The Science of Doom

With apologies to my many patient readers who want to cover more challenging subjects.

Many people trying to understand climate science have a conceptual problem.

I have written (too) many articles about the second law of thermodynamics – the real and the imaginary version. Resulting comments on this blog and elsewhere about those articles frequently contain comments of this form:

So if we take bucket A full of water at 80°C and bucket B full of water at 10°C, Science of Doom is saying that bucket A will heat up because of bucket B? Right! That’s ridiculous and climate science is absurd!

Yes, if anyone was saying that it would be ridiculous. I agree. To take one example from many, in The Real Second Law of Thermodynamics I said:

Put a hold and cold body together and they tend to come to the same temperature, not move apart in temperature.

Of course, it could be that I am inconsistent in my application of this principle.

One observation on the many contrary claims resulting from my articles – not a single person has provided a mathematical summary to demonstrate that the examples provided contradict the first or second law of thermodynamics.

It should be so easy to do – after all if one of the many systems I have outlined contravenes one of these laws, surely someone can write down the equations for energy conservation (1st law of thermodynamics) or for change in entropy (2nd law of thermodynamics) and prove me wrong. We aren’t talking complex maths here with double integrals or partial differentiation. Just equations of the form a + b = 0.

And here’s the reason why – the problem that people have is conceptual. It seems wrong so they keep explaining why it seems wrong.

Conceptual problems are the hardest to get around. At least, that’s what I have always found. Until a subject “clicks”, all the mathematical proof in the world is just a jumble of letters.

So with that introduction, I offer a conceptual model to help those many people who don’t understand how a cold atmosphere can lead to a warmer surface than would occur without the cold atmosphere.

And if you are one of those people in the “firmly convinced” camp, let me suggest this reason for making the effort to understand this conceptual model. If you understand why others are wrong you can help explain it to them. But if you just don’t understand the argument of people on “the other side” you can’t offer them any useful assistance.

Model 2 – Two bodies – The Boring One that Everyone Really Does Agree With

Very quickly, to “warm everyone up”, and to once again state the basics – if we have two bodies in a closed system, and body A is at temperature 80°C and body B is at 10°C, then over a period of time both will end up at the same temperature somewhere between 10°C and 80°C. It is impossible, for example, for body A to end up at 100°C and body B at 0°C.

Everyone is in agreement on this point.

Note that the “period of time” might be anything between seconds and many times the age of the universe – dependent upon the circumstances of the two bodies.

Model 3A – Three Bodies with the Third Body Being Quite Cold

Where’s Body 1? This picture is the view from Body 1, also known as “Chilly Earth”, which is a spherical solid planet.

To make the problem much easier to solve we will state that the heat capacities of Body 2 and Body 3 are extremely high. This means that whether they gain or lose energy, their temperature will stay almost exactly the same. Body 1, “Chilly Earth”, has a much lower heat capacity and will therefore adjust quickly to a temperature which balances the absorption and emission of radiation.

“Chilly Earth” doesn’t have an atmosphere.

However, for the purposes of helping the conceptual model, “Chilly Earth” reflects 30% of shortwave radiation from the Sun but at longer wavelengths absorbs 100% (reflects 0%). This means its emissivity at longwave is also 100%.

“Chilly Earth” has a very high conductivity for heat, and therefore the whole planet is at the same surface temperature. (See note 1).

“Sun” is 150M km away from “Chilly Earth”, and “Chilly Earth” has a radius of 1,000 km (a little different from the planet we call home).

Let’s calculate the approximate equilibrium temperature of “Chilly Earth”, T₁

How do we do this? By calculating the energy absorbed from Body 2 and from Body 3, and calculating the temperature of a surface that will radiate that same energy back out.

The method is simple – see below.

Energy Absorbed from Body 2, “Sun”

Radiation from “Sun” at 5780K = 6.3 x 10⁷ W/m² – near the surface of the sun. By the time the sun’s radiation reaches earth, because of the inverse square law (the radiation has “spread out”), it is reduced to 1,369 W/m². Remember that 30% is reflected, so the absorbed radiation = 958 W/m².

