the hotter). None of your quotes in any way undermines the accounts of the books SOD cites– as I said, if they did, the engineers would be getting the wrong results (or the physicists would be wrong and we’d need a new thermodynamics). If you keep reading physics books for brief quotes drawn out of context (contrast with the images of multiple pages from SOD) instead of understanding, you’re not going to get very far.

76. Tell me if I have this wrong, both SoD and Bryan. If you were standing three miles from the surface of the sun (you being Superman), and you looked toward Earth, could you see it?

My point is obvious. The earth does radiate toward a warmer body, yes? The radiation doesn’t stop at you and refuse to continue on toward the sun. This longwave radiation will add energy to a body that is hotter. It will not add energy at the rate that the hotter body is losing it, though. Yes?

77. DSL on October 20, 2010 at 1:44 am:

Correct.

• Sod/DSL, wouldn’t the Earth’s longwave radiation add energy to the 2.7K space that surrounds it, not to the distant 6000K sun? That’s what the entropy law would require.

• Not as I understand it. See the WISE project. If infrared radiation can travel across millions of light years of space without being absorbed (well, without very much being absorbed by interstellar bodies), you’d think it could travel a mere 92 million miles without being absorbed. And the sun can certainly absorb at the wavelengths that it emits at.

Infrared radiation from Earth doesn’t heat up the sun, though. It just replaces an incredibly tiny fraction of the energy being lost by the sun.

78. Bryson Brown

Still choosing to be blind, I see.

……”SOD has been very clear”…….

SoD clearly does not have a physics background.
He refuses to disclose what his area of expertise is in.
I wonder from time time time whether he is being devious or in fact is genuinely misinformed

However you should get yourself a physics book and read it carefully.
Work through the thermodynamics sections carefully.

University Physics Young and Freedman

page 470

Energy transfer that takes place solely because of a temperature difference is called heat flow or heat flow transfer and energy transferred in this way is called heat

pg 559

Heat always flows from a hot body to a cooler body never the reverse.

This is a standard textbook currently used throughout the world.

There is no Net Heat between two bodies only Heat travelling from the hotter to the colder.

I cannot type whole chapters to give you a bigger picture.

You will have to do some work for yourself or stay gullible and misinformed.

• Bryan, you seem to be using language as your defence rather than science. Anyone can quote lines from books that describe the same thing and different people will interpret what is being described differently.

The key is whether a person understands what is going on.

SoD (and others) has spent a lot of time explaining the science and you have done basically nothing.
Lets see your work and logic. If you are unwilling to publish it, then your motives are suspect.

79. Bryan can claim away.

The textbooks I cited provide equations which prove that hot bodies absorb radiant energy from cold bodies.

There is no confusion with equations – unless you can’t understand equations..

For those wondering, check back in the article. I provide a simple contrast of the maths between the two “points of view” under the sub-headings:
Radiation Exchange under The Real Second Law
Under The Imaginary Second Law

The textbook writers confirm the equations for the point of view I have explained many times – and which Bryan disputes:

Enet1 = -Enet2 (where Enet1 is the energy gained by one body, and Enet2 is the energy gained by the other body)

or, using the terms from most of the books, q1 = -q2

This is not about “careless definitions” by the writers. This is maths (extremely simple maths), which is very specific and only capable of mis-interpretation by people who don’t understand maths.

All of the six textbooks – the first six textbooks I took off the shelf in the university library – have the same formula.

This is not surprising, this was only cutting edge stuff sometime in the 1800s. Therefore, all the textbooks have the same formula.

These formulae clearly refute Bryan’s ideas. But Bryan will continue to cite vague statements unrelated to the specifics of radiant heat transfer.

If Bryan was correct in this claim:

You must have searched through hundreds of textbooks to find the few above who have been a bit careless with their thermodynamic definitions.

I think the authors would be appalled at your use of their lack of clarity namely to prove that heat flows from a colder object to an object at a higher temperature.

Then it will be an easy task for Bryan – or the other people who are reading this blog who support Bryan’s point of view – to find six textbooks which cover radiant heat transfer and have a different form of the equation, one which matches their point of view:

Enet1 ≠-Enet2

We will wait in vain for support for this specific claim by Bryan.

And “physics-degree” Bryan can claim away on his incredible grasp of the subject. For anyone new to this blog who might be impressed by these claims, read the discussion in:

And also the still unanswered issue on the very simple equation for temperature change.

Or Bryan’s original claim about “back radiation” – search for April 17th, 2010 at 9:26 pm within the link:

..The downward radiation must in large part be caused by Rayleigh Scattering (the effect that gives us blue sky’s)..

I write this comment because the blog is always getting new readers..

80. scienceofdoom

….”– the first six textbooks I took off the shelf in the university library – “…….

Are you for real?

All six are by Mechanical Engineers!

Why didn’t you go to the Physics shelf?

I think I know the reason why!

Thermal Radiation heat Transfer, by Siegel&Howell(freely down loadable set of files on the NASA , 3 volumes 1968-1971, NASA Ref SP-164. Check http://ntrs.nasa.gov/search.jsp

Or the lecture notes by Rodrigo Cabellero often recommended by DeWitt Payne again freely available.

These authors can make their points without getting their definitions confused.

…….”And also the still unanswered issue on the very simple equation for temperature change.”……

I answered the question in a previous post however here is an expanded explanation
The simple equation you use is derived from the 1st Law of TD (conservation of energy).

A typical example of the use of this equation would be to work out how long it would take an electric kettle to raise the temperature of a mass of water say by 60c.

Electrical energy supplied = Heat energy gained

PxT = CMdT with the usual meaning for the symbols.

The assumption here is that all the electric energy (highest quality) is used to raise the temperature of the water with no heat loss.

What you want to do with the equation is to use it for a situation forbidden by the 2nd Law.

You want say radiation of 200J from an object at say 100K to raise the temperature of an object at 500K.

To do so the radiation would have to be upconverted to radiation characteristic of 500K plus.

It just doesn’t happen!!!!

81. Bryan, you still haven’t answered either of my questions. They’re both pretty simple.

Btw, by your definition, wouldn’t radiation at 500K have to be downconverted to radiation characteristic of 100K in order to be absorbed by the 100K object? I mean, you can’t have it two ways. It sounds like you’re saying that radiation can only be absorbed if it’s coming from an object at the same temperature. Is this wrong?

• DSL

Look again at the equation above only this time let the input energy be thermal from the hot water.

Heat Energy Supplied = Electrical energy Gained

CMdT = Pxt

Now as far as the Ist law is concerned, everything is all right.
However as we all know this reverse situation just does not happen .

Similarly with thermal energy transfer between a higher and a lower temperature object.
The Ist Law allows transfer in either direction but the 2nd law insists that the difference between the thermal energy fluxes which we call Heat is always from the hotter to the colder object.

• Bryan:
“…but the 2nd law insists that the difference between the thermal energy fluxes which we call Heat is always from the hotter to the colder object.”

Oh good grief!
You’re not a complete dunce then.
Do you enjoy disagreeing with everyone for days then agreeing with them later, just for the fun???

• Bryan: “The Ist Law allows transfer in either direction but the 2nd law insists that the difference between the thermal energy fluxes which we call Heat is always from the hotter to the colder object.”

Just to be clear, you mean the “net difference,” as I can’t make any other sense out of the word “difference” in that context.

Both objects “flux,” but the hotter object will never see its temperature increase. Its temperature may decrease more slowly based on the amount of thermal energy (“flux”) it receives from the cooler object, but it will never receive enough net thermal energy from the cooler object to increase in temperature.

82. Bryan,

Yes, your 5 cm2 thought experiment is physically impossible.

83. Warmcast

….”Do you enjoy disagreeing with everyone for days then agreeing with them later, just for the fun???”…..

No what I have said over the last few days has been quite consistent but if you now agree with it I’m happy.

• Incorrect.
You are now contradicting your past statements and are agreeing with SoD and many others here.

• Warmcast

• From what I can make out, Bryan, you have accepted what SoD has been stating about radiation.

You aren’t contradicting your previous statement, you just haven’t fully acknowledged the basic point SoD and others have made.
Being pedantic may be fine from your perspective, but it doesn’t work well at all if you want to communicate effectively.

SoD is doing a brave job of communication, far more than you or anyone else has tried.

• Warmcast

Which one?

• Can’t you work out basic logic, sequences and language Bryan?

You accuse others of not being precise, yet you can’t read my posts in a logical order.

84. DeWitt Payne

Minor point

My original specification was “5cm square surface” i.e. 25cm2.
However perhaps it makes little difference to the practicalities of the situation.

However it does not seem impossible in principle to focus a large parallel beam of 50um radiation to produce 200J/s on such an area.
Why someone would find a practical use for such an arrangement is a different question.

Perhaps picking two more realistic radiations say 0.1um and 5um might make the same point more realistically.

It was raised as a way to highlight SoDs apparent belief that the wavelength of radiation is irrelevant.
Only the magnitude need be considered in his opinion as he has written above

85. I said on October 20, 2010 at 8:02 am:

Then it will be an easy task for Bryan – or the other people who are reading this blog who support Bryan’s point of view – to find six textbooks which cover radiant heat transfer and have a different form of the equation, one which matches their point of view:

Enet1 ≠-Enet2

We will wait in vain for support for this specific claim by Bryan.

Bryan responds:

That is not support for Bryan’s specific claim.

Bryan, please identify the section where Siegel and Howell derive or cite the equation for radiant exchange between two surfaces. The place where they support your claim:

Enet1 ≠-Enet2

– Of course, I know that Bryan hasn’t actually read this book..

86. If only these writers about radiant heat transfer weren’t such heretics. Here are a couple of extracts from the 1969 version of Siegel & Howell:

Confusing that they claim that a perfect absorber absorbs all energy. Bryan will explain that this is not true. If the source is colder than the surface under consideration – NOTHING will be absorbed. What a shame that Siegel & Howell forgot to mention it.

As with all of these heretics, they confuse their readers by not mentioning that the hotter body CANNOT absorb energy from the colder body. In fact, they imply (misleadingly and NOT their intention) that this actually happens.

They claim the same relationship as the not-to-be-trusted heat transfer experts cited in the article.

87. Sod to Bryan:

“Enet1 = -Enet2 (where Enet1 is the energy gained by one body, and Enet2 is the energy gained by the other body)

or, using the terms from most of the books, q1 = -q2”

I think this is the source of Sod’s misunderstanding, doing in one step what the entropy law requires to be done in two steps, between each body and the common surrounding medium. Assume the medium’s temperature to lie between the two bodies’. Then the hot body loses heat denoted as “-q1” and the medium gains “q1” where “q” is proportional to the difference in the fourth powers of the temperatures of the body and the medium (as shown by the textbook extracts). Similarly, though conversely, the cold body gains “q2” and the medium loses “-q2”. The conservation law requires that gains = losses or
(q1 + q2) = – (q1 + q2). But note that -q1 is not necessarily equal to q2, contrary to Sod’s assertion. It is equal only if the temperature differences between each body and the medium are equal. In the general case the heat lost by the hot body would equate to the sum of the heat gains by the other body and the medium. Sod’s result comes from assuming the absence of a medium.

It occurs to me that had K&T, in compiling the global energy budget, followed the textbook examples provided by Sod, they would have shown a much lower value for the upward surface flux, reflecting the small temperature difference between the surface and the boundary layer of air surrounding it. This flux, together with the upward non-radiative fluxes, would equate to the downward SW flux at the surface in accordance with the conservation law. There would be no need for back-radiation, no issue with the 2nd law of thermodynamics and we could all move on.

Another casualty of ignoring what Sod’s textbook extracts teach is the claim that GHGs slow the rate of surface cooling. Rather it is the temperature of the boundary layer surrounding the surface that does that. What GHGs do is determine the proportions, of that (minor) radiative heat loss from the surface, transmitted to space through the admosphere and absorbed by it for subsequent radiation to space.

• What medium are you talking about??
You seem to be the only person here that is including a third body between the two bodies.

If you are saying that their is some ‘atmosphere’ between the two bodies, then you have added a body that isn’t in the example.

• I take that back, one of the text books (Frank Kreiths) mentions a medium!
The Frank text book example states a medium that does not absorb radiation appreciably.

One assumes the example in the book is an attempt to be realistic rather than idealistic (a vacuum).

It does distort the whole issue by having a medium between the two bodies though.

A quick glance of the first two books suggests they do not appear mention what is between two surfaces or bodies.
One must assume a vacuum in an ideal situation.

I think that is what the majority would assume and what follow normal practice of creating ideal situations in order to do the basic math and to understand what is going on.

88. scienceofdoom

Once again we are at cross purposes.

I was saying that the term Net heat Flow as used above and sometimes highlighted by you was a term better better avoided.

Why?

Because it implies there is heat flowing from cold to hot as well as heat flowing from hot to cold.
I think that this is an unfortunate term as it implies a violation of the 2nd Law of TD.

Now I think that all of the authors if pressed would say that they did not actually think that heat flowed from cold to hot.
Another term much used by Mechanical Engineers is Centrifugal Force and this also is a term best avoided.

My guess now is you had some training as a Mechanical Engineer.
Am I correct?

Now the fairly innocent use of net “heat” flow by the engineers has been seized on by some in the AGW camp and has become “heat” flow then “heat” then simply heat from cold atmosphere to warmer Earth Surface.

Diagrams were produced showing arrows labelled HEAT from Atmosphere to Surface.

Since the advent of the G&T paper lots of the diagrams and papers displaying this mistake were removed.

Indeed DeWitt Payne challenged me to show that any serious scientist could be stupid enough believe that heat flows from cold to hot.
…………………………………………………………………………….

I thought that you were concerned mainly about;

ΔT = ΔQ/mc,

For this one

Enet1 ≠-Enet2

Or in a sentence

…”In the case of the real second law, the net energy absorbed by body 2 is the net energy lost by body 1.”…

Why should there be an absolute imperative that body 2 must absorb all the radiation emitted by body1?

…………………………………………………………………………….

You say “Of course, I know that Bryan hasn’t actually read this book..”

Now that’s the kind of comment that we would expect to find in Deltoid or similar sites; please return to a more civil exchange.

Of course I have read both the book and the lecture notes however the point I was making was that you do not need to use “loose” or inexact definitions to make your point.
For instance Nick Stokes probably has much the same general view of AGW as yourself but I would be surprised if he would express these views in the way you choose to do .

89. Warmcast

I like to reply to your post however my previous two posts are being held up for “moderation”

It does seem a bit odd that SoD would throw down a challenge to myself and other skeptics and yet refuse to screen the reply.

• That is complete junk.
SoD is using WordPress and a standard template.

This indicates more about you than anything else you have written.

90. Warmcast

Bryan

…….”This indicates more about you than anything else you have written.”………

What does that say about you?

• That I know WordPress and how it works.

• Warmcast

Bryan

scienceofdoom

Once again we are at cross purposes.

I was saying that the term Net heat Flow as used above and sometimes highlighted by you was a term better better avoided.

Why?

Because it implies there is heat flowing from cold to hot as well as heat flowing from hot to cold.
I think that this is an unfortunate term as it implies a violation of the 2nd Law of TD.

Now I think that all of the authors if pressed would say that they did not actually think that heat flowed from cold to hot.
Another term much used by Mechanical Engineers is Centrifugal Force and this also is a term best avoided.

My guess now is you had some training as a Mechanical Engineer.
Am I correct?

Now the fairly innocent use of net “heat” flow by the engineers has been seized on by some in the AGW camp and has become “heat” flow then “heat” then simply heat from cold atmosphere to warmer Earth Surface.

Diagrams were produced showing arrows labelled HEAT from Atmosphere to Surface.

Since the advent of the G&T paper lots of the diagrams and papers displaying this mistake were removed.

Indeed DeWitt Payne challenged me to show that any serious scientist could be stupid enough believe that heat flows from cold to hot.
……………………………..

I thought that you were concerned mainly about;

ΔT = ΔQ/mc,

For this one

Enet1 ≠-Enet2

Or in a sentence

…”In the case of the real second law, the net energy absorbed by body 2 is the net energy lost by body 1.”…

Why should there be an absolute imperative that body 2 must absorb all the radiation emitted by body1?

……………………..

You say “Of course, I know that Bryan hasn’t actually read this book..”

Now that’s the kind of comment that we would expect to find in Deltoid or similar sites; please return to a more civil exchange.

Of course I have read both the book and the lecture notes however the point I was making was that you do not need to use “loose” or inexact definitions to make your point.
For instance Nick Stokes probably has much the same general view of AGW as yourself but I would be surprised if he would express these views in the way you choose to do .

• Warmcast

That I know WordPress and how it works.

I hope your knowledge of other items is a bit better.
What are you going to do when Sod sorts out this this muddle?

91. Bryan

scienceofdoom

Once again we are at cross purposes.

I was saying that the term Net heat Flow as used above and sometimes highlighted by you was a term better better avoided.

Why?

Because it implies there is heat flowing from cold to hot as well as heat flowing from hot to cold.
I think that this is an unfortunate term as it implies a violation of the 2nd Law of TD.

Now I think that all of the authors if pressed would say that they did not actually think that heat flowed from cold to hot.
Another term much used by Mechanical Engineers is Centrifugal Force and this also is a term best avoided.

My guess now is you had some training as a Mechanical Engineer.
Am I correct?

