Guest post by Willis Eschenbach
There is a lot of misinformation floating around the web about how the greenhouse effect works. It is variously described as a “blanket” that keeps the Earth warm, a “mirror” that reflects part of the heat back to Earth, or “a pane of glass” that somehow keeps energy from escaping. It is none of these things.
A planetary “greenhouse” is a curiosity, a trick of nature. It works solely because although a sphere only has one side, a shell has two sides. The trick has nothing to do with greenhouse gases. It does not require an atmosphere. In fact, a planetary greenhouse can be built entirely of steel. A thought experiment shows how a steel greenhouse would work.
Before we start, however, a digression regarding temperature. The radiation emitted by a blackbody varies with the fourth power of the temperature. As a result, for a blackbody, we can measure the temperature in units of radiation, which are watts per square meter (W/m2). For objects with temperatures found on the Earth, this radiation is in the range called “longwave” or “infrared” radiation. See the Appendix for the formula that relates temperature to radiation.
This means that we can denote the temperature of a blackbody using W/m2 as well as the traditional measures (Fahrenheit, Celsius, Kelvin). The advantage is that while temperature (degrees) is not conserved, energy (W/m2) is conserved. So we can check to see if energy lost is equal to energy gained, since energy is neither being created nor destroyed by the climate.
For our thought experiment, imagine a planet the size of the Earth, a perfect blackbody, heated from the interior at 235 watts per square meter of surface area. How is it heated from the interior? Doesn’t matter, we’ll say “radioactive elements”, that sounds scientific.
The planet is in interstellar space, with no atmosphere and no nearby stars. The equilibrium surface temperature of this planet is, of course, 235 W/m2. To maintain the equilibrium temperature, it constantly radiates this amount of energy out to space. Coincidentally, this is the amount of solar radiation that makes it past the clouds to warm the Earth. If we convert 235 W/m2 to one of our normal temperature scales, it is -19 Celsius (C), or -3° Fahrenheit (F), or 254 Kelvins (K). It’s a possible temperature that the Earth might have if there were no greenhouse effect. That is to say … cold.
Now imagine that the planet gets completely surrounded by a thin black steel shell, located a few thousand meters above the surface, as shown in a cutaway view in the picture above, and in Figure 1 below. What happens to the surface temperature of the planet? (To simplify calculations, we can assume the shell has the same outer surface area as the surface. For an earth-sized planet with a shell two kilometers above the surface everywhere, the difference in area is only six-hundredths of one percent. This assumption makes no difference to the argument presented.)
In order to maintain its thermal equilibrium, including the new shell, the whole system must still radiate 235 W/m2 out to space. To do this, the steel shell must warm until it is radiating at 235 watts per square meter. Of course, since a shell has an inside and an outside, it will also radiate 235 watts inward to the planet. The planet is now being heated by 235 W/m2 of energy from the interior, and 235 W/m2 from the shell. This will warm the planetary surface until it reaches a temperature where it radiates at 470 watts per square meter. In vacuum conditions as described, this would be a perfect greenhouse, with no losses of any kind. Figure 1 shows how it works.
Figure 1. Building a steel greenhouse. (A) Planet without greenhouse. Surface temperature is 235 W/m2 heated from the interior. (B) Planet surrounded by steel greenhouse shell. Shell radiates the same amount to space as the planet without the shell, 235 W/m2. It radiates the same inward, which warms the planet to 470 W/m2 (29 C, 83°F, 302 K). [Clarification added] Note that the distance from the shell to the planet is greatly exaggerated. In reality, it is close enough to the planet that the difference in the areas can be neglected in practice.
The trick can be repeated again, by surrounding the planet and the shell with a second outer shell. In this two shell case, the planetary surface temperature (in W/m2) will be three times the initial surface temperature.
In nature, planets have atmospheric shells, of course, rather than steel shells. The trick works the same way, however. Rather than being warmed from the inside, the Earth receives 235 W/m2 from the sun. Solar radiation passes through the atmosphere. Outgoing longwave radiation is absorbed by the atmosphere, just as it is absorbed by the steel shell shown in Fig. 1.
So that’s the trick of the greenhouse. It has nothing to do with blankets, mirrors, or greenhouse gases. It works even when it is built out of steel. It depends on the fact that a shell radiates in two directions, inwards and outwards. This radiation keeps the planet warmer than it would be without the shell.
