You explain these things better than I thought possible! ]]>

Jan: 239 W/m2 is used because it was the best estimate of the average amount of radiation being absorbed (post-albedo). And emitted by the planet assuming a steady state before rising GHGs.

Most of what you have written is correct. There are multiple ways to answer this question.

A doubling of CO2 was calculated by Myhre to reduce LWR radiative cooling to space by 3.7 W/m2. The calculation assumes nothing else changes and is sometimes referred to as an instantaneous doubling. Under this hypothetical situation, the law of conservation of energy demands that temperature rise (somewhere below the tope of the atmosphere or TOA). That rise will continue until the planet emits 3.7 W/m2 and no radiative imbalance at the TOA exists.

To determine how much warming is needed to eliminate this imbalance, we need to know how much more radiation the planet emits for each degC its temperature rises. This is called the climate feedback parameter (W/m2/K).

As you noted, one can postulate that the Earth emits 239 W/m2 because it behaves like a blackbody at 255:

W = -oT^4

dW/dT = -4oT^3 = -3.76 W/m2/K

According to this model, the 3.7 W/m2 radiative imbalance from 2XCO2 will be negated by 0.98 K of warming. (By convention, the negative sign represents heat lost by the planet, though this convention is sometimes ignored.)

As you noted, others (presumably including this post) postulate a graybody model for the Earth, where emission of 239 W/m2 is the result of a surface temperature of 288 K and an emissivity of 0.615 (or slight variants of these values).

W = -eoT^4

dW/dT = -4oeT^3 = -3.33 W/m2

According to this model, the 3.7 W/m2 radiative imbalance from 2XCO2 will be negated by 1.11 K of warming.

A few postulate a model where the Earth behaves like at blackbody at 288, but this model doesn’t explain why the Earth emits 239 W/m2. Most people would call this “wrong”, but technically one is free to postulate any model one wants.

Climate models divide the planet up into roughly a million grid cells with realistic temperatures and calculate how much more radiation they absorb and emit. From their output, climate scientists have determined that a warming planet emits 3.21 W/m2/K (+/-0.05 W/m2/K) more radiation to space – assuming nothing else changes. According to this model, the 3.7 W/m2 radiative imbalance from 2XCO2 will be negated by 1.15 K of warming. This is what most people call the no-feedbacks climate sensitivity.

If you want to be more sophisticated, the following model I devised may be useful. Others explain this differently. The planet’s radiative imbalance (I) is:

I = (S/4)*(1-a) – eoTs^4

where a is albedo, S is incoming solar radiation, Ts is surface temperature and e is the planet’s effective emissivity given an average Ts of 288 K.

dI/dTs = -4eoTs^3 – (oTs^4)*(de/dTs) – (S/4)*(da/dTs)

dI/dTs is the climate feedback parameter. Here emissivity hasn’t been treated as a constant. The first term is called Planck feedback and is -3.3 W/m2. The third term is the change in reflection of SWR by clouds and the surface, called cloud SWR and ice albedo feedbacks. The sum of the first and second terms is called LWR feedback. The second term includes: a) the effect on emissivity of rising humidity (water vapor is a GHG reducing emissivity) with temperature, b) the change in emissivity due rising humidity causing more warming higher in the atmosphere than at the surface, and c) the change in emissivity of clouds due to a change in their altitude/temperature or composition. These are called water vapor, lapse rate and cloud LWR feedbacks. (And while we are being more sophisticated, the composition of atmosphere predicted by climate model changes with warming and those changes result in a doubling of CO2 producing an effective forcing (reduction of radiative cooling to space) from 2.4 to 4.4 W/m2.)

No one knows the correct values for de/dTs, da/dTs, or feedbacks, but they can be abstracted from the output of climate models. Most climate models predict dI/dTs is around -1 W/m2/K. Current forcing is about 2.5 W/m2 and ocean heat uptake is about 0.7 W/m2, so current warming of roughly 1 degK is sending about 1.8 W/m2 more radiation to space (emission of LWR and reflection of SWR). That’s -1.8 W/m2/K; almost half as much warming at steady state as predicted by climate models. This approach is called an energy balance model.

Assuming I haven’t made any mistakes, all of these answers begin with different models/assumptions and are as “correct” or “wrong” as the assumptions on which they are based. Confusion arises because different people use different assumptions without making them clear.