Now the fairly innocent use of net “heat” flow by the engineers has been seized on by some in the AGW camp and has become “heat” flow then “heat” then simply heat from cold atmosphere to warmer Earth Surface.

Diagrams were produced showing arrows labelled HEAT from Atmosphere to Surface.

Since the advent of the G&T paper lots of the diagrams and papers displaying this mistake were removed.

Indeed DeWitt Payne challenged me to show that any serious scientist could be stupid enough believe that heat flows from cold to hot.
…………………

I thought that you were concerned mainly about;

ΔT = ΔQ/mc,

For this one

Enet1 ≠-Enet2

Or in a sentence

…”In the case of the real second law, the net energy absorbed by body 2 is the net energy lost by body 1.”…

Why should there be an absolute imperative that body 2 must absorb all the radiation emitted by body1?

……………

You say “Of course, I know that Bryan hasn’t actually read this book..”

Now that’s the kind of comment that we would expect to find in Deltoid or similar sites; please return to a more civil exchange.

Of course I have read both the book and the lecture notes however the point I was making was that you do not need to use “loose” or inexact definitions to make your point.
For instance Nick Stokes probably has much the same general view of AGW as yourself but I would be surprised if he would express these views in the way you choose to do .

92. SoD

Are you aware that I have tried several times to send you a reply to your last post to me?

93. SoD

I will try to split the comment into parts to see if that works

………………………..
Bryan

scienceofdoom

Once again we are at cross purposes.

I was saying that the term Net heat Flow as used above and sometimes highlighted by you was a term better better avoided.

Why?

Because it implies there is heat flowing from cold to hot as well as heat flowing from hot to cold.
I think that this is an unfortunate term as it implies a violation of the 2nd Law of TD.

Now I think that all of the authors if pressed would say that they did not actually think that heat flowed from cold to hot.
Another term much used by Mechanical Engineers is Centrifugal Force and this also is a term best avoided.

94. SoD

Second part of post
…………………………………………………………….

My guess now is you had some training as a Mechanical Engineer.
Am I correct?

Now the fairly innocent use of net “heat” flow by the engineers has been seized on by some in the AGW camp and has become “heat” flow then “heat” then simply heat from cold atmosphere to warmer Earth Surface.

Diagrams were produced showing arrows labelled HEAT from Atmosphere to Surface.

Since the advent of the G&T paper lots of the diagrams and papers displaying this mistake were removed.

Indeed DeWitt Payne challenged me to show that any serious scientist could be stupid enough believe that heat flows from cold to hot.
…………………

I thought that you were concerned mainly about;

ΔT = ΔQ/mc,

For this one

Enet1 ≠-Enet2

Or in a sentence

…”In the case of the real second law, the net energy absorbed by body 2 is the net energy lost by body 1.”…

Why should there be an absolute imperative that body 2 must absorb all the radiation emitted by body1?

……………

You say “Of course, I know that Bryan hasn’t actually read this book..”

Now that’s the kind of comment that we would expect to find in Deltoid or similar sites; please return to a more civil exchange.

Of course I have read both the book and the lecture notes however the point I was making was that you do not need to use “loose” or inexact definitions to make your point.
For instance Nick Stokes probably has much the same general view of AGW as yourself but I would be surprised if he would express these views in the way you choose to do .

95. SoD

Second part of post
…………………………………………………………….

Now the fairly innocent use of net “heat” flow by the engineers has been seized on by some in the AGW camp and has become “heat” flow then “heat” then simply heat from cold atmosphere to warmer Earth Surface.

Diagrams were produced showing arrows labelled HEAT from Atmosphere to Surface.

Since the advent of the G&T paper lots of the diagrams and papers displaying this mistake were removed.

Indeed DeWitt Payne challenged me to show that any serious scientist could be stupid enough believe that heat flows from cold to hot.
…………………

I thought that you were concerned mainly about;

ΔT = ΔQ/mc,

For this one

Enet1 ≠-Enet2

Or in a sentence

…”In the case of the real second law, the net energy absorbed by body 2 is the net energy lost by body 1.”…

Why should there be an absolute imperative that body 2 must absorb all the radiation emitted by body1?

……………

You say “Of course, I know that Bryan hasn’t actually read this book..”

What am I to make of a comment like that?

Of course I have read both the book and the lecture notes however the point I was making was that you do not need to use “loose” or inexact definitions to make your point.
For instance Nick Stokes probably has much the same general view of AGW as yourself but I would be surprised if he would express these views in the way you choose to do .

96. Sod, tried several times to send second part but failed.

Any suggestions as to why

97. I can understand people’s frustration when their posts don’t appear. But I don’t moderate posts – as should be clear, because almost every comment appears immediately.

However, the WordPress moderation filter has been setup in a certain way which unfortunately catches more innocent comments from time to time.

Comments in the moderation queue then emerge when I get the opportunity to check. This isn’t a 24/7 service..

98. Bryan has claimed all sorts of nefarious activities on my part, and when I made the obvious statement that he hasn’t read the book he puts up as a counter-argument, says:

You say “Of course, I know that Bryan hasn’t actually read this book..”

Now that’s the kind of comment that we would expect to find in Deltoid or similar sites; please return to a more civil exchange.

Let’s examine Bryan’s claims.

From October 7, 2010 at 8:16 am:

You have obviously tried to cherry pick passages(which perhaps could have been more carefully written) to back up your claim that radiation from a colder body can increase the temperature of an object at a higher temperature.

Not very civil.

From October 7, 2010 at 8:33 am:

You are opening new posts at a record rate.
Hopefully this is not done so as to avoid answering awkward questions like this one unanswered from

Absorption of Radiation from Different Temperature ..

Not very civil.

From October 19, 2010 at 2:01 pm:

You must have searched through hundreds of textbooks to find the few above who have been a bit careless with their thermodynamic definitions.

I think the authors would be appalled at your use of their lack of clarity namely to prove that heat flows from a colder object to an object at a higher temperature.

What happened to the 99.999% of textbooks which give correct definitions that you must have discarded to pick your chosen sample.

We notice that there are no Physics Textbooks in your sample.

Why do you think that is?

I don’t know what you hope to gain by misleading people who might not know enough to see through a bogus scam.

Not very civil.

From October 20, 2010 at 12:30 pm:

All six [books] are by Mechanical Engineers!

Why didn’t you go to the Physics shelf?

I think I know the reason why!

Thermal Radiation heat Transfer, by Siegel & Howell

Not very civil.

Also, and more importantly, Bryan continues to make a complete ass of himself.

But it’s necessary to point this out for the many readers who are pre-disposed to think Bryan is onto something because they have also studied the imaginary second law of thermodynamics.

If you do a search in a university library catalogue for “Heat Transfer” you come to section with a number of books. Pretty much all of them are written by “Professors of Mechanical Engineering”.

For example, exhibit A for the defence:

Thermal Radiation Heat Transfer, by Robert Siegel and John R. Howell.

This is the 1981 edition.

For those not paying attention, this is the text put forward by Bryan.

John R. Howell is Professor of Mechanical Engineering.
Robert Siegel is from the Fluid Mechanics and Acoustics Division.

Slight diversion for those who like a certain type of humor:

Back to the subject.

Bryan says that I must have found lots of textbooks that disagreed with “my idea” (the standard physics idea) and thrown them away.

Bryan says that I must have chosen books by professors of mechanical engineering because I will know that “real science” by professors of physics prove me wrong.

Bryan holds out a book for us to study. We will see what the authors of this excellent book say in future comments or articles.

In the meantime, if Bryan had any integrity he would state exactly what his position was as far as:

Does a hotter body absorb the radiant energy from a colder body?

1. Bryan muddied the waters with his question from October 14, 2010 at 8:17 pm which I answered on October 15, 2010 at 7:19 am with a reference from Bryan’s earlier comment. but Bryan is still to clarify.

Hedging bets.

2. Bryan indicates that the many writers I have chosen have been quoted out of context. So this shows he is not in agreement with the specific point I made, despite his hedging.

3. Bryan so far refused to answer my question asked twice in different ways:

I seemed to have missed it, but do you think that the writers of these textbooks believe that energy from colder bodies is absorbed by hotter bodies?

Which shows an uncomfortable position.

4. Bryan suggests that I have discarded many textbooks with the correct answer.

5. Bryan suggests that if I chose “physics” textbooks the result would be different.

I don’t expect Bryan to defend his unscientific point of view, or even explain exactly what it is. Especially as he has promoted a book by professors of the same disciplines and with exactly the same understanding of the subject.

For the many who read and wonder about this subject, does it look like anyone who is an expert in the field of heat transfer agrees with Bryan?

99. Sod

You obviously missed my point entirely when I recommended that you should have a look at the two references I gave you

“Of course I have read both the book and the lecture notes however the point I was making was that you do not need to use “loose” or inexact definitions to make your point.
For instance Nick Stokes probably has much the same general view of AGW as yourself but I would be surprised if he would express these views in the way you choose to do.”

So I will isolate the main part of the paragraph

…the point I was making was that you do not need to use “loose” or inexact definitions to make your point…….

The two references were careful not to get their definitions muddled up and so avoided needlessly confusing readers.
Not that I agree with all the points in each reference.

“I seemed to have missed it, but do you think that the writers of these textbooks believe that energy from colder bodies is absorbed by hotter bodies?”

I think SoD would answer this with (in my words!): ‘Yes, as the photons emitted from the colder bodies are absorbed by the warmer bodies. I.e the IR photons of the colder atmosphere are absorbed by the warmer water body. Besides, that is what they explicitly write.”

But then you continue:
“For instance Nick Stokes probably has much the same general view of AGW as yourself but I would be surprised if he would express these views in the way you choose to do.”

As luck has it, Nick Stokes has made such a statement in another thread. Almost as if he responded to your confusion directly, albeit that he responded to another misconception:

“But that’s the misconception – the top layer is losing, not gaining heat. It emits. And it doesn’t freeze.

The down IR just makes up some of the heat the top layer is losing by radiating. The rest comes by conduction and advection from below, with no great difficulty.”

• cynicus

I have rephrased my response to this question.
There is no doubt that both the surfaces radiate to each other.
The higher temperature surface radiated more energy at every wavelength and the higher temperature surface will radiated at a higher average wavelength than the lower temperature surface.
For example 2000 photons of 10um from hotter are emitted and 500 photons of 10um photons from the colder surface are absorbed how should we express this?
In reality 1500 from the hotter surface are transferred to the colder surface or so we postulate .
The backradiation some people describe as heat but they are wrong.
Strangely enough back convection currents nobody calls heat.
Heat is that part of the thermal energy interchange that is available to be converted into WORK or higher grade energy in the given situation.
So heat can be transferred spontaneously form a higher temperature object to a lower temperature object.
But the thermal energy from the colder surface cannot be described in this way.
To sum up
For the hot and colder objects;
Radiation is transferred in both directions.
Energy is transferred in both directions
Heat is only transferred spontaneously from a higher to a lower temperature.

Your quote from Nick Stokes is saying much the same thing.
..”The down IR just makes up some of the heat the top layer is losing by radiating. The rest comes by conduction and advection from below, with no great difficulty.”…..

100. SoD

What do you make of this quotation from a very popular current Physics Textbook

University Physics Young and Freedman

page 470

Energy transfer that takes place solely because of a temperature difference is called heat flow or heat flow transfer and energy transferred in this way is called heat

pg 559

Heat always flows from a hot body to a cooler body never the reverse.
…………………………………………………..
Now when there is uncertainty or dispute about the effects of absorption of radiation from a colder body on reaching the hotter body the best method to resolve the matter is by experiment.

You must remember Gord and his Solar Heater!

When pointed at the night sky the radiation from the sky lowered the temperature of a sample of water placed at the focus of the parabolic reflector to the extent that it formed ice!
Now if instead of forming ice the water temperature rose I would take your theory seriously.

However I strive to have a flexible outlook at all times.
If you can come up with evidence of a colder object increasing the temperature of a hotter object then don’t hold back and please let all your viewers witness this unique event.

101. Could Gord see the colder glass of water in the warmer mirror? I wonder how that might be possible . . .

Further, if your theory were true, we would not be able to see objects colder than our eyes. I don’t think we see by conduction or convection.

“If you can come up with evidence of a colder object increasing the temperature of a hotter object then don’t hold back and please let all your viewers witness this unique event.”

You’re changing your tune. Now you use “increasing.” It was never about increasing. It was about net exchange and the possibility of a hotter body absorbing energy from a colder body–not increasing its temperature.

102. DSL

I think we are in almost total agreement.
The Hotter and Colder objects are in dynamic equilibrium transmitting and absorbing radiation from each other.
The radiation from the hotter has a greater magnitude and is of higher quality than the radiation it absorbs from the colder.
Thus the Heat which is the difference between the thermal fluxes always goes from the higher temperature object to the lower temperature object.
Heat has the thermodynamic capacity to do work so it would be incorrect to call the radiation from the colder object Heat.

……..”Its temperature may decrease more slowly based on the amount of thermal energy (“flux”) it receives from the cooler object, but it will never receive enough net thermal energy from the cooler object to increase in temperature.”…….

Is spot on and well expressed.

I think that if we look on the radiative aspect as being part of an “insulator effect” then the whole thing becomes less mysterious.

103. DSL

Your post that I agreed with is your earlier one not the one above.

…”Further, if your theory were true, we would not be able to see objects colder than our eyes.”

We see things because they reflect light not in the main because they radiate.
Our eyes are sensitive only to electromagnetic radiation in region 400 =>700 nm.

Gords solar heater is not a theory it actually freezes water when pointed to the night sky.

104. DSL

…….”You’re changing your tune. Now you use “increasing.” It was never about increasing. It was about net exchange and the possibility of a hotter body absorbing energy from a colder body–not increasing its temperature.”…….

If you look over my previous posts you will find examples where I explicitly say that the hotter object can absorb radiation.
Whether it is 100% absorption is another matter.
Absorption depends on the absorbing surface and the wavelength of the radiation.
There is no physical law that states any surface must absorb all radiation falling on it quite the opposite.

105. I think there are problems when trying to impose the same rules on photons that we see at the macroscopic level. To say that a photon that only hits colder materials than its source must therefore be “intelligent” is flawed thinking. Photons are not like tiny billiard balls, even if we sometimes think of them that way. At the level of photons, if the maths says it happens, then we must accept that it does, even if it makes no logical sense within our mental model.

Whenever it has been tested, it’s always the maths that’s right and the analogy that’s wrong:
http://news.sciencemag.org/sciencenow/2007/02/16-04.html

Do the photons really detect their environment and change their behaviour depending on how they are being observed. Or is it that our view of a photon as being a tiny billiard ball is an incomplete analogy that doesn’t quite encompass all of its behaviour.

We know from experiment that the second law of thermodynamics holds true. And I think we can be fairly sure that photons are part of the mechanism that makes it work. But it stretches the analogy too far to imagine a single photon transmitting energy from a cold material to a hotter one and present this as proof that the second law therefore only holds true on average.

Only a real experiment can settle a question like this. Thought experiments won’t help; neither will ridicule; and nor will appeals to the behaviour of everyday objects at the macro scale.

106. “You must remember Gord and his Solar Heater!

When pointed at the night sky the radiation from the sky lowered the temperature of a sample of water placed at the focus of the parabolic reflector to the extent that it formed ice!
Now if instead of forming ice the water temperature rose I would take your theory seriously.”

Both theories (exchange of photons and exchange of the difference in photons) predict the same result for this experiment. From an exchange of photons point of view, the addition of the mirror, leaves the water with the same amount of radiation from above as it had before, but replaces its view of the radiation from below with a lower amount from the night sky. If it emits the same but receives less then its temperature drops under this theory.

• What experiment?

Some amateur sticking a parabolic reflector out at night doesn’t qualify as an experiment and as has been pointed out, IR radiation emitted in random directions towards the ground isn’t a good source for a parabolic reflector to work with.
Plus no one has stated that there is a lot of IR at night, only that increasing CO2 will result in warmer nights than before.

107. Paul

I am in agreement with your post.
Prevosts theory of exchanges (1792) explained exactly what happened when a cold object is brought near another at a higher temperature.
There is a dynamic exchange of photons between the objects resulting in the colder getting warmer and the warmer getting colder.
At all times the heat moves from the hotter to the colder until equilibrium when there is no heat exchange even though now equal numbers of photon are exchanged between the bodies.

• My problem with that, and with the way others describe “heat” is that it makes “heat” sound like something other than energy that is being exchanged. When the two objects exchange photons, are they sharing the highway with some other substance called “heat”? Are both heat and photons being exchanged? Or is one object losing heat and the other gaining it through the exchange of energy at levels the objects are capable of radiating at?

108. DSL

…….”My problem with that, and with the way others describe “heat””……

The way so called Climate Scientists misuse the word “heat” shows the know very little about thermodynamics.
If you want to know the proper definition of heat look up an appropriate physics textbook.
Once you have a firm grasp of the scientific meaning of heat you will be disappointed by the level of ignorance shown by some self appointed experts on this topic.

• I have worked in places where some people were obsessively pedantic about terminology. Everyone else ignored them or took the ‘mickey’. Amazingly because everyone was relatively intelligent, they could understand and see through the mis-use of language and understood each other.

It is a form of intellectual egotism, I think, when two intelligent people argue about what word to use. But it is also a sign that they don’t understand language and the fact that there is no correct way of using it.