The surface area that “captures” this radiation = πr² = 3.14 x 10⁶ m².

Energy absorbed from body 2, E_r2 = 958 x 3.14 x 10⁶ = 3.01 x 10⁹ W.

Energy Absorbed from Body 3, “Space”

Radiation from “Space” at 3K = 4.59 x 10^-6 W/m². Apart from the very tiny angle in the sky for “Sun”, the entire rest of the sky is radiating towards the earth from all directions in the sky.

The surface area that “captures” this radiation = 4πr² = 1.26 x 10⁷ m².

Energy absorbed from body 3, E_r3 = 4.59 x 10^-6 x 1.26 x 10⁷ = 57.7W.

So energy from body 3 can be neglected which is not really surprising.

Energy Radiated from Body 1, “Chilly Earth”

For thermal equilibrium (energy in = energy out), “Chilly Earth” must radiate out 3.01 x 10⁹ W, from its entire surface area of 1.26 x 10⁷ m².

This equates to 239 W/m², which for a body with an emissivity of 1 (a blackbody) means T₁ = -18°C.

So we have calculated the equilibrium temperature of “Chilly Earth”.

Now, if we change the model conditions – the reflected portion of solar radiation, the emissivity of the earth at longwave, or the conductivity of the planet’s surface – any of these factors would affect the result. They wouldn’t invalidate the analysis, they would simply lead to a different number, one that was slightly more difficult to work out.

But hopefully everyone can agree that with these conditions there is nothing wrong with the method. (I realize that a few people will not agree..)

Model 3B – Three Bodies with the Third Body Being Somewhat Warmer

So now we are going to perform the same analysis with our new Body 1, “Warmer Earth” (a wild stab at an appropriate name).

The only thing that has really changed about the environment is that Body 3, “Crazy Background Radiation”, is now at 250K instead of 3K.

Note that the temperature of Body 3 is higher than before but lower than the equilibrium temperature of 255K calculated for “Chilly Earth” in the last model. As before, body 3 has an emissivity of 1 for longer wavelengths.

Body 1, “Warmer Earth”, still reflects 30% of solar radiation and is the same in every way as “Chilly Earth”.

What are we going to find?

We will do the same analysis as last time. Repeated in full to help those unfamiliar with this kind of problem.

Energy Absorbed from Body 2, “Sun”

Radiation from “Sun” at 5780K = 6.3 x 10⁷ W/m² – near the surface of the sun. By the time the sun’s radiation reaches earth, because of the inverse square law (the radiation has “spread out”), it is reduced to 1,369 W/m². Remember that 30% is reflected, so the absorbed radiation = 958 W/m².

The surface area that “captures” this radiation = πr² = 3.14 x 10⁶ m².

Energy absorbed from body 2, E_r2 = 958 x 3.14 x 10⁶ = 3.01 x 10⁹ W.

Energy Absorbed from Body 3, “Crazy Background Radiation”

Radiation from “Crazy Background Radiation” at 250K = 221 W/m². Apart from the very tiny angle in the sky for “Sun”, the entire rest of the sky is radiating towards the earth from all directions in the sky.

The surface area that “captures” this radiation = 4πr² = 1.26 x 10⁷ m². (See note 2).

Energy absorbed from body 3, E_r3 = 221 x 1.26 x 10⁷ = 2.78 x 10⁹ W.

In this case, energy from body 3 is comparable with body 2.

Energy Radiated from Body 1, “Warmer Earth”

Body 1 absorbs E_tot= E_r2 + E_r3 = 5.79 x 10⁹ W

For thermal equilibrium (energy in = energy out). “Warmer Earth” must radiate out 5.79 x 10⁹ W, from its entire surface area of 1.26 x 10⁷ m².

This equates to 460 W/m², which for a body with an emissivity of 1 (a blackbody) means T₁ = +27°C.

Discussion

Our two cases have revealed something very interesting.

A very very cold sky led to a surface temperature on our slightly different earth of -18°C, while a cold sky (colder than the original experiment’s planetary surface temperature) led to a surface temperature of 27°C.

Well, and here’s the thing, strictly speaking the temperature is actually caused primarily by the bright object in the middle of the picture, “Sun”. The energy absorbed from the sky just changes the outcome a little.