Now, it is tempting to think that we could model the Earth in this manner, as a sphere surrounded by a single shell. This is called a “two-layer” model, with the two layers being the surface and the atmospheric shell. In fact, a number of simplified climate models have been built in this way. Unnoticed by their programmers, however, is that is not physically possible to model the Earth as a two-layer system. Figure 2 shows why. Note that in all cases, the system has to remain in thermal equilibrium. This means that the surface must radiate as much as it absorbs, and the shell must also radiate as much as it absorbs. In addition, radiation from the shell to space (upwelling longwave radiation or ULR) must equal radiation from the shell to the Earth (downwelling longwave radiation, or DLR)
Figure 2. Single-shell (“two-layer”) greenhouse system, including various losses. S is the sun, E is the Earth, and G is the atmospheric greenhouse shell around the Earth. The height of the shell is greatly exaggerated; in reality the shell is so close to the Earth that they have about the same area, and thus the small difference in area can be neglected. Fig. 2(a) shows a perfect greenhouse. W is the total watts/m2 available to the greenhouse system after albedo. Fig. 2(b) is the same as Fig. 2(a) plus radiation losses Lr which pass through the atmosphere. Fig. 2(c) is the same as Fig. 2(b), plus the effect of absorption losses La. Fig. 2(d) is the same as Fig. 2(c), plus the effect of thermal losses Lt.
Figure 2(a) shows the same situation as Figure 1(B), which is a perfect planetary greenhouse. In this case, however, it is heated by an amount of energy “W”, which is coming from the sun. The planet receives solar radiation in the amount of “W” from the sun, and longwave radiation “W” from the atmospheric shell. The surface temperature is thus 2W. All energy flows are in Watts/square metre (W/m2).
Figure 2(b) adds two losses. The first is the reflection from the Earth’s albedo (Wo – W). This is energy which never enters the system and is reflected back into space. We are still getting the same energy entering the system (W). The second loss Lr is from radiation which goes from the surface to space through the “atmospheric window”. Because of the second loss, the surface of the Earth does not get as warm as in a perfect system. In a perfect system, the temperature of the surface is 2W. But including the radiation loss Lr, the surface temperature drops to 2W – 2Lr.
Figure 2(c) adds another loss La, which is the solar radiation which is absorbed by the shell. This cuts the radiation hitting the surface down to W – La. Including this loss, the surface temperature is 2W – 2Lr – La.
Figure 2(d) adds the final loss. This is Lt, the thermal loss. It is the sensible energy (energy you can feel) and latent energy (evaporation) which is transported from the surface to the shell by convection. Including all of these losses, the surface temperature of the Earth is 2W – 2Lr – La – Lt.
Now, why can’t the Earth be modeled in this manner? A look at the size of the various losses shows why. Here is the canonical global energy budget, called the “Kiehl/Trenberth” budget, or the K/T budget.
Figure 3. The Kiehl/Trenberth Global Energy Budget. This is a “two layer” representation, with the surface and the atmosphere being the two layers. Lr, the radiation loss, is the 40 W/m2 labeled “Atmospheric Window”. La, the absorption loss, is the 67 W/m2 labelled “Absorbed by Atmosphere”. Lt, the thermal loss, is 102 W/m2. This is the sum of the 24 W/m2 labelled “Thermals” and the 78 W/m2 labelled “Evapo-transpiration”. W, the energy from the sun, is the incoming solar radiation of 342 W/m2 less the 107 W/m2 that is reflected by the surface and the clouds. This means that W is 235 W/m2. SOURCE
What’s wrong with this picture? Note that the temperature of the Earth is 390 W/m2, labeled as “Surface Radiation”. This is 15 C, or 59°F. But from Fig. 2(d), we know that the surface temperature of a greenhouse system with a single shell, as shown in the drawing, is 2W (470) – 2Lr (80) – La (67) – Lt (102) = 221 W/m2. This is far below the known surface temperature of the Earth. In other words, a single shell greenhouse system simply isn’t efficient enough to give a surface temperature which is warm enough to allow for the known losses.
So where is the problem with the K/T budget diagram? The hidden fault is that the upward radiation from the atmospheric layer does not equal the downward radiation. There is 195 W/m2 going to space from the atmospheric shell, and 324 W/m2 going down to the surface.
In order to get enough energy to allow for the known losses, the simplest model requires two atmospheric shells. A perfect greenhouse with two shells would give a surface temperature of 3W, or 705 W/m2. This is enough to allow for the known losses, and still give a surface temperature which matches that of the Earth. Figure 4 shows one such possible system.