109. Warmcast

Where did you work — The Tower of Babel?

There is absolutely no doubt about the technical meaning of HEAT.
If it is incorrectly used by someone it gives the impression that they are muddled headed.

Therefor whatever else they wish to say about thermodynamics is taken with a pinch of salt.

• The fact that you even misunderstand my comment and can’t read the meaning, just shows how pedantic you are.
The fact is in the real world, people work in teams with different knowledge and backgrounds. Someone who is to obsessive, often doesn’t fit in because they fail to communicate with the wider audience of the team they work in.

You also seem to accuse everyone on the planet of being ‘muddled headed’.

110. Warmcast

Heat as used in thermodynamics.

Heat as used by SOD and more widely by some others in the Climate Science fraternity.

What we have here is called “equivocation of meaning.”

It is a reliable way to create muddle and confusion; it is listed in books on logic as a kind of fallacy.

Eliminated as far as possible in real science.

Frequently found in Climate Science

• No one is confused.
You have spent a lot of time here wasting your time by being pedantic. The science hasn’t changed.

As I said, in your case ego is a bigger issue.

111. DSL on October 28, 2010 at 7:00 pm
I think what you are asking directly relates to entropy, its hard to explain conceptually, but i shall take a stab
(this first para i posted elsewhere, c&p)

The second law is, entropy increases or stays the same. Entropy is the nature of energy/chaos. A high T is showing confined chaos, that chaos will increase, disperse. But it cant go from a more spread out state to a more confined state, without work being performed(and you still have increasing chaos here, because it will take as much energy to compress air say, than what you gain out of the compression, negative entropy can only happen if you look at part of a process. If energy is locked in chemical bonds, entropy stays the same, chaos dosnt spread, until it is released from its bonds, then its chaos will spread.

Now, obviously temperature is dependent on the material being energized, but basically temperature is chaos, energetic excitement. Some materials have higher thermal capacities, because they have more mass/it takes more energy to excite it. But the bottom line is, that the flow of energy is restricted by the concentration of chaos of the energy adjacent. Not the total amount of energy present.

So “heat”, the net exchange of energy, is really talking about what direction the chaos is increasing/spreading in, and increasing the concentration of chaos in this direction.

112. Since the second law is about entropy why not state that in your discussion.

dS> or = dQ/T

If you are saying that a hotter body increases it heat content because a colder body transferred heat to it then you are in violation of the second law. Absorption is a different matter.

Absorption will only take place if the object has the physical capibility to absorb at that wave length.

I.E. CO2 at 15 micro will not absorb at 8 micro or 52 micro. So CO2 does not gain heat or absorb from a cooler object at 52 micro.

Further, if 2 objects at the same temperature radiate at each other there is no heat exchanged even if they both absorb the others radiation.

So CO2 does not heat the surface of the earth.

• If you are saying that a hotter body increases it heat content because a colder body transferred heat to it then you are in violation of the second law.

Well the colder body can increase the thermal state of the hotter body, from transferring energy through radiation, resulting in a reduction in the net energy movement from the hotter body, or by slowing conductive losses by a decreased differential in T’s between the bodies. This is not a violation of the second law. But a consequence of it. The hotter body can only move as much energy across its boundary that it takes to move the colder body to equilibrium with it, at which stage no “net” energy would be being transferred. And thus obviously the closer to equilibrium the bodies are the less net energy transfer can take place. If the hotter body is receiving a supply of energy, energy will accumulate, until its differential in T is sufficient to move as much energy as it receives.

CO2 dosnt heat the surface of the earth, the sun does. The radiative properties of the atmosphere however mean that more energy is retained in the system than there otherwise would be.

113. mkelly:

Since the second law is about entropy why not state that in your discussion.

dS> or = dQ/T

Already covered in much detail in The Real Second Law of Thermodynamics.

Further, if 2 objects at the same temperature radiate at each other there is no heat exchanged even if they both absorb the others radiation.

Everyone agrees.
How about if there are 3 objects at different temperatures, what happens then?

• SoD

Now that you have access to the 8 books on heat transfer and you think yourself quite an expert, try the following simple problem.

Use an Ideal Carnot engine to transfer 200J of heat at 278K up to 200J at temperature of 330K.

Hint the answer is quite definitely not zero.

This little problem should illustrate the difference between the first law and second law of thermodynamics.
I will check, and correct your answer in 3 days, on my return from a short break

114. […] see Amazing Things we Find in Textbooks – The Real Second Law of Thermodynamics for six textbooks on heat transfer which all say, yes it does. Actually, seven textbooks, thanks to […]

115. I have a question about absorption= emission

During the day the land and sea absorb short wave radiation from the sun and emit the same amount of energy as long wave radiation.

However during the night the land (and sea?) apparently continue to emit long wave radiation while no longer receiving the short wave radiation from the sun.

In order to satisfy the absorption= emission balance during the night where does the absorption come from?

If there is some kind of time lag process going on then perhaps it needs to be discussed too?

In this case it would become integral(absorption) = integral(emission) however the bounds from the integration could be the source of a lot of debate (1 diurnal cycle, a week a year? the choice must be rigorously justified)

Maybe this point has already been discussed elsewhere? If this is the case maybe someone could redirect me to the explanation.

Many thanks.

• I don’t know what level you are approaching this. I apologise if this comment points out anything obvious to you.

At night (on the opposite side that is facing the Sun) the Earth will in total be emitting radiation (LW).
On the ‘sunny’ side it will be absorbing and emitting.
To add more complexity, there will be some horizontal mixing and movement between the night side and the day side. The night side would effectively be cooling (although clouds and greenhouse gases would delay it).

Then you have the issue of the Earths tilt and the fact that the equator getting the bulk of the insolation during the day.

The basic physics really only takes into account simple scenarios. If you try and go from the micro level and apply those basics to a macro level, you are going to get poor results, because you are leaving out to many other inputs to the system. It’s OK to keep it simple for educational purposes, but at some point you have to move on and consider the wider implications of other effects.

In theory if you took away the Sun (so it was night time 24/7), then the Earth would keep cooling. Gravitation pull from other objects might create sources of heat and emission, as would the Earths core.

116. Richard:

I’m not quite sure I understand where the confusion is.

I’ll take a stab at it, but feel free to ask further because it’s quite likely I haven’t addressed the underlying question.

In order to satisfy the absorption= emission balance during the night where does the absorption come from?

The solar radiation during the day is very high. During the night it is zero. On average it is “average”.

During the night the solar radiation to the surface is zero. But there is lots of stored heat in the surface from when the sun was shining.
During the night the atmospheric radiation to the surface (“back radiation” is still significant.

Why is the back radiation still significant? Because the atmosphere doesn’t cool to absolute zero because it has a heat capacity (like the earth’s surface).

And during the night the surface continues to radiate because it has retained heat within its heat capacity (as well as receiving some radiation from the atmosphere).

For a different perspective but a similar problem, take a look at Lunar Madness and Physics Basics.

In this case it would become integral(absorption) = integral(emission) however the bounds from the integration could be the source of a lot of debate (1 diurnal cycle, a week a year? the choice must be rigorously justified)

On average – if the earth is not heating or cooling – then the average emission of radiation to space will balance the absorbed solar radiation.

This is a fundamental result of the first law of thermodynamics and so doesn’t really need to be rigorously justified because it has already 150 years of evidence.

If you want to know whether one part of the earth over one time period is heating or cooling then you would need to know all of the specific heat fluxes into that region over that time period.

But maybe I haven’t understood your question/point.

I will try and clarify the question:

If we integrate the entire earth radiation balance at a given instant in time should it be zero or is there some lag due to the redistribution process?

• by lag I mean imbalance between total absorption and total emission at the given instant in time

117. Richard,

The first law of thermodynamics says that energy cannot be created or destroyed.

What usually happens in dynamics situations is that energy in does not equal energy out.

This means that the body or bodies under consideration are heating or cooling.

For example, let’s take a large body of water in some part of the ocean. It has a large heat capacity (thermal mass) – which means it takes a lot of energy to change its temperature.

Over any short period – say 1 hour – it is extremely unlikely that the heat fluxes into and out of this body of water balance. The ocean is heating up or cooling down. But the temperature changes are small because it has a large thermal mass.

This particular body of water might never be in equilibrium. So over 1 day or 1 week or 1000000 years the temperature might be in a continual state of movement.

If you look at the desert instead you find that the temperature changes are much more extreme – because the thermal mass of the desert surface is quite low.

If we integrate the entire earth radiation balance at a given instant in time should it be zero or is there some lag due to the redistribution process?

Is “No, it will almost never be zero”.

Have a look at Radiative Forcing, Thermal Lag and Equilibrium Temperatures.

• So surely it should be important to estimate the amount of energy and time delay for the given amount of absorbed energy to be emitted?

Is it possible to estimate/evaluate this and if so are you able to say to what degree of certainty the resulting values might be?

“Of course, calculations of feedback effects in the real climate might find thermal lag parameters to be extremely important.”

What I would like to get a feel for is the level of uncertainties that we are dealing with for the thermal lag of the real climate. (I guess this must have been already published somewhere?)

118. Richard,

This is down in the “absolute basics” section of thermodynamics – and climate science.

In any system you have to know the heat capacity if you want to calculate the temperature change.

The heat capacity of the earth’s climate system at any given location is one of the least challenging of all of the variables. Heat capacity is simply mass times specific heat capacity – which is a material property. E.g. water has a specific heat capacity of 4.2kJ/kg.K.
This means if you have 1kg of water it takes 4.2kJ to increase its temperature by 1K.

More challenging problems are those like the changes of water vapor in the atmosphere (see Clouds and Water Vapor, the large scale ocean currents that move heat around, and 100s of other difficult problems..

• I am familiar with Latent Heat and I’ve used Incropera and similar sources for quite a few years now.

My PhD was in turbulence simulation and since then I have had quite a few years post doc experience in turbulence modelling, heat transfer and combustion. I am currently a CFD and heat transfer research engineer for a big energy company.

My professional interest in heat transfer has predominantly been with convective and diffusive heat transfer although I have studied radiative transfer within the context of CFD of combustion for a couple of years during one of my numerous post doc jobs.

I apologise if this is a return to absolute basics for you and I appreciate the time that you have already given to responding to my questions.

I notice from the “clouds and water vapour” link, that Ramanathan 1981 seems to be the main reference paper. I imagine someone must have attempted to relook at the problem with more significant computing resources since then?

I wonder if you are able to put any figures to the thermal lag we might be able to expect from the real climate?

Many thanks.

119. Richard:

I think I have been misunderstanding your real question because it sounded like the concept of heat capacity was quite new to you. Clearly it’s not, from your background.

The thermal lag is primarily in the oceans. A lot of heat is also consumed in melting ice.

The challenge is understanding how quickly heat is moved into the ocean depths. This requires knowledge of ocean circulation which isn’t simple especially as density is determined from temperature and salinity. Salinity varies with evaporation less precipitation & ice melt.

Since you are familiar with fluid flows you will understand the challenge and why modeling ocean currents is the key to understanding the real thermal lag.

120. […] mangle of the second law of thermodynamics. See Science Roads Less Travelled and especially Amazing Things we Find in Textbooks – The Real Second Law , The Real Second Law of Thermodynamics and The Three Body Problem. And for real measurements of […]

121. Among all this discussion, and I have not read it all, I did not find where the problem caused by the Apollo temperature data on the moon mentioned. It was claimed to have created problems with the S-B theory for planets. What say you?

Willi

122. WilliMc:

You can find it explained in Lunar Madness and Physics Basics.

123. on December 14, 2011 at 10:55 pm | Reply Pierre R Latour

There is a basic flaw in SoD Radiation Exchange under The Real Second Law.
You start by defining Enet1 = E2 – E1, which is your prerogative, and then define Enet2 = E1 – E2, which is also your prerogative. You then use simple arithmetic to prove your second definition is the negative of your first definition, Enet2 = E1 – E2 = – Enet1. That is ok. There is no involvement of the Second Law, Boltzmann relation or any physics. All you proved is you are able to define something as the negative of your first definition and deploy arithmetic to prove you can to that.

I can do it too: I took 5 apples from that bag; then I returned 2. I define Enet1 = E2 – E1 = 2 – 5 = – 3. I also wish to define Enet2 = E1 – E2 = 5 – 2 = 3. Guess what? Enet1 = -Enet2 = – 3. Correct but not very complicated. Didn’t prove anything beyond I am left holding the bag.

You did not prove the hot plate must absorb E2, let alone reradiate it at higher E1. I do not see how you prove back-radiation does not violate the Second Law. I remain skeptical of back-radiation as enshrined in GHG theory.

Your second basic flaw is in SoD Under The Imaginary Second Law.
Your first equation ok, assumes all incident radiation absorbed, absorptivity of plate 2 = 1.0. You then define Enet2 = E1 – E2, which is your prerogative. Therefore, I find from arithmetic Enet2 = – Enet1 – E2. Then I claim E2 = 0 according to Second Law, and Enet2 = – Enet1 follows, in agreement with the First Law. So your “Imaginary Second Law” is quite real after all. QED.

I conclude your Radiation Exchange under The Real Second Law ignores the Real Second Law and your Under The Imaginary Second Law is actually in harmony with both the first and second. As you say, it is all in the English definitions and logic.

All the texts you cited assume the radiators are black bodies (or grey); that is they absorb all incident radiation. The texts are fine under their stated black body assumptions but they do not prove all radiators are black; when experiments prove none are. (These texts duplicate what I learned from chemical engineering radiant energy transfer texts in 1960’s, Bennett, C O & J E Myers, “Momentum, Heat, and Mass Transfer”, Mc-Graw Hill, 1962, pg 371-394; Bird R Byron, Warren E Stewart & Edwin N Lightfoot, “Transport Phenomena” John Wiley, 1960, pg 426-455; and Perry, John H & Hoyt C Hottel, “Chemical Engineers’ Handbook”, 1950, pg 483-498.) Therefore it is not surprising they conclude all incident radiation is absorbed. Concluding an input assumption is not very hard. Duh. Note the blog by Frank 7Oct10 is OK but does not refute my analysis.

Rather surprising none of your 199 bloggers caught either of your two basic flaws of logic. None proved hot radiators like bulb filaments and Earth surface absorb cold radiation from adjacent bulbs or cold CO2, and hence radiate more intensely. Because it just doesn’t add up. There is a basic flaw on the right side of Kiehl-Trenberth GHG theory energy flow diagram.

124. Pierre R Latour:

There is a basic flaw in SoD Radiation Exchange under The Real Second Law.

You start by defining Enet1 = E2 – E1, which is your prerogative, and then define Enet2 = E1 – E2, which is also your prerogative. You then use simple arithmetic to prove your second definition is the negative of your first definition, Enet2 = E1 – E2 = – Enet1.

That is ok. There is no involvement of the Second Law, Boltzmann relation or any physics. All you proved is you are able to define something as the negative of your first definition and deploy arithmetic to prove you can to that.

The writers of the textbooks claim that Enet2 = -Enet1.

The only way this can be true is if the radiation from the colder body is absorbed by the warmer body. (Or if the radiation emitted from the colder body is zero. This proposition is also contradicted by the writers of the textbooks).

My writing down of some simple equations is just to demonstrate this point mathematically. I don’t use simple arithmetic to prove that Enet2 = -Enet1, I use simple assumptions about radiation emitted and absorbed. It is not a mathematical definition unrelated to the physics, it is a mathematical definition derived from some assumptions about physics.

I would like you to write down the equations of radiative transfer and show how they match what the professors of heat transfer write.

All the texts you cited assume the radiators are black bodies (or grey); that is they absorb all incident radiation. The texts are fine under their stated black body assumptions but they do not prove all radiators are black; when experiments prove none are.

Textbooks on radiative theory start out with the simplest of assumptions – blackbody radiation and blackbody exchange. Then they move to greybody radiation and greybody exchange. Then they move onto yet more complex examples.

The textbook examples scanned include greybody exchange (a non-unity emissivity).

The textbook writers are not trying to prove that “all radiators are black”. They don’t believe that.

You did not prove the hot plate must absorb E2, let alone reradiate it at higher E1. I do not see how you prove back-radiation does not violate the Second Law. I remain skeptical of back-radiation as enshrined in GHG theory.

The writers of the textbooks, professors of heat transfer, claim that the hot plate absorbs E2 (the radiation from the colder plate).

I don’t claim that the hot plate “reradiates it at higher E1“. The hot plate radiates at higher E1 because it is hotter to start with. Not because it is creating energy.

Back radiation (downward longwave radiation from the atmosphere) exists. It is measured.

Just clarify which group you belong to:

1. The group that claims that back radiation doesn’t exist and all pyrgeometers and FT-IRs on the ground pointed up at the sky give totally false results?

2. Or the group that claims that back radiation exists, reaches the ground and then …? (be wonderful if you could fill in the dots because none of the adherents of the second group has yet explained what happens to this radiation)

125. Pierre R Latour

As you know there are two consistent ways of looking at radiative heat transport.

1. Using Maxwells equations and perhaps Claes Johnstons modifications.
This gives a single Poynting Vector from higher to lower temperature.
This gives the same answer as 2 (below) for radiation of wavelengths >10um.

2. Using radiative exchange of photons between higher and lower temperature objects.
This would be considered as the present orthodox physics method.
The radiative exchange is calculated using the SB equation

Anyone who uses two way radiative interactions ( by logic) thinks back radiation exists.