In both cases we calculated the equilibrium temperature by using the first law of thermodynamics (energy in = energy out).

If we do the calculation of entropy change we will find something interesting.. but first, let’s consider the conceptual model and what exactly is going on.

It’s very simple.

In a 3-body problem the temperature of the coldest body still has an effect on the equilibrium temperature of the body being heated by a hotter body.

I could make it more catchy, more media-friendly, but that would go against everything I stand for. I will call this Doom’s Law.

Entropy

The second law of thermodynamics says that entropy can’t reduce. The many cries of anguish that will now arise will claim that Model 3B has broken the Second Law of Thermodynamics. But it hasn’t.

See The Real Second Law of Thermodynamics for more on how to do this calculation. And even clearer, the article by Nick Stokes:

Change in entropy, δS = δQ / T

where δQ = change in energy, T = temperature

We will consider both models over 1 second.

Model 3A

Body 2, “Sun”, δS₂ = -3.85 x 10²⁶ / 5780 = -6.66 x 10²² J/K

Body 3, “Space”, δS₃ = 3.85 x 10²⁶ / 3 = +1.28 x 10²⁶ J/K

And finally, Body 1, “Chilly Earth”, δS₁ = 0 / 255 = 0 J/K

Total Entropy Change = δS₁ + δS₂ + δS₃ = +1.28 x 10²⁶ J/K  :a net increase in entropy.

Model 3B

Body 2, “Sun”, δS₂ = -3.85 x 10²⁶ / 5780 = -6.66 x 10²² J/K

Body 3, “Crazy Background Radiation”, δS₃ = 3.85 x 10²⁶ / 250 = +1.54 x 10²⁴ J/K

And finally, Body 1, “Warmer Earth”, δS₁ = 0 / 255 = 0 J/K

Total Entropy Change = δS₁ + δS₂ + δS₃ = +1.47 x 10²⁴ J/K   :a net increase in entropy.

Important points to note about the entropy calculation

Both scenarios increase entropy – by transferring heat from a high temperature source, “Sun”, to a low temperature source, “Space” in 3A, and “Crazy Background Radiation” in 3B (which is really also Space at a higher temperature).

The earth-like planet is sitting in the middle and doesn’t have a significant effect on the entropy of the universe.

In both cases the entropy of the system increases, so both are in accordance with the second law of thermodynamics.

The earth cools to space, but just at a slower rate when the background temperature of “space” is higher.

If we replaced “crazy background radiation” by an atmosphere that was mostly transparent to solar radiation, the analysis would be a little more complex but the result wouldn’t be much different.

Reasons Why It Might be Wrong

Just to be clear, these aren’t true..

1. The hotter body can’t absorb radiation from the colder body

a) 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 commenter Bryan identifying his “non-cherrypicked” textbook by “real physicists” which also agreed.

b) see The Amazing Case of “Back Radiation” – Part Three which includes the EBEX experiment as well as a brief explanation of fundamental physics

c) see Absorption of Radiation from Different Temperature Sources – clearing up a few misconceptions on this idea

2. It’s not a real situation because the atmosphere isn’t a black body

It is true that the atmosphere is not a blackbody. But look back at model 3B. It doesn’t matter. Body 3 in this model could be a 250K body with an emissivity of 0.1 and the temperature would still increase over model 3A.

In fact, if the claim is that a colder body can never increase the temperature of a warmer body – all we need is one counter-example to falsify this theory. Now, if you want to modify your theory to something different we can examine this new theory instead.

Reasons Why It Has to be Right

1. The First Law of Thermodynamics. This neglected little jewel is quite important. Energy can’t disappear (or be created) or be quarantined into a mental box.

There is a reason why all the people disputing these basic analyses never explain where the energy goes (if it “can’t” go into changing the temperature of the hotter body that might have absorbed it). The reason – they don’t know.

2. The Second Law of Thermodynamics. This law says that in a closed system entropy cannot decrease. Despite angry claims about “no such thing as a closed system” – that’s what the second law says. Entropy is often simple to calculate.