Figure 4. Simplest greenhouse global energy budget capable of representing the Earth. Note that all major flows in the K/T energy budget have been maintained. There are two shells, which must be physically separate. These are the lower troposphere and the lowest stratosphere. They are separated by the tropopause.
This budget fulfills all of the requirements for thermal equilibrium. The same amount is radiated upwards and downwards by each layer. The amount absorbed by each part of the system equals the amount radiated. Further verification of this energy budget is the 147 W/m2 emission from just above the tropopause. This converts to a Celsius temperature of about -50 C, which is a typical temperature for the lowest part of the stratosphere.
I have written a simplified Excel radiation/evaporation/convection model which encapsulates the above system. It is available here. Click on the “File” menu on the webpage and select “Download”.
I invite people to play with the model. It is what I term a “Tinkertoy” model, which is what I call the simplest possible model that can represent a particular reality. One of the interesting results from the model is that there can be very little thermal loss between the inner and the outer atmospheric layers. If there is more than a trivial amount of leakage, the surface cools significantly. Among the other insights yielded by the model is that a change equivalent to a doubling of CO2 (an increase of 3.7 W/m2 downwelling radiation at the top of the atmosphere) can be canceled by a 1% increase in the upper and lower cloud reflections.
Experiment with the values, it is an interesting insight into the energy flows in the simplest possible climate model that can represent the Earth’s greenhouse system.
APPENDIX
The formula that relates the temperature to radiation is called the “Stefan-Bolzmann” equation:
R = sigma * epsilon * T^4
where r = radiation (W/m2)
sigma = the Stefan-Boltzmann constant, 5.67 * 10^-8
epsilon = the emissivity of the body, which for a blackbody = 1
T^4 = absolute temperature in Kelvins (which is Celsius plus 273.15) raised to the fourth power
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“In order to maintain its thermal equilibrium, the whole system must still radiate 235 W/m2 out to space…” – from paragraph just above fig.1
In other words, by definition, the steel shell is arbitrarily assigned the same temperature as the earth? If so, then, from thermodynamics, there will be no heat transfer between them, because what drives radiant heat exchange (or any other, for that matter) is the difference in temperature (here it’s proportional to [T1^4 – T2^4]). So, all things being equal, the shell’s output per unit area should really be lower, because the rate at which the earth generates heat remains a constant. And, because the shell has a larger surface area, and the Net Power (see below) is constant, the per unit area output can’t be the same as it was for earth alone.
Now, if you insist that the per unit area output remain the same as it was for the smaller area of the earth, your problem would appear to be as follows:
The equation for a blackbody is P = esA(T1^4 – T2^4), where P (Net Power) has units of Watts, not Watts/squared meter. And because of the larger area of the shell (it’s radius is greater than that of the earth), for the radiation emitted by the shell to be maintained at the same Watts/square meter, the “Net Power” of the shell would then be greater than what the earth’s was alone. That will only happen if the earth increases it’s untill-now-fixed output.
Of course, even without the increase in earth’s output, it’s temp would rise slightly because it was no longer radiating to empty space at T~0, but now to the Shell at T2′. Nevertheless, the slight increase in earth’s temp., due to encasing the earth in the shell wouldn’t be sufficient to drive the shell’s output at the same level per unit area as the earth’s alone, the earth would have to increase it’s net power output for this scheme to work. But wasn’t it fixed?
“Bridge to engine room, Scotty . . . ?!”
NOTE – The addition of another shell would again necessitate another call to the engine room, otherwise the requirement that the new shell be in thermal equilibrium with the rest of the system is not going to happen.
What would happen without the unrealistic escalating demands on the earth’s ability to generate heat would be that each successive shell gets colder and colder, while the earth would warm slightly each time, with each successive warming less than that of the previous one. Compare that result with what’s expected for CO2, where for each fixed increment of CO2 added, the temp goes up less than it did for the previous one.
http://www.junkscience.com/Greenhouse/co2greenhouse-X4.png.
Ian Schumacher: The reason you are struggling so much with the blackbody idea from the old physics textbook is that you are misinterpreting what it means. The rule is simply: at a given temperature, nothing can emit more radiation than a blackbody. Perhaps, stated mathematically: the emissivity of a surface cannot be greater than one. Willis is setting the emissivity of his earth to one, so there is no problem. That doesn’t keep it from changing temperature, given changing conditions around it.