To say that photons cannot be incident on and absorbed by the colder ground is incorrect.
Heat is the net of the energy streams and always flows from higher to lower temperatures.
quite often an electric light bulb is analysed and conclusions drawn.

Where SoD has a weakness is his insistence that radiation does not have a quality.
In other words he thinks 1000J of black body radiation centred around 5um is exactly equivalent in all respects to 1000J of bb radiation centred around 15um

The big difference is in the complete inability of thermal energy to change spontaneously into work or even thermal energy of a higher quality in a given situation.

Electrical energy is supplied to the bulb and is of the highest quality.
This is transformed into thermal energy with some 5% being emitted as radiation characteristic of the temperature of the tungsten filament.

Lets say some of this radiation strikes a highly reflective reflector and is directed back to the filament.
What are the consequences?

1. If the bulb was switched off before the radiation arrives back.
Then all the back radiation would do would be to slow the rate of cooling of the bulb.
If a graph was taken of filament temperature against time it would show:
No reflector – a steep drop
Reflector present – a less steep drop.

So it is correct to say that the backradiation cannot heat the filament further.
Nor could any rearrangement of the details work because of the second law.
The incident radiation cannot be spontaneously transformed into higher quality energy.

2. Bulb left on .
The backradiation acts like an insulator and is exchanged for exactly the same quality of radiation from the bulb.
This is best thought of a kind of cancellation process for calculating purposes.
However the electric current is continuing to supply high quality energy and the surplus energy increases the temperature of the bulb .

The temperature rises not because of backradiation but because of continuing to supply energy at a fixed rate when extra insulation is added.
Now the increase in temperature might be very small but is real in my opinion.

126. Bryan,

Please show me a citation from any well accepted physics textbook or peer-reviewed paper published in a major journal that refers to the quality of EM radiation. EM radiation is defined by its intensity and spectrum. A CO2 laser emits light at narrow bands centering at 9.4 or 10.6 μm. That’s in the thermal IR. A blackbody emitting radiation with a peak wavelength of 10.6 μm would have a temperature of 273.4 K. A CO2 laser can cut steel. A microwave oven uses a frequency near 2.4 GHz. A blackbody with a peak frequency of 2.4 GHz would have a temperature of 0.041 K. You can cook food in a microwave oven.

You can do these things because the generating mechanism for the radiation isn’t thermal and the intensity of the radiation is very high over a very narrow bandwidth and, for the CO2 laser, is concentrated in a small area. In a quick search the only reference to radiation quality I could find was for the penetrating power of x-rays through aluminum. That’s actually a reference to the wavelength which is controlled mainly by the voltage applied to the x-ray tube to accelerate the electrons into the target. The metal used as the target also affects the x-ray emission spectrum of the tube. A target with high atomic number like tungsten will produce more intense radiation at short wavelengths for a given excitation voltage than a target with a lower atomic number like chromium.

You don’t do your credibility any good by using Claes Johnson as a reference.

127. DeWitt Payne

I can hardly believe you wrote this:

….”Please show me a citation from any well accepted physics textbook or peer-reviewed paper published in a major journal that refers to the quality of EM radiation.”…..

You will find literally thousands of references to the quality of radiation and heat.

I really thought that a chemistry degree would insist on an analysis of the Carnot Cycle.

quality of radiation, Kelvin , degredation of energy, Carnot cycle and you will find literally thousandsof hits.

You will have to work hard to resore the previous high regard I had for your opinions.

• Bryan,

Since you claim that energy quality is being ignored by everyone. Give me a quantitative analysis using energy quality that proves that, for example, the TFK09 energy balance is incorrect.

• DeWitt Payne says

…….”Bryan, Since you claim that energy quality is being ignored by everyone. Give me a quantitative analysis using energy quality that proves that, for example, the TFK09 energy balance is incorrect.”……

I have never said
“that energy quality is being ignored by everyone.”

Anyone who has studied the second law would not make that mistake.
SoD however has as a reading of posts above will indicate.
If all the energy flows were reversed the Ist Law would be satisfied however the second law would be violated.
Long wave input and solar wavelengths output just cannot happen

128. Newcomers impressed by Bryan can review earlier exchanges on this subject:

Comment from October 15, 2010 at 7:19 am

Comment from October 18, 2010 at 10:33 am

Comment from October 20, 2010 at 8:02 am

• Same old scienceofdoom.

I gave the first link (above) that I came to out of thousands of similar links.

Instead of agreeing or taking issue with the link you move swiftly along to avoid having to admit that on this point you are plain wrong.

129. Same old scienceofdoom – try and get Bryan to answer a basic question.

If it gets absorbed does it change the internal energy?
Does internal energy affect temperature?
(answer in all physics textbooks = yes)

The earlier links establish the difficulty of this exercise (getting Bryan to answer). No more from me on this unless Bryan answers these questions, apart from periodic pointers to the old questions on this subject for newcomers.

I gave him a link out of thousands.

Not only does radiation (or more generally energy) have QUALITY they now have a word that quantifies the quality.

This word is EXERGY.

It would be refreshing if just for once people could admit they were wrong and thank me for my patient efforts to rectify their mistakes

http://www.mdpi.com/1099-4300/3/3/116/pdf

• It would be refreshing if you understood language and the words quality and exergy. Stop defending your misuse of language.

• Paul D

Just read the links I provided and make a rational point or stop wasting everyone’s time.

130. on December 16, 2011 at 5:18 pm | Reply Pierre R Latour

SoD still proves nothing. SoD now admits he just copied some stuff down from text books. Failed to rebut my entry.

It is not my job to teach SoD. I already did SoD a favor by explaining what SoD actually did. Hint, matter does not necessarily absorb all incident radiation; neither does it reradiate all absorbed with same spectrum and higher intensity.

I am independent of any group; I think for myself and engineer things from valid science. I disapprove of attempts to classify people into ill-defined groups.

Brian is correct.

131. Pierre R Latour,

“Declaring victory” is something different from science.

I doubt you even understand what you think you are trying to refute, as your “hints” are not what I have claimed.

I defined some assumptions, wrote down corresponding equations and proved that the result matched what was in all the textbooks.

You have not demonstrated anything, except an ability to declare victory.

If you want to demonstrate something of interest to readers here:

– Write down your assumptions clearly
– Write down the corresponding equations
– Show how these reach the same conclusion as the professors of heat transfer.

I doubt you can do this simple task.

132. I’m failing to understand the basic arguments put forth that the GHE violates the 2nd law. The GHE is not about energy traveling from cold to warm through a conduction process, so I just don’t get the objections.

Do any of you detractors out there actually think that a re-emitted photon cannot travel from the colder atmosphere toward the warmer surface? If this were true, how do the photons from the Sun travel through to colder upper atmosphere and reach the surface?

• I mean, I can even agree the net effect of additional CO2 could be zero or a wash, but not because of any second law violation. This is silly.

133. […] surprise to people familiar with the basics of radiative heat transfer. However, Kramm & Dlugi are apparently “in support of” Gerlich & Tscheuschner, […]

134. […] That’s my equation. I have provided six textbooks to explain this idea in a slightly different way in Amazing Things we Find in Textbooks – The Real Second Law of Thermodynamics. […]

135. […] That’s my equation. I have provided six textbooks to explain this idea in a slightly different way in Amazing Things we Find in Textbooks – The Real Second Law of Thermodynamics. […]

• Its a pity for you that the quote and equation was from a peer reviewed paper that states that the greenhouse effect is an empty false conjecture.

• That statement by Kramm and Dlugi is an assertion that was not supported by the evidence in the paper. Kramm was completely unsuccessful defending that assertion in the relevant threads here.

136. DeWitt Payne says

“That statement by Kramm and Dlugi is an assertion that was not supported by the evidence in the paper. Kramm was completely unsuccessful defending that assertion in the relevant threads here.”

Well that, as “Dude” Lebowski was fond of saying” is just a matter of opinion”

You will need to be a bit more specific that issuing a wide generalised sound off.
Make it easy, pick just one point that you claim was not substantiated and then we will examine your conjecture.

137. First, let me say that I have no problem with the principles of radiant heat transfer as described in the various texts. It’s what I have always understood and it agrees with my understanding of Stefan–Boltzmann law and so on.

But I’ve been asked where someone can go to find details of physical measurements that confirm these results. I would imagine that the first year of a physics degree course would involve such experiments but I certainly found myself unable to come up with an answer.

Could you help me come up with an answer as to where one can find description of physical measurements that confitm these relations? Or maybe they follow inevitably as the consequence of simpler more fundamental laws that have been confirmed by physical measurements?

• Martin A,

Good and interesting question. I wish I had the complete answer. The essence of radiant heat theory was worked out in the late 1800’s and early 1900’s and I’m sure that a reprint of some seminal papers from that era would be appreciated by many including me.

I have got Max Planck’s The Theory of Heat Radiation from 1914 which includes little gems like:

This law, which states that the volume density and the specific intensity of black radiation are proportional to the fourth power of the absolute temperature, was first established by J. Stefan[1] on a basis of rather rough measurements. It was later deduced by L. Boltzmann[2] on a thermodynamics basis from Maxwell’s radiation pressure and has more recently been confirmed by O. Lummer and E. Pringsheim[3] by exact measurements between 100’C and 1300’C, the temperature being defined by the gas thermometer..

[1] J.Stefan, Wien. Berichte, 79, p391, 1879
[2] L.Boltzmann, Wied. Annalen, 22, p.291, 1884
[3] O.Lummer und E.Pringsheim, Wied. Annalen, 63, p.395, 1897. Annalen d. Physik, 3, p.159, 1900

(This book is available from The Book Depository for only \$26, free delivery worldwide).

Generally the results are presented as “known facts” in physics and engineering courses.

If any readers have some early papers or later experimental work, please post them. I have lots of papers which show experimental results of material properties like emissivity which of course rely on the fundamental theory being correct.

It’s good to be skeptical of results presented without evidence, although I do generally work on the basis that when all the textbooks say the same thing and the science was worked out 100 years ago it’s probably correct.

• Martin A

An experimental method of testing Stefan’s Law is given on page 91 of
Experimental Physics for Students by Whittle and Yarwood.

Pub Chapman and Hall (1973)

138. SoD,

In all your textbook examples, I did not see one sample calculation that indicated T2 ( the higher temperature body) ending up at a greater temperature because of T1 (the cooler body). Nor did I see the statement that a cool body can cause a warm body to become warmer either. Did any of these texts have statements on the Second Law? Next time I am at U.C. Davis (in a few weeks) I will browse the thermo textbooks and report what I find as far as the statements for the 2nd Law. Just a quick browse on the internet, I pulled up the following thermo notes from a MIT course:

http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node37.html

The 2nd Law does not appear to have changed since the thermodynamics course I took in the early 80’s. There is nothing “imaginary” about what is stated in the above.

My prediction is that you will not find any scientific experiments (not thought experiments) that violate the Clausian statement of the 2nd Law either.

The burden of proof is on you.

• Jonathan,

Why should anybody claim that “the Clausian statement of the 2nd law” is violated. There’s nothing wrong with such a statement. You have just failed to understand the statement. Your ideas are so bizarre and unbelievable to any physicist that no-one has seen any need to counter them.

There must be tens of thousands of empirical measurements that confirm the influence of a colder surface on the temperature on a warmer one. Every practical calculation of radiative heat transfer in any application is based on formulas that include that effect and really many of those calculations have been verified in practice. That’s not only esoteric science, that’s also very practical and successful engineering.

• SoD,

You keep avoiding the issue. Your samples simply do not show any calculation whatsoever of the hotter body becoming hotter, because that is a violation of the Second Law of Thermodynamics, which has not changed since its inception. Show me a textbook that presents a calculation indicating the warmer body being raised to a higher temperature by the cooler body. You have not, and you will not find one, ever. Don’t put the burden of proof on me, when it is on you to since you are the one who disagrees with basically all thermodynamic textbooks that have been published. I guess MIT had it wrong in their course notes as well, and the SoD site is right. Sure. Go argue with them.

• Pekka,1983]

I am not asking for 10,000 measurements. All I am asking for is one. Give reference to one controlled scientific experiment where a cooler body raises the temperature of a warmer body.

By your logic, it would then follow that two bodies at the same temperature next to each other would cause each to raise in temperature because they are radiating infrared towards each other. Right? Each one is receiving IR radiation, which would heat them further. Right?

If a cooler body could cause a warmer body to become warmer, then the “warmed up” warmer body would then in turn cause the cooler body to become warmer. Then the warmed up cooler body, would then warm up the warmer body even further, and so on. Now that is what I call bizarre, and a violation of the First Law of Thermodynamics as well. Presto! Energy creation.

The Clausian statement is very simple: “Heat cannot of itself pass from a colder to a hotter body” [1850]; “No process is possible whose sole result is the transfer of heat from a colder to a hotter body” [Adkins, 1983]; “Heat of itself cannot pass from a colder body to a hotter one; work is required” [Haynie, 2001]; “No process is possible which consists solely in the transfer of heat from one temperature level to a higher one” [Smith, Van Ness & Abbott, 2005]. The statement is quite easy to understand.

• Jonathan,
I tried to explain, why nobody reports such experiments in a way that you seem to require. All laws of physics get their strongest evidence when they are put in practice and when they do that successfully. When that has happened the issue that remains is to understand those mechanisms, not to prove them.

For this phenomena the full confirmation was strong enough sometimes in the 19th century. It was built up gradually. Therefore it wasn’t necessary even then to do any specific experiment. One could, however, say that every single experiment that was done to study the Stefan-Boltzmann law, Kirchoff’s law and Planck’s law were also confirmation of the physics that includes as an inseparable component the warming influence of a cooler body on a warmer body.

• Jonathon,

The burden of proof was on me and I have proved my case.

To do this I used a very simple pair of equations.

If you can read equations you will see that there is a difference between the two sets of equations under the subheading Radiation Exchange under The Real Second Law and Under The Imaginary Second Law.

My contention is that this is the equation for rate of energy change for the hotter body:

Enet1 = E2 – E1 = εσT24 – εσT14

The contention of the proponents of what I like to call the “imaginary law” is that the equation for rate of energy change for the hotter body is:

Enet1 = –E1 = -εσT14

Then I show that the textbooks claim the first equation.

It’s pretty simple. And this is proof. At least, proof that all the heat transfer textbooks agree with my claim.

So now the burden of proof is on you to:

1. Derive an equation for Enet1 under your assumptions and show that your new equation matches what is in the textbooks – OR

2. Find some textbooks with different equations.

Obviously you won’t be able to do either.. but, over to you.

And please, just to convince yourself and help all the other confused people, why not put this out on one of the popular climate “skeptic” blogs and get the army of people who agree with you to come up with an equation different from what I came up with.

Just ask all of the firmly convinced “I know heat transfer” claimers to dig out their notes from when they studied radiative heat transfer and find the section on what happens to the thermal radiation from the colder body when it gets near the hotter body.

Oh, silence from them. What, no one did the course?

Surely it should be so easy.

And when you are done pondering that, why not check out the simple examples with calculations of entropy changes:

The Real Second Law of Thermodynamics – explaining how entropy is calculated with a few simple examples. How entropy increases – in accordance with this law – when energy is transferred by radiation from colder to hotter bodies, as well as the reverse, so long as net heat transfer is in the right direction.

The Three Body Problem – a simple example with three bodies to demonstrate how a “with atmosphere” earth vs a “without atmosphere earth” will generate different equilibrium temperatures. This includes entropy calculations to show the second law of thermodynamics is not violated.

If entropy increases with radiative exchange why do you think the second law of thermodynamics is violated?

Once again, the reason is simple. None of the people who think thermal energy from a colder body being absorbed by a warmer body violates the second law of thermodynamics actually understand this law.

If they did understand it, someone would have come along and provided their own entropy calculation.

Take a look through the many articles on the subject on this site – Confusion over the Basics under the subheading The Second Law of Thermodynamics.

Do you see many people convinced that I am wrong? Do you see any of them provide a calculation of their own? Doesn’t that make you wonder?

139. Jonathan on March 22, 2013 at 4:29 am
… I did not see one sample calculation that indicated T2 ( the higher temperature body) ending up at a greater temperature because of T1 (the cooler body).
———————–
Please have a look at a couple of experiments I did here (more to follow)
Both seem to show the additive properties of radiation from hot or cold source.

Please feel free to criticise and suggest improvements!

http://climateandstuff.blogspot.co.uk/2013/03/a-cool-object-reduces-energy-loss-from.html

140. Jonathon,

I assume your responding to my comment of 10:20pm. I ask some questions throughout my response and hope you will respond to them.

You keep avoiding the issue. Your samples simply do not show any calculation whatsoever of the hotter body becoming hotter, because that is a violation of the Second Law of Thermodynamics, which has not changed since its inception. Show me a textbook that presents a calculation indicating the warmer body being raised to a higher temperature by the cooler body. You have not, and you will not find one, ever. Don’t put the burden of proof on me, when it is on you to since you are the one who disagrees with basically all thermodynamic textbooks that have been published. I guess MIT had it wrong in their course notes as well, and the SoD site is right. Sure. Go argue with them.

I’m not avoiding it, I am explaining it.

This particular article demonstrates that all heat transfer textbooks agree that cold bodies radiate towards hot bodies and a proportion of that radiation (as defined by the absorptivity) is absorbed by the hotter body.