]]>I would like to discuss the following section:

Tnew4/Told4 = (239 + 3.7)/239

where Tnew = the temperature we want to determine, Told = 15°C or 288K

We get Tnew = 289.1K or a 1.1°C increase.

Why do you take the value of 239? The current temperature is not because of the solar radiation of 239 W/m2, its due to the solar radiation of 239 W/m2 + the current radiation forcing of 157 W/m2 which brings it to 396 W/m2. You are going to raise the total energy warming the planet from this value of 396 W/m2 to 399.7 W/m2.

So we get

Tnew4/Told4 = (396 + 3.7)/396

where Tnew = the temperature we want to determine, Told = 15°C or 288K

We get Tnew = 288.67K or a 0.67°C increase.

What you are doing is calculating the ratio by which an earth without an atmosphere would increase given an extra 3.7 W/m^2. Then you multiply that ratio with the temperature of the earth with an atmosphere to get to the end value.

Off course it requires way less energy to heat non greenhouse gas -18 degree Earth than it does to heat pre industrial revolution +15 degree greenhouse earth.

Put differently the 239 comes from how much the sun shines on the top of the atmosphere and how much the earth radiates into space from the top of the atmosphere. This value is in equilibrium as you explained in another post. Obviously when we get more greenhouse gasses this value will not change! It will be the 396 W/m^2 that will change. So this should show in your calculations.

I also looked at calculations I came across at other places that try to derive the climate sensitivity parameter from the Boltzmann equation.

I came across this: landa = Ts / 4*sigma*Te^4

Here: http://web.ma.utexas.edu/mp_arc/c/11/11-16.pdf

She used Ts = 288 and Te = 255, again I would argue that 255 (temperature of a blackbody radiation 239 W/m^2) has no place in these calculations and it should also equate to 288. (289 is closer if I take your value of 396 btw)

Then that equation simplifies to landa = 1 / 4*sigma*T^3

Similar to the one I found here:

clivebest.com/blog/?p=4923

Where he uses an emissivity factor to make the earth a greybody instead of a blackbody.

Now you wrote on another page that the earth could be seen as close to a blackbody. So if we take the simplified formula and T = 289 then that solves for landa to 0.1827. Significantly lower than the 0.3 commonly quoted.

This could only come close to 0.3 (but not quite, 0.285) if we use an emissivity of 0.64 which wikipedia claims of the earth taking cloud cover into account.

But would this not already be effectively a feedback loop of the water vapor?

And even when not half of the atmosphere by mass (pressure, same thing if you think about it) lies under 5000 meter altitude. So half of the radiative forcing should take place here under the clouds (assuming that the average cloud lies at 5000 meter) and thus the emissivity that matters for that half at least is that of the earth’s surface which is quite close to 1, lets say 0.95. So the total emissivity should be somewhere in between 0.64 and 0.95 then.

However clouds are vapor and does not absorb the same wavelengths as CO2 albeit there’s some overlap as you showed in part 1. Thus assigning a lower emissivity to the earth due to the clouds seems to be a mistake from the getgo. The source of the radiation is the Earth and giving it a lower emissivity (which would increase the effect of CO2) because it already has gasses surrounding it that traps radiation that the CO2 cant trap a second time seems illogical to say the least.

Thus the question remains. Where does the 1-1.2 degree no-feedback calculation comes from? Because I get different figures using two different, albeit it similar methods.

Thank you for your attention.

I hope you can shine a good light on this.

Regards,

Jan

Bruce asked: Is it possible to follow heat emitted using satellites over time? If so, is it possible to demonstrate that the “CO2 bite” has become larger as more CO2 has been added to the atmosphere?

Let’s do some simple calculations to determine if what you ask is possible. Our satellites in space show that the planet is emitting an average of about 240 W/m2 of heat (thermal infrared or LWR). CO2 is near 400 ppm and increasing about 2 ppm/yr or 5%/decade. At that rate it would take 14 decades to double (1.05^14 = 20) or 20 decades using linear rather than exponential growth. According to laboratory measurements of radiation, a doubling of CO2 will reduce radiative cooling to space by 3.7 W/m2 or 0.26 W/m2/decade. Requires measuring an 0.1% change in 240 W/m2/decade – from space.