If a solution uses simple radiation of energy (Stefan-Boltzmann’s law) and satisfies the first and second law of thermodynamics, and some people don’t like it, it suggests that the problem is with their conceptual model.

Conclusion

This is a conceptual model that is very simple.

The sun warms up the earth, and the earth cools to space. The colder “space” is, the faster the rate of net heat transfer. The warmer “space” is, the slower the rate of net heat transfer. And because the sun “pumps in” heat at the same rate, if you slow the rate of heat loss the equilibrium temperature has to increase.

The first law of thermodynamics is the key to understanding this problem. It is simple to verify that model 3A & 3B both satisfy the first law of thermodynamics. In fact, more importantly, a different result would contradict the first law of thermodynamics.

It is also easy to verify that in both 3A & 3B entropy increases.

Just to be clear on a tedious point, the earth and space do not have to radiate as a blackbody to have these conclusions. They just make the model simpler to explain, and the maths easier to understand. We could easily change the emissivity of the planet to 0.9 and the emissivity of space to 0.5 in both models and we would still find that Model 3B had a warmer planetary surface than Model 3A.

Many people will be unhappy, but this blog is not about bringing happiness. Clarity is the objective.

One more hopeless note of despair – this article uses simple theory to prove a point, which is actually a very valuable exercise. Next, some will say – “I don’t want that pointless over-theoretical theory, these people need to prove it with some experiments“.

And so I offer the series, The Amazing Case of “Back Radiation” as proof, especially Part Three. Result of Part Three was – “well, that can’t happen because it goes against theory“..

And so the circle is complete.

Notes

Note 1 – These strange conditions that don’t relate to the real world are to make the conceptual model simpler (and the maths easy). This is the staple of physics (and other sciences) – compare simple models first, then make them more complex and more realistic. If you can prove a theory with a simple model you have saved a lot of work and more people can understand it.

Note 2 – Solar radiation is from a tiny “angle” in the sky, and so the radiation is effectively “captured” by the earth as a flat disk in space. This area is the area of a disk = πr². By contrast, radiation from the sky is from all around the planet, and so the radiation is effectively captured by the surface area of the sphere. This area = 4πr². See The Earth’s Energy Budget – Part One for more explanation of this.

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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:

Radiant heat transfer

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 E_net1 as the change in energy experienced by body 1 (per unit area). And E_net2 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)

E_net1 = E₂ – E₁ = εσT₂⁴ – εσT₁⁴

E_net2 = E₁ – E₂ = εσT₁⁴ – εσT₂⁴

Therefore, E_net1 = -E_net2

Under The Imaginary Second Law

E_net1 = – E₁ = -εσT₁⁴

E_net2 = E₁ – E₂ = εσT₁4⁴ – εσT₂⁴

Therefore, E_net1 ≠-E_net2 ; 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, E_net1 = – E_net2. 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.

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Following discussions about absorption of radiation I thought some examples might help illustrate one simple, but often misunderstood, aspect of the subject.

Many people believe that radiation from a colder atmosphere cannot be absorbed by a warmer surface. Usually they are at a loss to explain exactly why – for good reason.

However, some have the vague idea that radiation from a colder atmosphere has different wavelengths compared with radiation from a warmer atmosphere. And, therefore, that’s probably it. End of story. Unfortunately for people with this idea, it’s not actually solved the problem at all..

The specific question I posed to one commenter some time ago was very specific:

If 10μm photons from a 10°C atmosphere are 80% absorbed by a 0°C surface, what is the ratio of 10μm photons from a -10°C atmosphere absorbed by that same surface?

It was eventually conceded that there would be no difference – 10μm photons from a -10°C will also be 80% absorbed. This material property of a surface is called absorptivity and is the proportion of radiation absorbed vs reflected at each wavelength.

Basic physics tells us that the energy of a 10μm photon is always that same, no matter what temperature source it has come from – see note 1.

Here’s an example of the reflectivity/absorptivity of many different materials just for interest:

Reflectivity vs wavelength for various surfaces, Incropera & DeWitt (2007)

Clearly materials have very different abilities to absorb /reflect different wavelength photons. Is this the explanation?

No.