That’s this article. If you don’t understand equations then you won’t be able to (usefully) comment.

1. Do you understand equations?
2. If so, which of the two equations do you believe is the correct equation for the net energy absorption rate of the hotter body (in my response of 10:20pm – reproduced from the article)?

Two other articles I have written (see links in my earlier comment) explain that even though radiation from colder bodies is absorbed by hotter bodies this is not a violation of the second law of thermodynamics. I prove this by calculating the entropy change.
More thermal energy is transferred from the hotter body to the colder body than is transferred from the colder body to the hotter body. It’s very simple. That’s because heat (the net flow of energy) moves from the hotter to the colder.

3. If entropy increases is the second law of thermodynamics violated?
4. Can you find a flaw in my entropy calculation?

In your comment to Pekka at 12:55am you helpfully explain that you have no idea how to solve a simple equilibrium equation. And why you have a conceptual problem with radiative heat transfer.

Probably you don’t realize this. Many equilibrium problems of heat transfer can also be ridiculed in the same Zeno’s paradox fashion. If A affects B and B affects A then there’s a thermal runaway.. no, there’s just a different equilibrium condition. I demonstrate how to solve this in another article inspired by an equally confused commenter as yourself. The calculation is a comparison of how the temperature changes when instead of a body being confronted with very cold deep space it is confronted with another body at the same temperature.

5. Can you explain the flaw in the equilibrium calculation in that article?

If you refer to the three body problem I linked earlier you see a very simple comparison.

The surface of the earth is warmed by the sun and cools to space. In the “no atmosphere” case the equilibrium surface temperature is lower than in the “with atmosphere” case. In both cases entropy increases (second law). In both cases energy is conserved (first law).

6. Can you produce your own calculation of the equilibrium temperatures in the two cases and demonstrate conservation of energy and increase in entropy?

Here’s where a welcome response will be answers to the questions I have posed and an attempt to understand the subject, rather than “so you’re saying MIT is wrong”. Because I’m saying that standard thermodynamics is correct, and MIT is correct, and this is standard thermodynamics explained here and instead of repeating a mantra that you don’t understand, please engage with the physics explained.

• SoD,

Please do not be so condescending. I have a B.S. in civil engineering and also hold a professional civil engineering license in the state of California. Of course I can solve equations. You can look at any ABET accredited university and see the rigorous coursework that is required for a B.S in civil engineering. In all your posts you keep avoiding the issue of temperature. I am not confused. You are confused because you do not seem to understand the 150 plus years of the Clausian statement of the Second Law. It is very specific in what it says. It does NOT say anything about NET energy, or NET anything. That is your fabrication. You have not shown me ONE calculation from a textbook that indicates that a cooler body will increase the temperature of a warmer body. If your “calculations” are so robust and obvious, then there should be ample samples in multiple thermo textbooks indicating that a cooler body does increase the temperature of a warmer body. But so far you have not. As I said the burden of proof is on you, You are just one insignificant website that is going against the grain of the history of thermodynamics and physics in general. And you still have not indicated any published scientific experiment that supports your contentions either.

141. Jonathan,

Please do not be so condescending. I have a B.S. in civil engineering and also hold a professional civil engineering license in the state of California. Of course I can solve equations..

My claim is clear. Heat transfer textbooks, written by professors of heat transfer, teach the same thing that I am explaining on this blog.

Anyone who can read an equation can look at the textbooks in this article and see that this is so.

People who claim otherwise can’t read equations or don’t want to try.
Your other option is to claim that all the heat transfer textbooks provided are wrong. One of our regular commenters claimed I had cherry picked but unfortunately provided an alternate textbook that said exactly the same.

This is not in dispute in the field of heat transfer and hasn’t been for over 100 years. It’s just in dispute by people who have never studied it.

..I am not confused. You are confused because you do not seem to understand the 150 plus years of the Clausian statement of the Second Law. It is very specific in what it says. It does NOT say anything about NET energy, or NET anything. That is your fabrication.

When Clausius wrote this statement the subject of thermal emission of radiation was not really understood. Radiative emission and absorption became clear in the late 1800s. Prior to that it wasn’t very clear that there was any such thing as a radiative exchange.

You have not shown me ONE calculation from a textbook that indicates that a cooler body will increase the temperature of a warmer body.

I showed 6 textbooks with exactly that. Strictly speaking you are correct I haven’t shown ONE, I’ve shown SIX. (Well, seven, if we add the contribution of the cherry-picking fame).

That’s why I asked my “condescending question” – can you read an equation. Many people can’t, that’s why I asked.

If your “calculations” are so robust and obvious, then there should be ample samples in multiple thermo textbooks indicating that a cooler body does increase the temperature of a warmer body..

Let me refer you to the first equation highlighted from the first scanned textbook on heat transfer.

This is equation 2.1 in the first scan.

q = σ(T14 – T24)

If T2 increases what happens to the rate of heat loss?

Is the equation wrong?

Can T2 change in temperature and not affect the rate of heat loss?

• SoD,

How many times do I have to state this? You have not shown ANY calculations (not 1, not 6, not 7) from a textbook that specifically show that the temperature T of the warmer body is elevated to a new and higher value due the presence of a cooler body. None. Nada. Zip. You are dreaming. The samples do not show this type of calculation. Because it has never happened and never will. Where are the specific before and after values shown anywhere? If T2 is the warmer body, and T1 is the cooler body, where are the samples that show the specific higher value for T2? And what are the final values for T1? Your samples simply do not address this at all. Can you not comprehend this simple question? Show me a specific calculation with actual values…from a textbook……get it? Like real numbers, with real temperature values (T2-initial, T2-final, T1-initial, T1-final).

The Clausian statement has not changed at all over 150 plus years. You are guilty of revisionism. It is still the same statement with the same meaning. Your revisionist explanation is not indicated anywhere in textbooks.

This magical feat should have been well documented experimentally. Please indicate a published real scientific experiment which confirms your hypothesis. That’s how the scientific method operates; 1. Ask a question (is the Clausian Statement true?) 2. Form a hypothesis. 3. Make a prediction. 4. Testing: conduct an experiment to determine whether the real world behaves as predicted by the hypothesis. 5. Analysis: determine what the results of the experiment show and decide on the next actions to take; for example, revising the hypothesis if the results do not confirm the hypothesis. So far you are somewhere between steps 2 and 3.

That is how real science operates.

• Jonathan,

You haven’t answered any of my questions. Do you accept the equation that I presented? Or not?

No commitment from you.

Using these equations (from the textbooks), accepted by everyone in heat transfer, it is trivial to demonstrate, as I do in The Three Body Problem that the colder atmosphere warms the surface (compared with the case of no atmosphere). You haven’t disputed these calculations or presented your alternatives.

Probably you haven’t even bothered to read it.

So far in our discussion, as with all passionate advocates of the imaginary second law of thermodynamics, you have not accepted or rejected any equations. You have not presented any equations.

So I am left in the dark as to what equation you would use to calculate a change in temperature.

This is because you don’t know.

I claim this, because after 3 years of claims from passionate advocates of the imaginary law of thermodynamics I have never seen one equation produced.

If you knew, you would either supply the equation with the results, or demonstrate the flaw in my calculations.

The three body problem makes it plain as day. But only for people who can read equations.

The passionate advocates of the imaginary second law of thermodynamics cannot explain what happens to the radiation from the colder body when it reaches the hotter body.

Does it disappear?

• SoD,

This is hopeless. You are not engaging in science. Show me from a valid textbook, then show me experimentally, all in accordance with the scientific method. So far you’ve given 7 textbook samples which do not in any way shape or form provide calculations indicating a cooler body increasing the temperature of a warmer one. Your three body “problem” is a device of your own creation, not from a thermodynamic textbook.

I stand on absolute firm ground, behind a “law” of physics. Not just some suggestion, or hypothesis. A law that has not changed since its inception over 150 years ago. You have not shown me one example of the (your) new and improved version of the Clausian statement in any textbook.

Now if you propose to amend the Second law, hey, be my guest. But in order to accomplish this scientifically, the strict tenets of the scientific method must be followed. You are not even close. Your burden of proof. It is not our responsibility to prove you incorrect. You have to prove to the world that your hypothesis is correct, not just to your army of followers, or some gullible individual who happens upon this site. And the methodology for that proof is the scientific method.

You have been working on this for three years?! And you are still on second base? I suggest you get cracking. Maybe I’ll check back in a few more to see if you’ve made any progress, or if you are still just spinning your wheels.

• Jonathan,

Development of physics didn’t stop with Clausius. His formulation of the Second law is still valid, but there are more detailed formulations that are also valid. These more detailed formulations of the Second law are consistent with the Clausius formulation, but they also add to that.

Heat as considered in the Classical Thermodynamics is by definition net flow of energy. None of the classical texts mentions net heat flow just because their definition is that heat flow is always net. Such a heat flow does not by itself say anything about the components of the energy flow. It could equally well be just unidirectional flow or it may be the difference of two opposite energy flows that are much larger than the net flow.

When we look at the heat transfer through radiation, we have the laws like the Stefan-Boltzmann law and we have the more detailed formula of Planck. Both these formulas describe emission from a body or surface. They don’t by themselves tell about the net transfer, i.e. the heat transfer using the language of Classical Thermodynamics. When we want to calculate the energy transfer from a body to another we must look also at the geometry of the bodies and we must take into account the emissivities and absorptivities. To get the heat transfer we must do all that and calculate the difference of the opposing flows.

One very important point is that emissivity and absorptivity of a surface or a volume of gas are equal for a given wavelength. This is often called Kirchhoff’s law although the formulation differs from the original one given by Kirchhoff. It’s possible to derive the modern formulation from the original one and vice versa. On a more fundamental level Kirchhoff’s law can be derived from Quantum Mechanics. Kirchoff’s law is also one way of expressing the Second law. If the emissivity would not equal absorptivity then the Second law might be violated.

All these laws can be combined to express the radiative heat transfer as a difference of two terms, one of which describes energy transfer from the warmer body to the colder one while the other describes energy flow in the opposite direction. The meaning of the two terms is obvious and the formula tells beyond doubt that the temperature of the colder body does affect the warmer body. Making the colder one less cold leads to a warming of the warmer one.

Where do the empirical confirmations enter?

They enter in confirmation of the full theory, i.e. of the formula that describes the energy transfer based on temperatures of both bodies, the geometry of the setup and the emissivities. This full theory is the one that has been verified tens of thousands of times by engineers who do heat transfer calculations and measure whether their calculations were correct at a level expected from the care they have put in doing the calculation.

142. Jonathan,

For radiative physics in the thermal region (wavelengths > 3um) two approaches are valid.

Quantum Theory (Planck) and the classical Rayleigh – Jeans Law.

That is both give the same answers to specific problems.

http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html#c4

Indeed there is no backradiation in the Rayleigh Jeans model.

For radiation < 3um the Quantum Theory must be used.

SoD and other supporters of IPCC science who frequent this site tie themselves in knots when asked to define HEAT.

Indeed some say that radiation is heat.

Heat is the net flux of the radiative exchange and always goes from the higher to the lower temperature object.
Heat is always capable of doing thermodynamic work and this is an excellent test of whether we are dealing with heat or not.

Backradiation is not heat and cannot of itself increase the temperature of the higher temperature object.

143. Jonathan,

I ask my same question again, because these equations do precisely and definitely result in something opposite to what you claim.

Let me refer you to the first equation highlighted from the first scanned textbook on heat transfer.

This is equation 2.1 in the first scan.

q = σ(T14 – T24)

If T2 increases what happens to the rate of heat loss?

Is the equation wrong?

Can T2 change in temperature and not affect the rate of heat loss?

Why do you not answer these simple questions?

Too risky?

I will answer because you are afraid to. Anyone who can read an equation can see that when T2, the colder temperature increases, the net loss in energy from the warmer body is reduced. This is what the equation says in English.

Therefore, from the first law of thermodynamics, if this body is receiving a constant supply of energy from elsewhere (the sun), the reduction in energy loss must result in a temperature increase.

Most maths textbooks don’t bother proving 13+7=20 because it follows obviously from definitions and basic rules. Likewise most heat transfer textbooks don’t pick out the problem of the sun, earth and atmosphere because there’s 18,000 other examples to choose from. They state the basic equations and give worked examples to show how the principles are applied.

As and when a defender of the imaginary second law of thermodynamics is prepared to state an equation of radiative transfer and follow it through it will be an amazing day. I am not holding my breath. On the other hand, I have stated equations and proved the obvious consequences. Imaginary second law proponents – please one of you identify which equation is wrong, or which result violates the second law by providing an entropy calculation. On this also I expect many years to go by with no actual response.

144. on March 23, 2013 at 12:11 pm | Reply Glenn Tamblyn

Jonathon

You are falling into a fundamental trap here, You are looking at the process behind the GH Effect and assuming that you can apply 2nd Law processes to just the temperature of the Earth;s surface and the atmosphere. But to do that you are fundamentally violating the 2nd Law itself. Consider SoD’s statement in their post:

1b. Entropy of a CLOSED SYSTEM can never reduce

This the MOST FUNDAMENTAL statement of the 2nd Law. Statements about different bodies warming or cooling are actually derivative consequences of this statement.

So can a colder body ‘warm’ a warmer body? No. IN A CLOSED SYSTEM!!!!

But the Surface/Atmosphere system being considered here IS NOT CLOSED!

A more accurate description of the system, of a closed system, is that it has 4 parts – T1, the Sun at 5700 Deg C, T2, the Earth’s surface. T3, the Atmosphere, T4, deep space at around 3 K (-270 Deg C)

So the closed system we need to consider contains all 4 parts. And the 2nd Law applies to the start and end of this chain. T1 (Sun) and T4 (deep space). Can deep space cause the Sun to be warmer? No. The 2nd Law prohibits that because the ‘system that includes T1 & T4 is a closed system. But can T2 be warmer as a result of T3? Yes. Those tow things aren’t a closed system.

Think of it this way. The 2nd Law controls the start and end temperatures for the system – Sun and Deep Space. However it says nothing about the intermediate temperatures that internal components WITHIN the system may have. The 2nd Law does not apply to the relationship between T2 & T3. (or for that matter between T1 & T2, T3 & T4 etc. It only applies to T1 vs T4)

2nd Law thinking does not apply to internal states within a closed system. It only applies to the system as a whole. T1 to T4.

• Glenn Tamblyn

Your definition of the second law is so wide that almost nothing contradicts it.

For instance would you find this an acceptable proposition.

The troposphere at an average temperature of 255K (say), heats the adjacent Earth surface at 280 K .

• Bryan,

The problem here is that I cannot get a clear definition of Mr. Doom’s hypothesis. Forget about the greenhouse effect. He keeps droning on and on that the 2nd Law does not really mean what it says, then provides no textual back-up for his revised Second Law, nor does he provide any experimental back-up either. I provided 5 different versions of the Clausian statement of the Second Law. All say the same thing. His version simply cannot be found.

So in effect, as far as science is concerned, he is not even on step 2 of the scientific method concerning the formation of a clear hypothesis for his revised Second Law. For three years he has just been asking questions, messing around with “equations” and spinning his wheels. Not making any scientific progress.

Kind of reminds me of some lyrics from that old Blood, Sweat and Tears song:

“What goes up must come down. Spinning wheel got to go round…”

I mean according to some of his followers, T1 (the sun), would be heated up slightly due to the small amount of IR it receives back from the earth (T2). Find that in any thermo book!

I can only shake my head.

• Jonathan,

I’m not “SoD’s follower”. I have in common with him that we both try to explain what science tells. Sometimes we succeed in getting the readers or other audience to understand the points that we make, sometimes we fail. We both are followers of what science tells in the way as all physics teachers are followers of what science tells.

When we discuss with someone who is totally convinced that his view of physics is the only right one and that view differs from the standard interpretation of textbook physics, we have little change in getting the message trough. We are bound to follow the present understanding of physics. When that’s not accepted, there’s little that we can do.

• Yes, the fundamental formulation of the Second law is:

The entropy of a closed system cannot decrease.

All other consequences of the Second law can be derived from that. One of those is:

The temperature of the hottest body of the closed system cannot increase and the temperature of the coldest one cannot decrease. The law by itself does not say anything on how the bodies in that closed system influence otherwise each other. To draw conclusions on that more detailed calculations are needed.

The law tells also that the net heat transfer between two bodies is always from the warmer to the colder. I included here the word net to make the sentence more general. It’s superfluous in the language of Classical Thermodynamics, because in that theory heat refers only to the net transfer of energy that’s not work. It’s, however, important, when we have a more concrete definition of heat as unorganized internal energy of bodies.

Physics is a theory that’s best described by formulas. A working formula is a valid statement of physics. Words are used mainly to define the meaning of the symbols used in the formulas. It’s important to understand the definitions consistently, i.e. with the meaning that they have when the formula was written and when it’s validity was verified. The same word may have had different meanings at different times. Heat is an example of a word that’s used even presently in somewhat varying meanings. One of the ways is that of the Classical Thermodynamics, but that differs from the casual use of both laymen and scientists. This difference causes sometimes confusion at least in net discussions, but hardly ever among scientists.

145. Always amazes me, how these armchair quarterbacks, many of which show evidence of a DEEPLY rooted hatred/misunderstanding/evangelistic distrust of 200+ years of SETTLED science, attempt to lay waste to said years of confirmed research, but NEVER publish anywhere but…a BLOG. Hmmm.