The cavity with the pinhole illustration is only given because it approaches the perfect blackbody: something with perfect absorption, as well as maximum and diffuse emission.
*Actually, this brings a slight clarification: Willis’s earth emits like a perfect blackbody, but it does not absorb like one. Which is perfectly fine.
John A
I have followed your arguments about the second law of thermodynamics, and after reading your reply to Joel Shore’s excellent explanation, I conclude that you are making the same mistake over and over and over.
The example you give of the ice surrounding the soup shows that you completely misunderstand the problem. The mistake you make is that you are not seeing the whole picture. You evisage ice surrounding soup and your experience tells you that the ice does not make the soup warmer, but quite the reverse. But in your thought experiment, your bowel of soup is in a nice warm kitchen, and then you surrounded it with cold ice. The result obviously is then the soup cools faster and the ice melts. But your mistake is to have imagined the soup to be in the warm kitchen in the first place. Instead, lets imagine the soup (in a pressure vessel of course) is placed in space, out of the sun, near absolute zero. The soup will cool down pretty damn quick on its own. But now surround it with the ice, and the rate of cooling is less.
Your reference to Maxwells demon, explains that the entropy of the system increases, since the Demon has to do work to make the warm partition hotter. But in the example with two radiating bodies at different temperatures that was given by Willis, the warmer body does indeed loose heat to the cooler body, in accordance to the second law of thermodynamics. However, some photons of energy from the cooler body will still strike the warmer body. Why does that violate the second law? Please try and think of a physical mechanism that would stop this from happening. And do not just say “it violates the second law” because that is a tautoligious argument. Why, cannot this photon from the cooler body strike the warmer?
The consequence of a photon from the cooler striking the warmer, is that the warmer will not cool down so quickly as it would have done if it was completely surrounded by empty space at absolute zero. So therefore, yes, the cooler body, will have reduced the rate of cooling of the warmer body.
Willis Eschenbach (23:46:55) :
“Radiation goes on within a gas, but not, as far as I know, within a solid”
To clarify: all matter radiates, no matter where it is. However, radiation emitted by a molecule within a solid will generally be absorbed by the immediately neighboring molecules: it doesn’t get very far. It may or may not be intuitive then, that within a given solid, we can neglect heat transfer by radiation.
Willis Eschenbach (20:31:04)
i’m trying to establish whether the optimum energy that this hypothetical planet emits is 235w/m2, of its own radiation in the first instance
looking at another analogy, the energy emitted by a 100 watt incandescent lightbulb, that emits heat and light across a wide range of frequencies, but lets just use heat and say that three feet from the bulb, it is IR, and we were to put a globe of aluminium foil around it to prevent convection, and in another simultaneous experiment we were to line the foil at the same distance with black paper or another blackbody material. After two days of constant emission, is the lightbulb hotter than if it were not surrounded by either of these two insulators, or is the interior of each globe at roughly a slighly lower temperature than the recorded energy leaving the surface of the lightbulb?
Willis Eschenbach (00:05:34)
“The Thermos bottle operates exactly like the steel greenhouse in my thought experiment, using a shell surrounding the inner chamber and separated from the chamber by a vacuum. This is why the Thermos is so much more efficient at keeping your coffee from losing heat than an insulated bottle. Half of the heat that is lost from the inner chamber to the shell is radiated back inward, keeping the contents warm and slowing heat loss.”
ok lets take this example as i’m still struggling here with the thermodynamic process. If there were a heat source in this flask emitting energy at a temperature that measured 25C over a period of time, how exactly does the flask increase this energy? If the flask were measured with a thermometer would the entire contents be stable at 25C, or would it increase overtime. Given that the optimum possible source was measured 25C, how can reflected energy increase it? I see how it is done in theory, but nothow it is borne out in practice?
It would be interesting to hear if you’ve done a heat experiment of this nature
To all of those arguing that a cooler object will not radiate towards a warmer object I have an example.
Consider two light bulbs, a 40W and a 100W bulb. If we place them in fixtures that are 1M apart and turn them on, they will both radiate. Even though the 100W bulb is “hotter” than the 40W bulb, the 40W bulb will still radiate on the side that is facing the 100W bulb.