The important point to understand is that even though radiation emitted from different temperature sources have different peak wavelengths, there is a large spread of wavelengths:

The peak wavelength of +10°C radiation is 10.2μm, while that of the -10°C radiation is 11.0μm – but, as you can see, both sources emit photons over a very similar range of wavelengths.

Scenarios

Let’s now take a look at the proportion of radiation absorbed from both of these sources.

First, with the case where the surface absorptivity is higher at shorter wavelengths – this should favor absorbing more energy from a hotter source and less from a colder source:

The top graph shows the absorptivity as a function of wavelength, and the bottom graph shows the consequent absorption of energy for the two cases.

Because absorptivity is higher at shorter wavelengths, there is a slight bias towards absorbing energy from the hotter +10°C source – but the effect is almost unnoticeable.

The actual numbers:

• 43% of the -10°C radiation is absorbed
• 46% of the +10°C radiation is absorbed

So let’s try something more ‘brutal’, with all of the energy from wavelengths shorter than 10.5μm absorbed and none from wavelengths longer than 10.5um absorbed (all reflected).

As you can see, the proportion absorbed of the energy from the hotter source vs colder source appears very similar. It is simply a result of the fact that +10°C and -10°C radiation have almost identical proportions of energy between any given wavelengths – the main difference is that radiation from +10°C has a higher total energy.

The actual numbers:

• 22% of the -10°C is absorbed
• 27% of the +10°C is absorbed

So – as is very obvious to most people already – there is no possible surface which can absorb a significant proportion of 10°C radiation and yet reflect all of the -10°C radiation.

And If There Was Such a Surface

Suppose that we could somehow construct a surface which absorbed a significant proportion of radiation from a +10°C source, and yet reflect almost all radiation from a -10°C source.

Well, that would just create a new problem. Because now, when our surface heats up to 11°C the radiation from the 10°C source would still be absorbed. And yet, the radiation is now from a colder source than the surface. Red alert for all the people who say this can’t happen.

Conclusion

The claim that radiation from a colder source is not absorbed by a warmer surface has no physical basis. People who claim it don’t understand one or all of these facts of basic physics:

a) Radiation incident on a surface has to be absorbed, reflected or transmitted through the surface. This last (transmitted) is not possible with a surface like the earth (it is relevant for something like a thin piece of glass or a body of gas), therefore radiation is either absorbed or reflected.

b) The material property of a surface which determines the proportion of radiation absorbed or reflected is called the absorptivity, and it is a function of wavelength of the incident photons. (See note 2)

c) The energy of any given photon is only dependent on its wavelength, not on the temperature of the source that emitted it.

d) Radiation emitted by the atmosphere has a spectrum of wavelengths and the difference between a -10°C emitter and a +10°C emitter (for example) is not very significant (total energy varies significantly, but not the proportion of energy between any two wavelengths). See note 3.

The only way that radiation from a colder source could not be absorbed by a warmer surface is for one of these basic principles to be wrong.

These have all been established for at least 100 years. But no one has really checked them out that thoroughly. Remember, it’s highly unlikely that you have just misunderstood the Second Law of Thermodynamics.

See also: The Real Second Law of Thermodynamics

Intelligent Materials and the Imaginary Second Law of Thermodynamics

The First Law of Thermodynamics Meets the Imaginary Second Law

The Amazing Case of “Back Radiation” – Part Three

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

Note 1 – Already explained in a little more detail in The Amazing Case of “Back Radiation” – Part Three – the energy of a photon is only dependent on the wavelength of that photon:

Energy = hc/λ

where h = Planck’s constant = 6.6×10^-34 J.s, c = the speed of light = 3×10⁸ m/s and λ = wavelength.

Note 2 – Absorptivity/reflectivity is also a function of the direction of the incident radiation with some surfaces.