146. “Give reference to one controlled scientific experiment where a cooler body raises the temperature of a warmer body.”

Um…OK. When I (a warmer body) go to bed, and I get cool, I put on a blanket (a cooler body.) There. I solved it for you. Unless you DON’T use blankets, too?

• Harry,

Put a blanket around a corpse. Does the corpse rise in temperature? Maybe it will spring back to life as well!

Hey! And all these years I should have just wrapped a blanket around my wallboard heater and produced extra heat! Energy crisis solved!

Gee Harry, I must have missed that published and peer reviewed study in Nature, or Science, regarding the blankets. Maybe you can give me the reference.

147. Jonathan says:

The problem here is that I cannot get a clear definition of Mr. Doom’s hypothesis. Forget about the greenhouse effect. He keeps droning on and on that the 2nd Law does not really mean what it says, then provides no textual back-up for his revised Second Law, nor does he provide any experimental back-up either. I provided 5 different versions of the Clausian statement of the Second Law. All say the same thing. His version simply cannot be found.

However, I have already referred Jonathon to the article The Real Second Law of Thermodynamics.

In that article I state the second law of thermodynamics in terms of entropy, provide an equation for entropy, and provide a worked example for radiative exchange.

And I have already requested Jonathan to confirm or deny that an increase in entropy can take place while the second law of thermodynics is violated.

If Jonathan says “yes entropy can increase while the second law is violated” then I will produce textbook refutations.

If Jonathan says “no entropy cannot increase while the second law is violated” then I will ask Jonathan to provide either:

a) an alternative calculation of entropy – if he does so I will provide textbooks to prove the standard calculation of entropy
OR
b) demonstrate that my calculation of entropy in the worked example is incorrect

The only alternative for Jonathan is to do what he is already ably doing – claim I am avoiding the subject while refusing to comment on the subject of entropy – probably because he has realized he doesn’t know what entropy is and doesn’t want to say so.

This is the only “successful” course of action on this topic, and most passionate previous believers in the imaginary second law followed the same pathway.

None stated what they believed about entropy, none provided an alternative entropy calculation.

The reason is clear.

• on March 24, 2013 at 12:22 am | Reply Glenn Tamblyn

Just to highlight SoD’s point, the opening sentence in the Wikipedia entry on the 2nd Law, which is quite detailed says:

“The second law of thermodynamics states that the entropy of an isolated system never decreases…”

• Glen,

Wikipedia is also clear on the Clausian statement:

“Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time” and,

“Heat cannot spontaneously flow from cold regions to hot regions without external work being performed on the system, which is evident from ordinary experience of refrigeration, for example. In a refrigerator, heat flows from cold to hot, but only when forced by an external agent, the refrigeration system.”

Wow!. What a total bummer.

148. SoD
You must have a bit of sympathy for the ordinary blogger with a standard laptop.
Setting out equations is very awkward and time consuming.

However this link to Rodrigo Caballeros excellent set of notes should keep you happy.

On page 36 Rodrigo with the aid of entropy equations proves that heat can only flow from a warmer to a colder body.

• Bryan,

Regular readers will expect no less from you than avoiding a direct question.

A. Which equation is correct for the net rate of energy change from the hotter body?

1. Enet1 = E2 – E1 = εσT24 – εσT14

2. Enet1 = –E1 = -εσT14

You will avoid the question.

B. Where is the error in my calculation of entropy in The Real Second Law of Thermodynamics ?

You will avoid the question.

The reason is clear.

• SoD says….” Bryan, Regular readers will expect no less from you than avoiding a direct question.

A. Which equation is correct for the net rate of energy change from the hotter body?

1. Enet1 = E2 – E1 = εσT24 – εσT14

2. Enet1 = –E1 = -εσT14

You will avoid the question. ”

The heat loss from hotter body 2

E = E2 – E1 = ε2σT2^4 – ε1T1^4

• Bryan,

Wonderful. I hear angels singing (temporarily).

Of course, we can agree that the emissivity of the two surfaces might be different, the above was a slight simplification (equal gray bodies) in that particular textbook to assist the student. Textbook 3, by M. Necati Özisik, has a more complete equation.

1. So perhaps you can help Jonathan in understanding why this does not violate the second law of thermodynamics. Your equation confirms that the warmer body absorbs thermal radiation from the colder body. Jonathan will now claim that you have abandoned the second law of thermodynamics for spurious reasons.

2. For housekeeping, please confirm that the entropy calculation for a radiative exchange (as linked) is correct.

3. So, with all other conditions remaining the same, if T2 increases, what happens to Enet1?

Jonathan was unable or unwilling to answer this point. It’s a particularly simple equation, much simpler than many civil engineering problems.

I contend, in common with maths textbooks everywhere, that when T2 increases Enet1 increases in value. That is, it starts negative, but the net rate of energy loss reduces, Enet1 gets closer to zero.

4. Therefore, if body 1 is supplied with a constant energy source, and T2 increases, what happens to the temperature of body 1?

I hope to have angels continue singing and for you to prove me wrong in my earlier comment. This would be a wonderful day.

• SoD

This apt comment was made yesterday, on Tallbloke’s site by wayne.

Who says,

“So, if a surface at exactly the same moment both emits and absorbs an identical sized quanta, zero time between the two, what has happened to that surface? Has the temperature changed? Did the two events even occur? Are you sure? Can you prove it? How?”

The quantum theory would lead us to believe that identical photons are exchanged leading to no temperature change in the warmer due directly to the colder .
However the warmer is radiating more intensively and with additional shorter wavelength photons leading to increased internal energy of the colder.
But we all agree with that!

To anticipate your next question ‘how does the warmer react to not having to ‘net’ emit the identical exchange photons’ ?

The reaction will depend on whether there is a continuing power input to the warmer.

If there is no additional power supplied there will be no increase in the temperature of the warmer indeed its internal energy will drop by exactly the same internal energy rise of the colder.

If there is a continuing power supply this loss will be made up and in some cases the internal energy of the warmer will rise.

However this is down to the power supply continuing.

The effect of the colder is best thought of as a kind of radiative insulation.

• Bryan

“So, if a surface at exactly the same moment both emits and absorbs an identical sized quanta, zero time between the two, what has happened to that surface? Has the temperature changed? Did the two events even occur? Are you sure? Can you prove it? How?”

Such a thing never occurs, i.e. it’s probability and frequency is zero. It’s not forbidden by any law but there’s so much freedom that it never happens in practice.

Otherwise your comment is essentially correct, and in full agreement with physics. Therefore I have often included in my comments expressions like “assuming that other energy flows are not changed for some other reason”. It’s, however, too cumbersome to include that every time. Mostly it’s better to assume that the readers understand that such an assumption is done even when it’s not stated explicitly.

If there’s no continuing power supply to the warmer body it keeps on cooling but at a rate influenced by the colder one, if there’s a constant energy input, the equilibrium temperature will change.

But then. The issue we are discussing is the one where solar SW provides continuing warming and the power of that is often assumed to be constant. Whether the power is constant or not, the GHE will influence very significantly the temperature of the surface and radiation from the atmosphere to the surface is one factor in this. That’s the only important additional source of energy on top of the solar SW that allows the surface to both radiate at a power that corresponds to its temperature and emissivity and to lose energy through other channels (mainly evaporation and convection).

• Bryan,

I pick up item 4, assuming from your response that you agree with item 3.

And further clarify item 4.

With T1, T2 and other parameters in the agreed equation, Enet1 (a negative value, i.e., a loss of energy) is balanced by a constant source of power Esolar.

That is, Enet + Esolar = 0.

Therefore, T1 is constant.

Esolar, an external source of power, does not change.

What happens to T1 when T2 increases? Does T1 reduce, stay the same, or increase?

“4. Therefore, if body 1 is supplied with a constant energy source, and T2 increases, what happens to the temperature of body 1?”

and again

“What happens to T1 when T2 increases? Does T1 reduce, stay the same, or increase?”

“If there is a continuing power supply this loss will be made up and in some cases the internal energy of the warmer will rise.

However this is down to the power supply continuing.

The effect of the colder is best thought of as a kind of radiative insulation.”

The reason that the temperature of the hotter increases is due to the external solar supply staying constant and not due to back radiation from the colder.
Remember also that power supplies are not perfect, even solar.

A qualification is in order therefor to say the rise in temperature depends on the reaction of the power supply to the increasing temperature of the hotter.

• Bryan,

And so, continuing the theme from your answer of March 24, 2013 at 9:35 pm..

It is not a violation of the second law of thermodynamics for a cold atmosphere to increase the temperature of a warmer surface (when this warmer surface is supplied from a constant solar energy source), compared with the case of an even colder atmosphere under the same conditions.

True or False?

• Bryan,

We have again the semantic issue that I have discussed in a couple on my above comments.

What is heat?

The statement that heat can flow only from the warmer body to the colder body is strictly true in the language of Classical Thermodynamics. In that language heat is not simply what almost everyone thinks the word to mean. The definition is very roundabout. My old textbook (Lee & Sears) has a couple of paragraphs introducing to the idea trough examples, and then reaches the point of defining heat:

We define the quantity of heat flowing into the system, in a nonadiabatic process, as proportional to the difference between the work W and the adiabatic work W_ad.

Heat is defined trough the flow as the residual between to measurable quantities. It’s kind of “black energy” whose further nature is unknown and beyond further analysis. Such a residual is always a net flow as there’s no way of looking at the component flows. In classical thermodynamics heat can flow but the heat content of a body cannot be defined. Only changes in the heat content are measurable, not the absolute value.

The sentence “Heat can flow only from the warmer body to the colder” is true in that language.

Even in that language, there’s no reason to conclude that the temperature of the colder body could not influence the flow of heat. The flow gets the weaker the warmer the cold body is. When the heat loss gets weaker the warmer body gets warmer assuming that other energy flows are not changed for some other reason. That’s the warming influence of the colder body.

In another language, that of radiative energy transfer, we can discuss simultaneously radiation from A to B and from B to A. Those are the two terms in SoD’s equation. Both terms are real. Both terms represent radiative energy flows. There’s radiative energy flow also from the colder to the warmer. The net radiative energy flow is always from the warmer to the colder. That’s the heat flow in language of Classical Thermodynamics.

There’s nothing in the Second law of any formulation that would say that radiation cannot transfer energy from colder to warmer, but that cannot occur without a stronger opposite flow.

There’s nothing in the Second law of any formulation that would say that the temperature of a colder body cannot affect the temperature of a warmer body. (I.e., a warmer cold body leads to a warmer warm body.)

• Pekka

I dont find fault in the substance of your post.

However it seems to be the case that for climate science in particular there are a number of proponents who are uncomfortable with heat as a concept and its relationship to the second law.

You will find evidence on this site of people who say;

The second law does not apply to climate.
The second law does not apply to radiation
Statistical thermodynamics supersedes and is somehow in conflict with classical thermodynamics
…..and so on.

Outside climate science I find no tendency in general physics literature to sidestep the use of the word heat or distort its meaning.

Indeed there are some within climate science like Rodrigo Caballero who can give a consistent account without bending the meaning of heat.

• Bryan,

There are no doubt people, who have misconceptions along the lines you list. There may be even some scientists who have such misconceptions, but none of them can work on questions where these issues are important. Only laymen can foster real misconceptions at this level. Going deeper in applying thermodynamics issues get often so complex that errors are made even by working scientists as they are made by engineers who apply thermodynamics in their work.

There are also people (including scientists), who understand these issues correctly, but who write erroneous messages related to them either as a mistaken attempt to make issues more understandable to laymen or just because they are not careful in expressing themselves. Such badly formulated messages are, unfortunately, rather common.

If there’s a tendency to see such deep misconceptions in climate discussion that’s due to the large number of laymen that contribute to this discussion. Among the laymen severe errors are made on both sides of the debate.

Looking at the list you present I cannot, however, see any reason for anyone defending main stream thinking to foster such ideas. Correct scientific understanding of the basics supports main stream thinking. No single point works in the opposite direction.

• Bryan,

I admire your patience. When Mr. Doom originated this post in 2010, he started out with a false premise, that the Clausian statement did not really say what it says. It’s been almost 3 years, and he has failed to provide one example of a thermodynamic or physics text that states his mangled version of the Clausian statement. He has had almost 3 years to provide experimental evidence to confirm his notion that a cooler body can indeed warm up a warmer body; but has failed to provide such evidence. You are dealing with someone who does not abide by the tenets of the scientific method, i.e., a non-scientist. He has had ample time and opportunity to clearly prove his position from a scientific standpoint, but has completely and utterly failed in that endeavor. So you are wasting your time. Just allow the good Mr. Doom his version or reality, and move along, because if you come back here another 3 years from now, nothing will have changed.

• Jonathan,

Your exhortations are too late. Bryan has just stated (March 24, 2013 at 9:35 pm) that he disagrees with you.

1. The radiation from the colder body is absorbed by the hotter body.
2. From a steady state position with a constant external source of energy, if we increase the temperature from the colder body the temperature of the hotter body increases.

149. Jonathan has failed to answer any of my questions, including this one which was in response to his claim:

[Jonathon] “If your “calculations” are so robust and obvious, then there should be ample samples in multiple thermo textbooks indicating that a cooler body does increase the temperature of a warmer body..”

[My response] Let me refer you to the first equation highlighted from the first scanned textbook on heat transfer.

This is equation 2.1 in the first scan.

q = σ(T14 – T24)

If T2 increases what happens to the rate of heat loss?

Is the equation wrong?

Can T2 change in temperature and not affect the rate of heat loss?

But Jonathan continues to be angry. I would be as well if I couldn’t answer a basic question like this.

The point is that the authors of heat transfer textbooks, professors in the field, disagree with Jonathan. Probably they never understood the second law either and no doubt Jonathan will be writing letters to the publishers of these textbooks.

Maybe when Jonathan has calmed down he can return and clarify what the right equation is for radiative exchange between a hotter and colder body.
We’ll see.

• SoD,

I am rather amused. Your 7 textbook samples show basic heat transfer equations. Those are not Clausian statements concerning the Second Law. And like I said, none provided worked-up samples that indicated a cooler body warming up a warmer body to a higher temperature.

The MIT thermo course notes that I linked to earlier gave the correct “non- Doom” version of Clausian Statement of the Second Law that you will find in any thermo textbook:

http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node37.html

The same MIT course notes also give the standard equations for radiative heat transfer between two bodies:

http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node136.html

MIT does not seem to be confused at all. I am not confused either. That leaves……..?

Like I said, scientific advancement and discovery throughout history follows a rigorous process known as the scientific method. So rather than the “Clausian” statement of the Second Law, you want to come up with the “Doom” statement of the Second Law. Fine. First, you have to clearly state your hypothesis. Then you have two options:

1. Find another scientist who has come up with the same hypothesis, and has already confirmed the hypothesis by experiment. (you have not done that yet)

2. Follow the scientific method yourself: State your hypothesis, make a prediction that follows the logical consequences of the hypothesis, devise (and conduct) an experiment to see if the real world behaves as predicted; modify your hypothesis as needed. (you have not done that either)

OK. So you are two for two, as in like a bad thing. There is actually a third option that is commonly known as the “Do Nothing” approach:

3. Do Nothing: This consists of not following the scientific method. It involves creating one’s own blog in an attempt to create an aura of authenticity; “name dropping” impressive equations; performing lots of hand waving; ignoring the laws of thermodynamics; creating a loyal cult-like following; expending a lot of time and resources without making any real “scientific” progress. (Bingo! You have performed quite well with this option.)

Nowadays the whole scientific process needs to be published in some well known and respected scientific journal. This roots out all the quacks through the peer review process, and publication allows other scientists to review and perhaps try to disprove the hypothesis.

Good luck!

• Jonathan,

Clausius lived in the 19th century. His statement on the heat transfer were correct, but he proceeded further as he gave also the mathematical formulation for entropy and through that allowed for a better way of formulating and discussing the Second law. He made that transition over years1850-62. You haven’t got that far by 2013.

And then you are arrogant enough to think that you are the one to judge the correctness of the writings of others.

150. In my experience many people with a rudimentary acquaintance with the second law of thermodynamics and no acquaintance with radiative heat transfer are shocked when they first encounter the claims of basic radiative physics, determined in the late 1800’s.

19th century physics is in the spotlight because of the prominence of climate science, which obviously uses the basic building blocks of physics.

These people immediately claim, with full conviction, that thermal energy radiated from a colder body can’t be absorbed by a hotter body. Or can’t change the temperature of a hotter body. Or something.

They believe this is a violation of the second law of thermodynamics because they have never properly understood the second law or radiative heat transfer.

What is fascinating is when I present a textbook reference on radiative heat transfer I usually cannot get the claimer to commit to:

a) whether the textbook is right or wrong or
b) whether the textbook agrees or disagrees with my statement/equation

And when I present a calculation of entropy, the “gold standard” for the second law of thermodynamics, I likewise cannot get a commitment as to whether entropy has increased or decreased, or what the relevance is of entropy to the second law.

It’s a fascinating insight into human psychology, but not much use for understanding physics.

Passionate people will continue to be passionate people. This blog is about climate science, specifically physics. And physics is most clearly explained through equations. Physics textbooks are full of equations. Physics papers are full of equations.

I realize and respect that many readers cannot understand equations. This is totally fine.