If we consider the same situation using perfect IR bulbs in a vacuum, we can see that it is reasonable that the 40W IR source will still radiate on the side that is facing the 100W IR source. However, if we measure the energy flow through an imaginary plane located between the two sources, we will see that the NET flow between the sources will be towards the 40W “cooler” IR source.
– Charlie K
@ur momisugly Anton Eagle (15:36:25) :
“However, how are 10 shells all 1 inch appart any different than 1 shell that is just 10 times as thick?”
Graeme W (18:59:18) touched on this point in his post. In the case of 1 shell that is 10 times as thick, you have changed the heat transfer from radiation to convection. The effectiveness of a heat shield is proportional to the number of layers, and not their thickness assuming that the layers are isolated from each other. See the Wikipedia article on Multi-Layer Insulation for a better description of how MLI works than I can give here.
http://en.wikipedia.org/wiki/Multi-layer_insulation
Sorry, I don’t know how to make the links in comments, so you’ll probably have to copy & paste the url.
– Charlie K
As a source of heat, I would not like to spend a cold winter’s night under a steel blanket!
Willis
You wrote that you answered all critics but you didn’t mine .
You stopped only at the question of units .
.
1)
The question of units .
The SB law is derived from the Planck law by integrating for all frequencies and OVER A HALF SPHERE
[I (nu,T) dnu domega] where I units are W/m²/sr .
As the sr (solid angle unit) disappeared in the integration , follows that SB law is only valid for convex bodies . The SB constant is also only valid for this case .
Whenever one has to calculate a radiation flux in a specific geometry one has to keep in mind that SB is only valid for half spheres . If I substituted a non convex shell in your model it would give wrong results from A to Z . I will show later why the solid angles are important .
.
2)
Irradiance (units W/m²) “conservation”
This one simply contradicts the laws of physics . As I have shown the irradiance of the shell is P.r²/R² where P (units W/m²) is the irradiance of the inner sphere , r and R corresponding radiuses of the sphere and the shell . Obviously P.r²/R² < P so irradiance (units W/m²) is NOT conserved . In other words the shell has a lower temperature and irradiance than the sphere .
You may say that when r~R then P.r²/R² ~ P but you may not say that irradiance is "conserved" .
Of course when you increase R , like John A is saying , this postulated "conservation" is violated much more . In the limit at R infinite the temperature of the shell is 0 K and it's irradiance 0 !
It doesn't mean that your model is "wrong" but it means that you are forbidden to talk about irradiance "conservation" .
This must be corrected .
3)
The solution of the apparent paradox that the shell receives 240 W/m² but SEEMS to emit 2 times "more" e.g 480 W/m² .
Here one has to recall what I said in 1) . The SB law is only valid for a half sphere . But the problem is that a 2 face object doesn't see only a half sphere , it sees the full sphere . So it will emit in a solid angle of 4Pi instead of only 2 Pi for which the SB law is valid .
So emitting 480 W/m² per 4Pi sr (in the whole space) is perfectly equivalent to emitting 240 W/m² per 2 Pi sr (in a half space) and the SB law is safe . What's more important is that Planck's law
(I units are W/m²/sr) is safe too .
Here I don't contradict your model , I give only a rigorous explanation of the "paradox" that perturbated most of the posters .
4)
Relevance to the atmosphere .
Here it is for me 0 and again for a solid angle problem .
Atmospheric radiation can be assimilated to ponctual sources .
But ponctual sources see the whole space . Moreover they form no convex surfaces . SB law doesn't apply . That should be enough .
Please note that this remark doesn't relate directly to the steel "model" itself . While it has a qualitative feature that LOOKS like one specific atmospheric process , it has no chance to show the dynamics that the atmosphere does .
The atmosphere can't clearly be reduced to a BB equilibrium radiation problem .
But it's true , there is a vague partial qualitative analogy .
.
5)
Synthetically as a physicist I range myself with the opinion of L.Motl .
You did no glaring errors (the irradiance "conservation" is wrong but doesn't impact much because you considered a case where r~R) , there is a qualitative similarity to the GHE but the model will not simulate even partly the atmospheric dynamics .
carrot eater,
You might be right, but here is my mental problem illustrated (very poorly I might add) http://www.ianschumacher.com/img/diagram.png
Imagine instead of photons, or whatever your favorite energy metaphor, we are dealing with energetic particles. Imagine there is a flow of these particles going in a very specific direction (like a water hose kind of thing) and these particles are being directed through a hole in a sphere. My problem is that the density of the energy of the particles (proportional to the pressure, assuming particles are all equal mass) is the same inside as outside.