Note 3 – For those fascinated by actual numbers – the energy from a blackbody source at -10°C = 272 W/m² compared with that from a +10°C source = 364 W/m² – the colder source providing only 75% of the total energy of the warmer source. But take a look at the proportion of total energy in various wavelength ranges:

• Between 8-10 μm  10.7% (-10°C)   12.2% (10°C)
• Between 10-12 μm  11.9% (-10°C)   12.7% (10°C)
• Between 12-14 μm  11.2% (-10°C)   12.4% (10°C)
• Between 14-16 μm   9.8% (-10°C)     9.5% (10°C)

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Many people often claim that the atmosphere cannot “heat” the earth’s surface and therefore the idea of the inappropriately-named “greenhouse” effect is scientifically impossible. The famous Second Law of Thermodynamics is invoked in support.

Let’s avoid a semantic argument about the correct or incorrect use of the word “heat”.

I claim that energy from the atmosphere is absorbed by the surface.

This absorbed energy has no magic properties. If the surface loses 100J of energy by other means and gains 100J of energy from the atmosphere then its temperature will stay constant. If the surface hasn’t lost or gained any energy by any other means, this 100J of energy from the atmosphere will increase the surface temperature.

I also claim that because the atmosphere is on average colder than the surface, more energy is transferred from the surface to the atmosphere compared with the reverse situation.

Let’s consider whether this violates the real second law of thermodynamics..

The Conceptual Problem

In Heat Transfer Basics – Part Zero a slightly off-topic discussion about the “greenhouse” effect began. One of our most valiant defenders of the imaginary second law of thermodynamics said:

An irradiated object can never reach a higher temperature than the source causing the radiation

I have demonstrated previously in The First Law of Thermodynamics Meets the Imaginary Second Law that a colder body can increase the temperature of a hotter body (compared with the scenario when the colder body was not there).

In that example, there was more than one source of energy. So, with this recent exchange in Heat Transfer Basics it dawned on me what the conceptual problem was. So this article is written for the many people who find themselves agreeing with the comment above. As a paraphrased restatement by the same commenter:

If the atmosphere is at -30°C then it can’t have any effect on the surface if the surface is above -30°C

Entropy Basics and The Special Case

Entropy is a difficult subject to understand. Heat and temperature are concepts we can understand quite easily. We all know what temperature is (in a non-precise way) and heat, although a little more abstract, is something most people can relate to.

Entropy appears to be an abstract concept with no real meaning – nothing you can get your hands around.

The second law of thermodynamics says:

Entropy of a “closed system” can never reduce

Before defining entropy, here is an important consequence of this second law:

Increasing entropy means that heat flows spontaneously from hotter to colder bodies and never in reverse

This fits everyone’s common experience.

• Ice melts in a glass of water
• A hot pan of water on the stove cools down to room temperature when the heat source is removed
• Put a hold and cold body together and they tend to come to the same temperature, not move apart in temperature

And all of these are easier to visualize than a mathematical formula.

What is entropy?  I will keep the maths to an absolute minimum, but we have to introduce a tiny amount of maths just to define entropy.

For a body absorbing a tiny amount of heat, δQ (note that δ is a symbol which means “tiny change”), the change in entropy, δS, is given by:

δS = δQ / T, where T is absolute temperature (see note 1)

It’s not easy to visualize – but take a look at a simple example. Suppose that a tiny amount of heat, 1000 J, moves from a body at 1000K to a body at 500K:

Example 1

The net change in energy in the system is zero because 1000J leaves the first body and is absorbed by the second body. This is the first law of thermodynamics – energy cannot be created or destroyed.

However, there is a change in entropy.

The change in total entropy of the system = δS₁ + δS₂ = -1 + 2 = 1 J/K.

This strange value called “entropy” has increased.

Notice that if the energy flow of 1000J was from the 500K body to the 1000K body the change in entropy would be -1 J/K. This would be a reduction in entropy – forbidden by the real second law of thermodynamics. This would be a spontaneous flow of heat from the colder body to the hotter body.

Updated note Sep 30th – this example is intended to clarify the absolute basics.

Think of the example above like this – If, for some reason, in a closed system, this was the only movement of energy taking place, we could calculate the entropy change and it has increased.

The example is not meant to be an example of only one half of a radiative energy exchange. Just a very very simply example to show how entropy is calculated. It could be conductive heat transfer through a liquid that is totally opaque to radiation.

The Special Case

The simplest example demonstrating the second law of thermodynamics is with two bodies which are in a closed system.