But it makes it difficult for such readers to evaluate competing claims. This is why I press people who are passionate about their competing claims to step up and provide an equation. Or provide a calculation. Or even pick which equation is correct. Or even state whether my equation matches the equation in the book.

The intent is to demonstrate to equationless readers that the passionate might have fine words but are unable – or unwilling – to engage in what is known as physics.

Hopefully this is clear to the equationless community.

• SOD: I would add the following questions: Aren’t gases made of molecules moving with a range of speeds in random directions? Won’t such molecules occasionally collide so as to transfer kinetic energy from a slower-moving molecule to a faster-moving one (thus maintaining the Boltzmann distribution of molecular speeds)? Why doesn’t this violate the 2LoT?

If a slower-moving molecule in the air can transfer kinetic energy via a collision to a faster-moving molecule in the air, why can’t a slower-moving molecule in the air transfer energy via photon to a faster-moving molecule on the ground? If one photon can do so, why not many?

Energy transfer in both cases BETWEEN MOLECULES is possible because the concept of temperature exists only for large groups of molecules, not for energy transfer between individual molecules by collisions or photons. (SInce entropy is dq/T, the concept of entropy also applies only to groups of molecules). The 2LoT only applies NET ENERGY TRANSFER between groups of molecules large enough (and colliding frequently enough) to have a defined temperature. Individual molecules obey the laws of quantum mechanics and radiation and the result of those laws always results in heat transfer between large groups of molecules from the hotter group to the colder one (and the rest of classical thermodynamics).

• on March 26, 2013 at 9:28 pm DeWitt Payne

Frank,

Indeed.

At the level of individual photons and molecules or atoms, there is no Second Law. In fact, the subject of why time is unidirectional has still not been definitively answered.

There are only three things that can happen when a photon interacts with matter. It can be reflected, absorbed or transmitted. Unless you vastly complicate the physics of photon interaction, there is no way a photon can ‘know’ the temperature of its emission or the temperature of the surface with which it interacts.

Significant reflection of ‘colder’ photons by a warmer object can be ruled out by spectrophotometry. The emission spectrum of a highly absorbing surface exposed to emission from the atmosphere would be measurably different from a black body spectrum if there were significant reflection of the highly structured spectrum of atmospheric emission.

Photons cannot simply disappear without violating the First Law. Absorption must increase the energy content of the absorber and emission must decrease the energy content.

Finally, English or any other language cannot be used to prove or disprove physical hypotheses. Language is too ambiguous. Words have multiple meanings and connotations depending on context. It’s too easy to create paradoxical statements in language. It’s not enough to quote Clausius without defining exactly what he meant by ‘heat’. And even then, you are still in the 19th century. Only mathematics is unambiguous enough to properly state physical concepts. That’s why SoD insists on equations.

• I agree on what Frank and DeWitt write and continue on that.

The Second law and entropy have their basis in probabilities. They are so good guidelines, because the huge number of molecules in any easily observable volume make the probability distributions of all summarizing characteristics very narrow, i.e. the values are in practice always so close to the expectation value that observing any deviation is either impossible or requires at least very precise measurements.

The flow is always from the warmer to the colder, because the number of microscopic events of energy transfer is larger in that direction than in the opposite direction. That’s the whole content of the statement on the direction of heat transfer, there’s nothing more in that. That applies similarly to heat conduction and to radiative heat transfer.

Entropy is a measure of the likelihood of a macroscopically defined state. Every macroscopically defined state can be realized by a huge number of different microscopically different states. The maximum entropy is obtained for a macroscopic state that’s most likely taking into account both the number of microscopic states that correspond the same macroscopic state and the likelihoods of each of these microscopic states. To have an entropy we must have a system that allows for a number of different microscopic states. The concept gets useful when the number of such possible states is large. The strong laws like the Second law require that the number of possible states is very large.

In Quantum Mechanics it’s possible to define entropy in a meaningful way even for a very small system, because the concept of probability is so inherent in all QM considerations.

There are many difficult questions in the fundamental basis of statistical thermodynamics and entropy. How does a time reversal invariant theory lead to the unidirectional Second law is not an easy issue to handle. It’s intuitively rather easy to understand, but trying to make that formally correct leads to all kind of apparent paradoxes. The problem is not very different from the related problem on the fate of information in black holes that Susskind and Hawkins argued about.

151. SoD

The increase in temperature of the hotter body is due to the continuing power supply to it.
Cut of the power
No increase, in fact a decrease.

I’m interested in the physics of the problem.

How you want to describe “the cause” in words is up to you.

Others might differ. Change only B, as a result A changes. No! The cause of A changing is the unchanging C. And so on. All exciting stuff for the poets.

I’m interested in your claim on the second law of thermodynamics for this effect.

And of course I absolutely agree with the consequences of the new disturbance you highlight (cut the power..). And I have explained the same in different words on numerous occasions on this blog.

• Yes.

For some reason there was a computer page incompatibility which stopped me replying at the actual question.
Its good that we all agree with Clausius that a colder object will not spontaneously heat a warmer one.
A common agreed language would resolve many apparent conflicts.

• Bryan,

Sorry to press you on this point, for clarity, it would just help the readership to know your view.

On March 25, 2013 at 8:17 am I asked:

It is not a violation of the second law of thermodynamics..

..True or False?

Your response to my subsequent question was “Yes.” The Yes part is not clear.

“Yes, it is not a violation of the second law of thermodynamics”, or

• Bryan,

Re my question of March 25, 2013 at 9:37 am, why no response?

• SoD

Your top diagram shows T1 and T2 without any power supplies to either.
So the radiation from the colder to hotter cannot heat the hotter by second law.
All it can do is reduce its heat loss perhaps depending on the surroundings.

In the case of A and B the hotter is supplied with a power source.
Its temperature can rise if power is greater than the heat losses from the hotter object.
But this is due to the power source continuing rather than the radiative insulative properties of the colder object

• Bryan,

I have never read a single comment where anyone would claim that the warming influence of GHE would not be just the one that you agree upon.

On the other side I have read tens or hundreds of skeptic messages that depend directly or indirectly on rejecting the influence on temperature trough reducing heat losses in the way you accept.

The problems related to misunderstanding this point are 100% on skeptic side of the discussion. The only fault that many “warmist” have is that they cannot explain clearly what they think.

• Bryan,

Given the amount of effort everyone has put in to specifically define scenarios and request each side to define their position, why should anyone assume “the answer from Bryan is obvious“.

If there’s one constant in the world of this blog, it is that the answer from Bryan is not obvious. (Except to Bryan).

I don’t believe I have the right to assume anything about your response and that is why I specifically requested you to answer this specific point.

You seem to be hugely concerned with cause and effect. Cause A has not changed, cause B has changed but the reason for the resulting effect on condition C is the unchanging A and not the changing B. Good for you. You can wax lyrical on your wording for the physics.

My only interest is the second law of thermodynamics in the defined situation. Is the second law violated?

Therefore, on March 25, 2013 at 8:17 am I requested:

Bryan,

..It is not a violation of the second law of thermodynamics for a cold atmosphere to increase the temperature of a warmer surface (when this warmer surface is supplied from a constant solar energy source), compared with the case of an even colder atmosphere under the same conditions.

True or False?

Please confirm. It is NOT obvious what you think.

• Such is the way of things with the many claimed advocates of the second law of thermodynamics.

Requests that will tie them down to a falsifiable claim are almost always unanswered.

As we see with my request on March 25, 2013 at 9:37 am, repeated on March 28, 2013 at 10:14 am and still not answered.

• SoD says

“if there’s one constant in the world of this blog, it is that the answer from Bryan is not obvious. (Except to Bryan).”

How many ways can you set out something which seems obvious?

Object A is at a temperature of 300K

Object B is at a temperature of 200K
Object C is at a temperature of 350K
Object D is at a temperature of 100K
Object E is at a temperature of 2.7K

The only object that can ‘heat’ or ‘warm’ object A is object C.

Climate Science explanations often follow the logic of ;

‘since object A would lose more heat to entity E than to B then we can say that B has warmed A’

I feel that this is a misuse of language and can give a misleading impression.
Whether this is intentional or not is a genuine question.

152. Look up uncooled microbolometer arrays.
Then without the calling the use of “cold rays” explain how an array at 50C can show the difference between an object at -40C and -39C and 49C
These temperatures are less than the array and according to some should have no effect on the warmer sensor.
Obviously cameras based on this science work. So 49C radiation warms the plate more than -39C which warms the plate more than-40C. The difference in temperature of the bolometers produces a detectable output!
http://www.climateandstuff.blogspot.co.uk/2012/05/cool-body-can-transfer-measurable-heat.html.

153. Pekka, (March 24, 2013 at 7:18 am)
“When the heat loss gets weaker the warmer body gets warmer assuming that other energy flows are not changed for some other reason. That’s the warming influence of the colder body”

I can’t see how this can be right. Isn’t it the case that heat gain by (or work done on) a body is necessary for it to warm? The way I see it, weaker heat loss is not a cause: it’s a consequence of convergence of temperatures of the body and its surroundings.

Later you write “a warmer cold body leads to a warmer warm body”. As I see it, your “warmer warm body” is not a contemporaneous body; for that would imply that its temperature had risen. It is a state-shifted, new steady-state one. This result stems from an independently warmed cool body. The increment of energy injected into the system causes the warmer steady state. In a closed system, on the other hand, by virtue of the First Law, warming of the cool body occurs at the expense of cooling of the warm body. The steady state warm body would always be cooler than it was initially, irrespective of the temperature of the cooler body. It would be the warmer the independently warmer the initial cool body. Have I got this right?

• When we are discussing Global Warming we are not discussing a closed system. Therefore most of the meaningful arguments are not about closed systems.

Comparatives like “warmer” refer to an alternative. Usually it’s easy to deduce what the point of comparison is – assuming that there’s willingness to deduce rather than to confuse.

154. Pekka, (March 24, 2013 at 7:18 am)
“There’s nothing in the Second law of any formulation that would say that radiation cannot transfer energy from colder to warmer, but that cannot occur without a stronger opposite flow”
Clarification, please. You seem to be saying that radiation from the surroundings cannot transfer energy to the warmer body in the absence of stronger opposite flow? This would seem to contradict radiative exchange theory which posits that radiation from a body and from its surroundings are independent phenomena. Or is it the case that while the radiation is independent, absorption/thermalisation of it by the warmer body is conditional on stronger emission by it? If so, wouldn’t that beg the question of what happens at thermal equilibrium? In the absence of a stronger opposite flow absorption/thermalisation would cease; and the absence of absorption would imply that radiation had ceased? Relatedly, Pekka, doesn’t your answer to Bryan refute the theory of radiative exchange which, at thermal equilibrium between a body and its surroundings, requires simultaneous absorption and emission in equal quantities?
March 24, 2013 at 6:04 pm Pekka Pirilä
Bryan
“So, if a surface at exactly the same moment both emits and absorbs an identical sized quanta, zero time between the two, what has happened to that surface? Has the temperature changed? Did the two events even occur? Are you sure? Can you prove it? How?”
Such a thing never occurs, i.e. it’s probability and frequency is zero. It’s not forbidden by any law but there’s so much freedom that it never happens in practice.

• John,

Each emission of a photon occurs essentially independently of every other emission and absorption. Similarly the absorption is not affected by other photons. The rate of energy transfer is a statistically determined quantity with very small variability because the number of independent events is so huge. This expectation value depends on the temperature of the emitting body, but not on that of the absorbing body. As the rate of emission grows with temperature there are more photons going from the warmer to the colder than in the opposite direction. Each case is independent, there are just more in one direction. This is one formulation of Kirchhoff’s law.

155. on April 1, 2013 at 3:10 am | Reply John Millett

Scienceofdoom
October 11 2010 at 10:16 am | scienceofdoom
“John Millett
You can make up whatever ideas you like to support your belief in the imaginary second law of thermodynamics.
I invite new readers to read what the textbooks writers actually say”.

Thus ended our conversation back then. On the reactivation of the thread I want to take you back to your original formulation where, for two grey bodies with equal emissivites, “e”, facing each other and where “E” denotes emission, you argue:

E1net = -E1 + E2
E2net = -E2 + E1
Therefore E1net = -E2net

Allowing for the (overlooked) less than full absorption by each body of the emission from the other, wouldn’t the more correct argument be:

E1net = -E1 + eE2
E2net = -E2 + eE1
Therefore E1net is not generally equal to minus E2net ?

Isn’t it also the case that E1net = -E2net when -E1 + eE2 + eE1 – E2 = 0 which simplifies to:
e(E2+E1) = (E2+E1), that is, when e =1, contradicting the grey body premise? More later on the reason for this unsatisfactory result, ignoring reflected radiation.

• John,

The quantities E1 and E2 do already contain the emissivities and absorptivities of both bodies. You should not add it one more time. You can find more complete formulas for E1 and E2 in the book excerpts of the original post.

A fundamental law is that the emissivity is always equal to absorptivity for every type of radiation, i.e, for a precisely defined direction, wavelength and polarization, which may also have an influence. Thus the combined factor is always the same for such a radiation from A to B and from B to A. There’s also a geometric factor that tells, how well the bodies phase each other, and this factor is also the same for both directions of the radiation. The only factor that differs is the temperature dependent Planck’s formula.

I refer above to Planck’s formula rather than to Stefan-Boltzmann’s law, because it’s the one that tells what happens at a given wavelength. SB law tells the overall effect for a black or gray body, we need Planck’s law when the emissivity depends on the wavelength as it does always to some extent.

The equality of emissivity and absorptivity as described in the second paragraph is really fundamental. It’s the most detailed formulation of Kirchhoff’s law and it’s directly linked to the validity of the Second law. Kirchhoff realized the connection to the second law and formulated his law in a very different way. it’s, however, possible to derive the detailed formulation from the original one. While Kirchhoff realized that this law must be true to have the Second law in force, he could not explain it from other properties of interaction of radiation with matter.

That explanation has been given by Quantum Mechanics and in particular by the fact that the microscopic laws of QM don’t make any distinction for the direction of time. That property of QM tells that the strength of the interaction where a photon is emitted and the material system goes from state 1 to state 2 is equal to the strength of interaction that allows for the absorption of a photon while the system goes from state 2 to state 1.

156. on April 1, 2013 at 3:38 am | Reply John Millett

DeWitt Payne

“Significant reflection of ‘colder’ photons by a warmer object can be ruled out by spectrophotometry”

As you imply earlier, a photon on arrival at a surface has no history. Nor does it have unconditional entry to that surface. As a different energy species it must meet a compatibility test between its radiative energy quantum and the kinetic energy structure of the surface atoms it encounters. The probability that it meets the test, one surmises, would be less than one and the probability of it being reflected greater than zero. Over a small interval of time photons encountering a surface presumably exhibit a range of energy quanta whereas the atoms’ energy structure would remain practically constant. This hypothesis suggests significant reflection. Can you cite a spectrophotometry exhibit rejecting the hypothesis, please?

“Finally, English or any other language cannot be used to prove or disprove physical hypotheses”.

Quite so, physical experiment is needed to do that. The lack of experimental validation of some hypotheses presented as scientific certainty continues to trouble me. The proposition that some theories are so well founded as to make experimental validation redundant begs the question: Who defines “well founded” and separates the exceptional from the ordinary? Also worth noting is the predominance of words over equations in the texts displayed by SOD. The authors describe for students in words, whose meanings they already understand, the meaning of the abstract symbols they are yet to understand. Scientific discourse is effected mainly in words – as evidenced in scientific journals. Compared to words, equations make for economy – and rigour – of expression. In practice, equations, like words, may also be subject to loose usage.

• John,

SoD, DeWitt and myself have all the time referred to basic textbook physics. All formulas and laws described in basic physics textbooks have been very thoroughly tested by innumerable empirical observations. Furthermore new tests are made every time a formula is used for a practical application and it’s found that the outcome is as predicted by the formula.

Nobody keeps track anymore on every experiment that’s done to verify the laws presented in basic textbooks, that’s totally unnecessary. There are more esoteric areas of physics where the laws are less certain. Using the laws to derive results for practical applications is also often difficult, but there’s no space for doubt on the basic laws. We have also described the thought experiments in such a way that using the laws to make the predictions is simple and not prone to errors. Therefore we are fully certain that the conclusions are right. All that is based on empirical confirmation, but that confirmation is used trough the formulas.

You ask: “Who defines well founded“. In the case of science the answer is the scientific community. That’s again a concept without precise definition. The concepts are not precisely defined and by the nature of science can never be precisely defined. They are, however, clear enough in most cases. Again there are borderline cases where the question you present is very relevant, but basic textbook physics is an example of the opposite. If anything in this world is well known and certain, the basic textbook physics belongs to that.

When I state the above I don’t claim that every formula is infinitely precise, only that they are so precise that the possible missing details have no effect on any ordinary application of the formula. Modifications like the one Einstein introduced to Newton’s laws are possible, but as special relativity changes the results significantly only for very high speeds, any future development can affect only some extreme conditions. That we know just because all that physics has been so thoroughly tested empirically.

• John,

The kind of considerations that you discuss in your first paragraph are just those that influence the absorptivity (and the emissivity as it’s equal). They are not dismissed or forgotten, but calculating the absorptivity of a solid or liquid surface from theory is very difficult, often essentially impossible. Therefore the absorptivity of a surface is usually measured empirically rather than calculated. Theory tells then only that absorptivity and emissivity are equal and either one can be used in the empirical determination of their common value.