In order for them to not be the same, we need a special hole. one that lets particles in more easily then it lets them out. I don’t believe such a thing exists in the physical world and as I and others have mentioned, this would be equivalent to Maxwell’s demon. A thought-experiment mischief maker that allows us to violate the second law of thermodynamics.
It is my belief that there is no special filter that can be made for energetic particles of any kind.
Basically you can not get the ‘pressure’ inside higher than the pressure flow of the particles through the hole.
carrot eater,
Just a follow up, you said
“*Actually, this brings a slight clarification: Willis’s earth emits like a perfect blackbody, but it does not absorb like one. Which is perfectly fine.”
I don’t think that is true. I believe at equilibrium emissivity and absorption must be equal. Kirchoff’s Law I think.
Vincent (04:14:30) :
these demonstrations also incur the 2nd law, but at different magnitudes. The soup will lose its heat and the ice will reach the temperature in this lost in space region you refer to, although the change in rates would be related to the pressure. They will both thermalise to near absolute zero. Remove the pressure chamber altogether and throw the soup into space.
Are you arguing that space will reach soup temperature, and soup will reach this modified space temperature? They both thermalise to around the hypothetical absolute zero in a short period of time.
Vincent
“However, some photons of energy from the cooler body will still strike the warmer body.”
I’ll put my eggs in the fridge, and see if they are cooked tomorrow morning.
So many who do not understand positive feedback? Try this:
You put a shell around the planet. Using 256 because that gives nicer numbers.
The shell receives 256 W/m2 and emits 128 from the outside and 128 from the inside.
1st iteration: the surface receives 256+128=384 W and must emit 384 W
The shell now receives 384W and will emit 192 out and 192 in.
2nd iteration: surface receives 256+192=448 W and emits the same, shell emits 224 inward.
3rd iteration: surface receives 256+224=480 W
4th iteration: surface receives 256+240=496 W
So the runs add 128 W, then 64, then 32, 16, 8, 4, 2 …
and after an indefinite number of times this adds up to 256 W i.e the double of the initial.
Willis Eschenbach (00:05:34) :
You skipped over the main advantage of having a vacuum layer in a thermos bottle: the elimination of conduction and convection as possible modes of heat transfer. If all you have is radiation (or conduction leakage out the top; the vacuum can’t go all the way around), the heat transfer is slower.
Willis, thanks for your apparently infinite patience. Some people have really exposed their misconceptions on this post — the strawmen have piled up even after you stated that it’s a simplified “toy” model right at the start (and I can’t see any basic conceptual problems w/it, as it is). Kudos for extending it to a better representation than the original Kiehl/Trenberth budget.
Michael Mann is supposedly a physicist too. 🙂
In reply to: Willis Eschenbach (23:46:55).
On the subject of radiative transfer and transmission within solids…
I thought about this more as I went to sleep last night. Thinking of the example of glass or clear water, I decided that it must happen, because a clear substance is surely capable of letting radiation through. So I am guessing that in an opaque substance the radiative transmission happens but it also gets stopped very quickly – in the same way that it is stopped at the surface of the opaque substance. Now the question is whether my ideas about heat transfer as vibrational are backward. I want to say that that way of thinking is still valid because I was shown Brownian motion in freshman biology. I also think I remember that Einstein studied Brownian motion and made some mathematical calculations concerning them.
John A: Curt and Vincent have answered you very well so I will just make a few more comments here on what you have said.
I don’t see what Maxwell’s Demon has to do with what I said; in fact, if anything it tends to re-enforce what I said as Maxwell’s Demon essentially serves to counter the statistics by making an “intelligent decision” as to which molecules to let through and which to block. Furthermore, I said pretty much the opposite of what you seem to think I said. What I said was that the statement that cold does not flow to hot is just a statistical one, albeit one that becomes completely statistically overwhelming for any sort of macroscopic objects. And, my point is that it applies to the NET heat flow because for any sort of macroscopic objects it becomes ridiculously improbable to have NET heat flow from cold to hot even though there are exchanges of heat going in both directions.