Let’s say that we have a gas at 273K (Body 1) and a solid (Body 2) surrounded by the gas. The solid starts off much colder.

What is the maximum temperature that can be reached by the solid?

273K

Easy. In fact, depending on the starting temperature of the solid and the respective heat capacities of the gas and solid, the actual temperature that both end up (the same temperature eventually) might be a little lower or a lot lower.

But the temperature reached by the solid can never get to more than 273K. For the solid to get to a temperature higher than 273K the gas would have to cool down below 273K (otherwise energy would have been created). Heat does not spontaneously flow from a colder to a hotter body so this never happens.

This defining example is illuminating but no surprise to anyone.

It is important to note that this special case is not the second law of thermodynamics, it is an example that conforms to the second law of thermodynamics.

The second law of thermodynamics says that the entropy of a system cannot reduce. If we want to find out whether the second law of thermodynamics forbids some situation then we need to calculate the change in entropy – not use “insight” from this super-simple scenario.

So let’s consider some simple examples and see what happens to the entropy.

Simple Examples

What I want to demonstrate is that the standard picture in heat transfer textbooks doesn’t violate the second law of thermodynamics.

What is the standard picture?

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

This says that two bodies separated in space both emit radiation. And both absorb radiation from the other body (see note 2).

The challenging concept for some is the idea that radiation from the colder body is absorbed by the hotter body.

We start with Example 1 above, but this time we consider an exchange of radiation and see what happens to the entropy of that system.

Example 2

What I have introduced here is thermal radiation from Body 2 incident on Body 1. We will assume all of it is absorbed, and vice-versa.

According to the Stefan-Boltzmann equation, energy radiated is proportional to the 4th power of temperature. Given that Body 2 is half the temperature of Body 1 it will radiate at a factor of 2⁴ = 2 x 2 x 2 x 2 = 16 times less. Therefore, if 1000J from Body 1 reaches Body 2, then 62J (1000/16) will be transmitted in the reverse direction. However, the exact value doesn’t matter for the purposes of this example.

So with our example above, what is the change in entropy?

Body 1 loses energy, which is negative entropy. Body 2 gains energy, which is positive entropy.

δS₁ = -(1000-62)/1000 = -938 / 1000 = -0.94 J/k
δS₂ =(1000-62)/500 = 938 / 500 = 1.88 J/K

Total entropy change = -0.94 + 1.88 = 0.94 J/K.

So even though energy from the colder body has been absorbed by the hotter body, the entropy of the system has increased. This is because more energy has moved in the opposite direction.

There is no violation of the second law of thermodynamics with this example.

Now let’s consider an example with values closer to what we encounter near the earth’s surface:

Example 3

This isn’t intended to be the complete surface – atmosphere system, just values that are more familiar.

Surface:        δS₁ = -(390-301)/288 = -89 / 288 = -0.31 J/k
Atmosphere: δS₂ = (390-301)/270 = 89 / 270 = 0.33 J/K

Total entropy change = -0.31 + 0.33 = 0.02 J/K.

So even though the temperatures of the two bodies are much closer together, when they exchange energy, total entropy still increases.

Energy from the colder atmosphere has been absorbed by the hotter surface and yet entropy of the system has still increased.

Now, the example above (example 3) is an exchange of a fixed amount of energy (in Joules, J). Suppose this is the amount of energy per second (Watts, W) or the amount of energy per second per square meter (W/m²).

If the atmosphere keeps absorbing more energy than it is emitting it will heat up. If the earth keeps emitting more energy than it absorbs, it will cool down.

If example 3 was the complete system, then the atmosphere would heat up and the earth would cool down until they were in thermal equilibrium. This doesn’t happen because the sun continually provides energy.

The Complete Climate System

The earth-atmosphere system is very complex. If we analyze a long term average scenario, like that painted by Kiehl and Trenberth there is an immediate problem in calculating the change in entropy:

From Kiehl & Trenberth (1997)

[Note from Sep 28th – This section is wrong, thanks Nick Stokes for highlighting it and so delicately! Preserved in italics for entertainment value only..] If we consider the surface, for example, it absorbs 492 W/m² (δQ = 492 per second per square meter) and it loses 492 W/m² (δQ = -492 per second per square meter).