The situation is different for gases. In the case of gases theoretical calculations are also possible, but even for gases the spectral properties listed in Hitran-database are empirically determined for all important lines. Only weak lines with little overall influence are determined from theoretical models. Comparison of the empirically determined properties with those calculated from the models are a very good test on the Quantum Mechanical models of the molecules.

157. on April 1, 2013 at 4:22 am | Reply John Millett

Scienceofdoom,

Following my earlier comment:
This unsatisfactory result stems from ignoring reflected radiation. The general equation should therefore include a term being the sum of a sequence of absorptions of decreasing amounts of reflected radiation bouncing back and forth between the two bodies. Some of the textbook extracts deal with that issue. But none deals with the related possibility that, depending on emissivity values and distance between the bodies, a steady state may never be reached – a new series of reflections is initiated before the previous one has expired. But in practice a steady state is reached – in an enclosure, two bodies initially at different temperatures will eventually arrive at the same temperature somewhere between the initial ones. This suggests a flaw in the orthodox equation for Enet and in the concept of radiative exchange between bodies on which it is based.
An alternative concept of exchange would be between the bodies separately and the common surrounding medium (ignored in your graphic). It is worth noting that inter-body exchange involves two-way conversions between kinetic and radiative energies, emission being a conversion from kinetic to radiative and absorption a reverse conversion. However, emission in one body cannot be paired with absorption in the other; for, as DeWitt Payne writes, photons have no history. An emitted photon simply joins the multitude of photons comprising the bodies’ common surrounding medium; an absorbed one simply exits it. They enter and exit without identity tags. It seems to me, therefore, both more realistic and simpler to think in terms of one-way body-medium exchange than two-way body-body exchange. Equations to come.

• John,

It’s often possible, in principle, to do calculations following histories of individual particles – or photons – but that’s not the only valid approach and that’s often unnecessarily cumbersome. There’s nothing wrong in the alternative approach, where we look only at what happens at the two surfaces. In this approach we don’t care about the history of a photon hitting the surface. It may have been scattered from the other surface or it may be freshly emitted.

As it’s much easier to do the correct calculations setting the history aside, that’s always the natural choice. The other approach would surely give the same result, if all the effort is taken to make it complete.

As photons move with the speed of light, they have plenty of time for all scatterings back and forth before the temperature of either has changed significantly at all.

I have used the word scattering, as that is better applicable for single well specified photons than reflection. Reflection is a wave phenomenon. One of the most fundamental features of QM is that photons as well as material particles have simultaneously properties of waves and particles. When we discuss the nature as particle, scattering and absorption are better applicable, when we look at the wave aspects, we get reflection, refraction, and a somewhat different view of absorption. While even single particles have both aspects, starting to identify the single one leads to loss of many wavelike properties. This kind of issues make it so difficult to understand fully QM. I mention this, because you cannot avoid these problems, if you try to go deep in describing the detailed histories of photons.

158. on April 3, 2013 at 5:07 am | Reply John Millett

Pekka,

March 31, 10.57
I suppose I should have been smart enough to extract the full meaning of your elegantly parsimonious statement. On the other hand you might concede that it may not be the least unambiguous one to grace these pages? But rest assured, to confuse formed no part of the motivation for the comment.

The validity of the statement irrespective of how the cool body is warmed is open to question. Specifically, would the warm body be warmer if it warms the cool body? I don’t think so. I think the statement, under these conditions, would read “A warmer cool body, warmed by the warm body, leads to a cooler warm body”.

March 31, 11.12
My interest is the equilibrium case when there is no excess on one side. In that case there are equal numbers of photons per unit area per unit time entering and exiting the surface. In the limit of infinitesimal elements of surface and time, in a single surface atom there would be simultaneous transitions of electrons from lower to higher energy levels and from higher to lower ones. Is this what you say, in your reply to Bryan, never happens?

April 1, 7.40
“The quantities E1 and E2 do already contain the emissivities and absorptivities of both bodies. You should not add it one more time”.

One step at a time…
E1 = body1’s emission which takes emissivity less than 1 into account.
Likewise E2 for body2.
Body1’s absorption of body2’s emission must take into account body1’s absorptivity less than 1.
Body1’s absorptivity = its emissivity (Kirchoff).
Therefore, Body1’s absorption of body2’s emission is e1.E2.
Therefore, E1net = the excess of body1’s emission over its absorption = -E1 + e.E2
(where e1 = e2 = e)

April 1, 8.45
“As photons move with the speed of light, they have plenty of time for all scatterings back and forth before the temperature of either has changed significantly at all”.

Wouldn’t the lifetime of an atom’s excited state, 10^-9 sec, as well as the photon’s velocity, 3×10^8 ms-1, be relevant?

• The quantities E1 and E2 are amounts of unidirectional energy transfer. They include only photons that are both emitted and absorbed. Thus they include both the emissivity of the source and the absorptivity of the receiving body.

The temperature is always a property of an macroscopic body that contains perhaps 10^23 or 10^26 molecules or even many more. The lifetime of a molecular excited state tells only how often and excited state is typically formed and then de-excited. That tells also, how rapidly the energy is shared within a group of molecules.

It’s useful to understand as an example that about 7 % of CO2 molecules are typically in an excited state that can radiate at 15 µm. Each of the molecules goes to such an state hundred million of times every second and is naturally de-excited as often. Only one case in billion is related to absorption or emission of radiation, in all other cases the reason is a molecular collisions. (The numbers are rough but close enough to true values.)

Because collisions are so dominating there’s practically no connection between a case of absorption and a case of emission on molecular level. That applies equally to gases like CO2 and to solid or liquid surfaces. No emission is linked to a case of absorption or to another case of emission on a level that’s of any significance. (There are some effects but at the level of one part in billion).

159. on April 4, 2013 at 4:34 am | Reply John Millett

Pekka Pirila,
“The quantities E1 and E2 are amounts of unidirectional energy transfer. They include only photons that are both emitted and absorbed. Thus they include both the emissivity of the source and the absorptivity of the receiving body’.

I’m struggling – but it looks to me that you ascribe to E1 and E2 what SoD ascribes to Enet1 and Enet2?

Let’s go back to square 1 where SoD writes:

“E1 is the energy radiated from body 1 (per unit area) and we consider the case when all of it reaches body 2, E2 is the energy radiated from body 2 (per unit area) and we consider that all of it reaches body 1″. (my emphasis)

Energy radiated by one body reaching another must be absorbed for radiative energy exchange to have been effected. By definition a body that absorbs all the radiation incident upon it is a blackbody. What SoD has conjured up here are two-tone bodies – gray for emission and black for absorption. Then he writes:

“We define Enet1 as the change in energy experienced by body 1 (per unit area). And Enet2 as the change in energy experienced by body 2 (per unit area)”.

That is, in general, Enet = gray-body emission – blackbody absorption
And Enet1 = (gray-body1 emission) – (gray-body2 emission)
And Enet2 = (gray-body2 emission) – (gray-body1 emission)
= minus Enet1…….a result ensured by the premise.

Finally he writes:

“In the case of the imaginary second law, there is some energy floating around. No advocates have so far explained what happens to it. Probably it floats off into space where it can eventually be absorbed by a colder body”.

It’s floating around in the common surroundings, in general a mix of two gas species and in SoD’s example, implicit but ignored, a photon gas. One of the texts recognises the existence of a medium but deftly takes it out of play by declaring it “nonparticipating”.

• on April 4, 2013 at 6:23 am | Reply DeWitt Payne

John,

You’re neglecting reflection. You can have a photon gas with a blackbody spectrum and radiance at wall temperature T even if the walls reflect 99.99999% of the incident radiation and only absorb/emit 0.00001%. The emitted photons just bounce around until they are absorbed. At equilibrium each wall sees what looks like black body emission from the other wall(s) because, while they only emit a small amount, the sum of reflection and emission is identical to unity, provided that the system is closed. The radiance has to be that high so that emission equals absorption.

Btw, the reason why a cavity radiator does not have an emissivity identical to unity is because you have to have a hole the wall to observe the spectrum and photons leak in or out of the hole.

• on April 8, 2013 at 11:46 pm John Millett

Thanks for that, DeWitt. You have described what SoD has omitted from his model – cavity radiation. The quantity (energy density) and quality (spectrum) of that radiation depends only on the temperature of the enclosing walls and is independent of the size and shape of the enclosure and of the nature of the walls – it is blackbody radiation. A body placed in the cavity must come into equilibrium with it. A hot body would give up energy to the cavity; a cold one would draw energy from it. Two bodies would interact separately with the cavity, not mutually with each other, as SoD has it.

In the super-reflective enclosure you describe, the emitted and the reflected radiations are not synchronous, the latter lagging the former. Reaching equilibrium would logistically require that emission cease at some stage in order to allow absorption of laggard reflected radiation to catch up. That looks like a rejection of the theory of radiative exchanges.

5.67×10^-8×300^4 (J.s-1.m-2) x 6 (m2) x {1 (m) / 3×10^8 (ms-1)} (s) = 9.2×10^-6 J

• on April 9, 2013 at 1:25 pm DeWitt Payne

John Millet,

As Pekka points out below, the radiation field will be the same inside a perfectly reflective cavity containing objects at temperature T as for a cavity with walls at temperature T. The energy content of the field is determined by the characteristic length of the cavity. For a cube it would be the length of a side. Multiply the flux density by the total wall area and multiply by the length divided by the speed of light. Even the miniscule flux of the cosmic microwave background becomes significant when integrated over the volume of the observable universe.

• John,

The original post could have been more precise concerning the coefficient ε. That coefficient is in all cases considered there the product of three terms:

– emissivity of body A
– emissivity of body B
– a geometric factor that describes, how well the bodies face each other.

When SoD uses the word “reaches”, he means hits and gets absorbed.

Try to accept that this is what the ε is. Perhaps the whole post gets easier to understand. Every sentence makes sense when this is accepted. Some sentences allow for different interpretations but this is consistent with one of them in all cases that I have looked at.

• on April 4, 2013 at 6:27 am | Reply DeWitt Payne

As an exercise, you might want to calculate the energy in the photon gas inside a one meter cube with a wall temperature of 300K. Note that, as pointed out above, it’s unnecessary to specify the emissivity of the walls as long as it’s not identically zero.

• John,

I don’t understand what makes you think that

In the super-reflective enclosure you describe, the emitted and the reflected radiations are not synchronous, the latter lagging the former. Reaching equilibrium would logistically require that emission cease at some stage in order to allow absorption of laggard reflected radiation to catch up. That looks like a rejection of the theory of radiative exchanges.

The only consequence of the “super-reflectivity” is that it takes more time to reach the equilibrium. With the numbers of DeWitt the time constant that tells, how long it takes, is the linear size of the enclosure divided by 30 m/s. For a cavity of 1 m diameter we would have essentially an equilibrium after one second. Nothing else changes only the time to reach the equilibrium. Only if the external conditions change significantly in a required period, will some persistent lagging be observable.

Nothing this presents any rejection on the theory of radiative exchanges. That theory remains as valid as ever.

• The above was an answer to the message of
April 8, 2013 at 11:46 pm by John Millett.

• on April 9, 2013 at 11:04 am John Millett

The lag between absorptions of emitted and reflected radiation is not peculiar to “super-reflectivity”, although the comment does invite that inference – sorry. The lag applies generally for gray bodies. By lag I mean that absorption of reflected radiation occurs after that of emitted radiation in a series of “bounces” which take time to execute. Given the high frequency of emissions, absorption would never be complete unless emission ceased at some stage – unless the bodies were quite close together.

160. on April 4, 2013 at 11:01 am | Reply John Millett

Pekka,

So, I should rewrite SoD’s equations (assuming unity view factor) thus:

E1 = -e1.e2.sigma.T1^4

E1net = -e1.e2.sigma.T1^4 + e2.e1.sigma.T2^4

E2net = -e2.e1.sigma.T2^4 + e1.e2.sigma.T1^4

= -E1net

As I write I realise that the result E1net = – E2net has nothing to do with emissivities. Rather it is a consequence of SoD’s specification of the problem – two discrete bodies surrounded by nothingness.- for which the conservation law demands that one body’s energy gain must equal the other’s loss. Trouble is, the specification is unphysical. Nothingness exists nowhere, notime. SoD’s specification should include a common surrounding medium to the two bodies. Then, there is no requirement of offsetting gains and losses between the two bodies. The medium makes up the difference between them.

• on April 9, 2013 at 1:47 pm | Reply DeWitt Payne

John Millet,

The convention in SoD’s example in the post is usually for infinite parallel planes with a heat source behind one plane and a heat sink behind the other. In the textbook examples the planes are small elements of larger bodies. It’s a thought experiment.

• DeWitt
It is easy enough to misinterpret SoD’s graphic. However, it seems to me, infinite parallel planes constitute an enclosure, the space between the planes being the cavity. Then we have a cavity with walls at different temperatures. The energy content of the cavity (per square metre of plane surface area and for planes 1m apart) would be sigma.(Th^4 + Tc^4) x 1/speed of light. The energy density in the cavity would be less than it would be if the walls were at the same temperature, Th, and more than it would be if at Tc. Accordingly, the heat source would give up joules to the cavity and the sink would take joules from it. Eventually the system – source, sink and cavity- would reach an equilibrium temperature. But not by mutual exchange between the planes.

That is how I interpret SoD’s “ Incroptera and DeWitt” extract.

One of the amazing things we see in texts is the one worked example in those cited by SoD in which we want to know how long it would take a steel billet at 1000 degrees C to cool to 800 degrees, temperatures far removed from those prevailing in the climate system.

161. For those of you who insist that the GHE violates the 2nd law, how do you explain that the surface at is supplied with more power (about 390 W/m^2) than the amount of power that comes in and out from the Sun (about 240 W/m^2). What else could be the origin of the +150 W/m^2 of net power gained at the surface if not the absorption of upwelling radiation acting to cool that is absorbed and subsequently re-radiated back downward? It can’t be sourced from geothermal, because way more than about 240 W/m^2 would have to exit at the TOA if that were the case.

162. Scienceofdoom
The (added) medium would exhibit a temperature, T3. Equations for this model – heterodox, but within the frame of physics – would be:

Let T1 = 320K; T2 = 280K; T3 = 300K and e = 0.8 (the common emissivity of the two bodies)

E1net = E1 = – e.sigma.T1^4.[ 1 – (T3/T1)^4 ] = – 475.63 x 0.2275 = -108.22Wm-2
E2net = E2 = – e.sigma.T2^4.[ 1 – (T3/T2)^4 ] = – 278.81 x – 0.3178 = 88.61 Wm-2

In words, the hot body is losing energy to the surrounding medium at the rate of 108.22 Wm-2; the cool body is gaining energy from the medium at the rate of 88.61 Wm-2; and, by 1LoT, the medium is gaining energy at the net rate of 19.61 Wm-2 of the body’s area (assuming identical bodies – since 1LoT applies to energy in absolute terms, not fluxes, the case of bodies of different materials and size is considerably complicated).

Entropy:
dS1 = -108.22/320 = -0.3382
dS2 = 88.61/280 = 0.3165
dS3 = 19.61/300 = 0.0654
dStotal = 0.0437, that is, entropy of the system has increased in satisfaction of 2LoT

The medium here is a passive energy source/sink. Alternatively, consider it as an active agent:

E3net = -e.sigma.T3^4.[{1 – (T1/T3)^4} + {1 – (T2/T3)^4}]
= -367.42 ( -0.2945 + 0.2412)
= 108.20 – 88.62
= 19.58 Wm-2…………… as before.
Entropy:
dS1-3 = -108/320 + 108/300 = -0.3375 + 0.36 = 0.0225
dS2-3 = 88/280 – 88/300 = 0.3143 – 0.2933 = 0.0210
dStotal = 0.0435…………………….as before.

163. […] But I gave EVERYONE the link to the scienceofdoom page that did the research.. SEVERAL TIMES.. Amazing Things we Find in Textbooks ? The Real Second Law of Thermodynamics | The Science of Doom On that page you will find NOT ONLY the stuff I snipped but also the TITLE PAGE FROM EACH OF TEXTS […]

164. […] Amazing Things we Find in Textbooks – The Real Second Law of Thermodynamics […]

165. So if you are aware that the transfer or net transfer goes from hot to cold, why on earth do you insist that Tb can warm up Ta when they are both the same temperature? The “net” transfer is zero after you add up the individual waves, photons and molecules.

166. The basic error of many is that they confuse energy or heat With matter.

A stronger Source of heat or light or electromagnetic radiation does not stop…. and push back… or repel…. heat or energy or irradiation from a weaker or cooler Source.

It superposes og course.

Thus, take away the cooler or weaker source, and the temperature and radiation of the stronger Source will drop a bit, namely by what it gets from the weaker or cooler source in addition to its own effect.

Any enviroment that has got a temperature above absolute zero will heat up any white hot incadescent lamp or…. star… to still a bit higher brightness or temperature, and it is easily shown experimentally on small scale. The effect may be minute ,… depending,……… but it is necessarily real and there.

167. […] surface and absorbed by a hot surface. More examples of this principle, including equations, in Amazing Things we Find in Textbooks – The Real Second Law of Thermodynamics – scanned pages from six undergraduate heat transfer textbooks (seven textbooks if we include […]