Even worse, the planet is warming (because of that reduction in emission of heat to space). A warmer planet emits more heat to space, compensating with some lag for the reduction caused by CO2.

Even worse, the 240 W/m2 average we do measure varies with the seasons (almost 10 W/m2) and weather and phenomena like ENSO.

No, we can’t measure what you would like us to measure.

What we can do is use data from laboratory measurements to predict what we should “see” at various locations on the planet. “See” means measure the radiation intensity at all wavelengths coming down from the sky to the surface or up from the Earth to space. The former is easier to do. First, a radiosonde is sent to measure the temperature and humidity at all altitudes overhead. Then the spectrum of radiation reaching the surface is compared with what we calculate based on laboratory measurements. This has been done from Antarctica to the tropics, producing changes in downward radiation of more than 100 W/m2 due to temperature differences and large changes in the spectrum due to changes in water vapor. Agreement between theory and experiment with these large changes should give you confidence we can calculate small changes. SOD provided this link to compare theory and experiment.

https://scienceofdoom.com/2010/11/01/theory-and-experiment-atmospheric-radiation/

In general, it is a bad idea to expect to confirm the physics of climate change by observing the planet. The climate change you are seeking is changing extremely slowly, weather changes chaotically, large El Ninos and La Ninas are massive disruptions.

]]>Bruce,

It’s a good question but it’s not as simple as you might think to do this experimentally.

1. Satellites measuring outgoing longwave radiation (OLR) across the spectrum are rare. Well, over the last 15 years there has been the AIRS satellite, but prior to that it was just occasional measurements.

2. The total radiance at each wavelength actually depends on the surface temperature as well as how much the CO2 takes out of the spectrum. It also depends on the absorption by the water vapor continuum in this range of wavelengths (this is hard to explain in a sentence or two but more in the series Visualizing Atmospheric Radiation.

Now perhaps you are wondering whether the whole idea of CO2 absorption is quite speculative. Not at all. It’s very repeatable and boring to put radiation of different wavelengths through gases with different proportions of CO2 and get exactly the same amount of absorption each time. For example, if you want bedtime reading you can read *Journal of Quantitive Spectroscopy and Radiative Transfer* where there are decades of papers on this subject by people who do these kind of measurements.

Likewise, if we know the surface temperature, the atmospheric temperature profile and the concentration of CO2, water vapor and other GHGs, we can calculate the spectrum of radiation at the top of the atmosphere and compare it with satellite measurements. The results match, which is why the theory of radiative transfer is fundamental physics (and used in all kinds of measurements).

For example, this comparison of theory and experiment is shown in Theory and Experiment – Atmospheric Radiation:

]]>Here is my question: Is it possible to follow heat emitted using satellites over time? If so, is it possible to demonstrate that the “CO2 bite” has become larger as more CO2 has been added to the atmosphere?

]]>Erin,

On latent heat, have a read of Sensible Heat, Latent Heat and Radiation.

I don’t think I have anything useful about ocean currents, but a good introduction about the deep ocean currents is Thermohaline circulation.

]]>Let’s take the analogy of a small boat crossing the Atlantic.

(Analogies don’t prove anything, they are for illustration. For proof, please review Theory and Experiment – Atmospheric Radiation).

We’ve done a few crossings and it’s taken 45 days, 42 days and 46 days (I have no idea what the right time is, I’m not a nautical person).

We measure the engine output – the torque of the propellors. We want to get across quicker. So Fred the engine guy makes a few adjustments and we remeasure the torque at 5% higher. We also do Fred’s standardized test, which is to zip across a local sheltered bay with no currents, no waves and no wind – the time taken for Fred’s standarized test is 4% faster. Nice.

So we all set out on our journey across the Atlantic. Winds, rain, waves, ocean currents. We have our books to read, Belgian beer and red wine and the time flies. Oh no, when we get to our final destination, it’s actually taken 47 days.

Clearly Fred is some kind of charlatan! No need to check his measurements or review the time across the bay. We didn’t make it across the Atlantic in less time and clearly the ONLY variable involved in that expedition was the output of the propellor.

Well, there’s no point trying to use more powerful engines to get across the Atlantic (or any ocean) faster. Torque has no relationship to speed. Case closed.

Analogy over.

Read the article I cited.

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