Net energy change = 0. Net entropy change = 0.

Why isn’t entropy increasing? We haven’t considered the whole system – the sun is generating all the energy to power the climate system. If we do consider the sun, it is emitting a huge amount of energy and, therefore, losing entropy. But the energy generation inside the sun creates more entropy – that is, unless the second law of thermodynamics is flawed.

[Now the rewritten bit]

Previous sections explained that calculations of entropy “removed” (negative entropy) are based on energy emitted divided by the temperature of the source. And calculations of entropy “produced” are based on energy absorbed divided by the temperature of the absorber. In a closed system we can add these up and we find that entropy always increases.

So the calculation in italics above is incorrect. Change in entropy at the surface is not zero.

Change in entropy at the surface is a large negative value, because we have to consider the source temperature of the energy.

So as Nick Stokes points out (in a comment below), we can draw a line around the whole climate system, including the emission of radiation by the sun (see example 4 just below). This calculation produces a large negative entropy, because it isn’t a closed system. This is explained by the fact that the production of solar energy creates an even larger amount of positive entropy.

Example 4

The Classic Energy Exchange by Radiation

I was in the university library recently and opened up a number of heat transfer textbooks. All of them had a similar picture to that from Incropera and DeWitt (above). And not a single one said, This doesn’t happen.

In any case, for someone to claim that an energy exchange violates the second law of thermodynamics they need to show there is a reduction in entropy of a closed system.

But one important point did occur to me when thinking about this subject. Let’s reconsider our commenter’s claim:

An irradiated object can never reach a higher temperature than the source causing the radiation

As I pointed out in the The Special Case section – this is true if this is the only source of energy. Yet the surface of the earth receives energy from both the sun and the atmosphere.

If the colder atmosphere cannot transfer energy to a warmer surface, and the second law of thermodynamics is the reason, the actual event that is forbidden is the emission of radiation by the colder atmosphere. When the colder atmosphere radiates energy it loses entropy.

After all, the entropy loss takes place when the atmosphere has given up its energy. Not when another body has absorbed the energy.

Our commenter has frequently agreed that the colder atmosphere does radiate. But he doesn’t believe that the surface can absorb it. He has never been able to explain what happens to the energy when it “reaches” the surface. Or why the surface doesn’t absorb it. Instead we have followed many enjoyable detours into attempts to undermine any of a number of fundamental physics laws in an attempt to defend “the imaginary second law of thermodynamics”.

Conclusion

Entropy is a conceptually difficult subject, but all of us can see the example in “the special case” and agree that the picture is correct.

However, the atmosphere – surface interaction is more complex than that simple case. The surface of the earth receives energy from the sun and the atmosphere.

As we have seen, in simple examples of radiant heat exchange between two bodies, entropy is still positive even when the hotter body absorbs energy from the colder body. This is because more energy flows from the hotter to the colder than the reverse.

To prove that the second law of thermodynamics has been violated someone needs to demonstrate that a system is reducing entropy. So we would expect to see an entropy calculation.

Turgid undergraduate books about heat transfer in university libraries all write that radiation emitted by a colder body is absorbed by a hotter body.

That is because the first law of thermodynamics is still true – energy cannot be created, destroyed, or magically lost.

Other Relevant Articles

The Amazing Case of “Back Radiation” – Part Three

The First Law of Thermodynamics Meets the Imaginary Second Law

Intelligent Materials and the Imaginary Second Law of Thermodynamics

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

Notes

Note 1 – There are more fundamental ways to define entropy, but it won’t help to see this kind of detail. And for the purists, the equation as shown relies on the temperature not changing as a result of the small transfer of energy.

If the temperature did change then the correct formula is to integrate:

ΔS = ∫Cp/T. dT (with the integral from T1 to T2) and the result,

ΔS = Cp log (T2/T1),   this is log to the base e.

Note 2 – This assumes there is some “view factor” between the two bodies – that is, some portion of the radiation emitted by one body can “hit” the other. Just pointing out the obvious, just in case..

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