Sure it does. However, even more heat is radiated by the planet to the shell, so the net flow is from the warmer planet to the cooler shell. [What some people get hung up on is how this can then result in the planet being warmer…i.e., if the net flow is from the planet to the shell, why doesn’t the shell make the planet cooler? However, the answer to this is carefully considering the case you are comparing to which is the case where there is no shell at all. In that case, all of the heat that the planet radiates would simply escape out into space. By contrast, when you have the shell there, then some of the heat that the planet radiates (actually, in this simple example where the shell is a perfect blackbody, all of that heat) is absorbed by the shell which subsequently radiates part of it back to the planet. This means the planet will end up warmer than it does in the case where none of the heat that it radiates finds it way back.]
Sure, you are correct that the NET flow of heat is from the soup to the ice. However, that does not mean that there is no heat flow from the ice…just that there is more in the other direction.
There is nothing in Willis’s description that violates the laws of thermodynamics. In particular, all net flows of heat are from warmer objects to colder objects. The only part of your complaint here that is true is that it is not about the fluxes “bouncing backwards”. What happens is that the shell absorbs the radiation from the earth and then because it has a nonzero temperature (determined by the balance of the energy it receives to what it radiates), it itself also radiates. In this regard, here is the webpage of a retired atmospheric science professor who is sort of stickler for describing the greenhouse effect correctly: http://www.ems.psu.edu/~fraser/Bad/BadGreenhouse.html He is a bit militant in what the correct pedagogy is for my taste, but I think he does have a valid point that thinking about it in the wrong way can lead people astray.
Anders L. says:
The answer to that question is that it depends on how many Watts of power it takes to provide enough energy to raise temperatures back up to the point where radiative balance is restored. And, this means that one has to calculate what is called the “thermal inertia” or “thermal capacity” of the system. The atmosphere itself does not have that large a thermal inertia, but the oceans do…In fact, just the top ~2.5m of the oceans have as much thermal inertia as the entire atmosphere. (There is also some thermal inertia in the land but not as much as in the oceans.) So, it is the oceans that really slow down the re-establishment of radiative balance.
The oceans are complicated and can be thought of as having a mixed layer on top, which is generally about 100m in depth (but can vary quite a bit) that is in pretty good contact with the atmosphere. The deep ocean below that tends to have only slow exchange of heat with the mixed layer. How quickly most of the temperature rise occurs is pretty sensitive to assumptions regarding the deep ocean and its communication with the mixed layer. If the deep ocean were completely isolated so you just had to heat the mixed layer, you’d have an approximately exponential response with a the time scale for the of ~5-10 years. The effect of the deep ocean is to lengthen the time for re-adjustment and also make the approach to the new radiative balance more non-exponential.
Joel since you’re explaining it in absence of Willis, its still confounding, given that the source of emission – a black body planet surrounded by a black steel enclosure – if the optimum radiation is 235w/m2, then surely the optimum exchange will be no more than 235w/m2, since that is the optimum energy in the system. This could surely be verified by experimental procedure, since it seems to be universally agreed by consensus that 235w/m2 produces a temperature of -19C.
Thats not an impossible temperature to reproduce in a closed experiment.
Ian Schumacher (06:47:03) :
“I don’t think that is true. I believe at equilibrium emissivity and absorption must be equal. Kirchoff’s Law I think.”
Be careful. We have been to some extent been glossing over this, but emissivity and absorptivity are functions of wavelength, and even direction. Because such spectral dependence complicates matters, undergraduate classes generally don’t concentrate on them.
With some generality, you can say that at a given wavelength, the emissivity and absorptivity are equal. But remember that emissivity and absorptivity themselves can vary with wavelength. For the earth, we are looking at very different bands of wavelength: the earth is absorbing or reflecting visible light, while it is emitting infrared.
lgl (06:56:12)
its assumed that the emission is a constant process, so that at some stage the shell will always be receiving and emitting 256w/m2. What happens in reality? They go into a stable state of thermal equilibrium where there is no change of energy flow and both remain constant at the temperature that accords with 256w/m2.
Ian Schumacher:
I think my last comment might be a bit confusing, so let me put it thus:
Say the earth has an absorptivity around 0.7 in the visible wavelengths. That does not tell you anything about the absorptivity in the infrared wavelengths. Therefore, it also does not tell you anything about the emissivity in the infrared wavelengths. Thus, there is no problem in using an emissivity of 1 for the relevant band of infrared. If the actual emissivity in the appropriate IR band is measured to be 0.97 to 0.99 ( Willis Eschenbach (10:41:34) : ), well, for the purposes of a tinkertoy model, using 1 is good enough.