# The greenhouse effect and the 2nd law of thermodynamics

## What the science says...

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The 2nd law of thermodynamics is consistent with the greenhouse effect which is directly observed. |

## Climate Myth...

2nd law of thermodynamics contradicts greenhouse theory

"The atmospheric greenhouse effect, an idea that many authors trace back to the traditional works of Fourier 1824, Tyndall 1861, and Arrhenius 1896, and which is still supported in global climatology, essentially describes a fictitious mechanism, in which a planetary atmosphere acts as a heat pump driven by an environment that is radiatively interacting with but radiatively equilibrated to the atmospheric system. According to the second law of thermodynamics such a planetary machine can never exist." (Gerhard Gerlich)

Skeptics sometimes claim that the explanation for global warming contradicts the second law of thermodynamics. But does it? To answer that, first, we need to know how global warming works. Then, we need to know what the second law of thermodynamics is, and how it applies to global warming. Global warming, in a nutshell, works like this:

The sun warms the Earth. The Earth and its atmosphere radiate heat away into space. They radiate most of the heat that is received from the sun, so the average temperature of the Earth stays more or less constant. Greenhouse gases trap some of the escaping heat closer to the Earth's surface, making it harder for it to shed that heat, so the Earth warms up in order to radiate the heat more effectively. So the greenhouse gases make the Earth warmer - like a blanket conserving body heat - and *voila*, you have global warming. See What is Global Warming and the Greenhouse Effect for a more detailed explanation.

The second law of thermodynamics has been stated in many ways. For us, Rudolf Clausius said it best:

"Heat generally cannot flow spontaneously from a material at lower temperature to a material at higher temperature."

So if you put something hot next to something cold, the hot thing won't get hotter, and the cold thing won't get colder. That's so obvious that it hardly needs a scientist to say it, we know this from our daily lives. If you put an ice-cube into your drink, the drink doesn't boil!

The skeptic tells us that, because the air, including the greenhouse gasses, is cooler than the surface of the Earth, it cannot warm the Earth. If it did, they say, that means heat would have to flow from cold to hot, in apparent violation of the second law of thermodynamics.

So have climate scientists made an elementary mistake? Of course not! The skeptic is ignoring the fact that the Earth is being warmed by the sun, which makes all the difference.

To see why, consider that blanket that keeps you warm. If your skin feels cold, wrapping yourself in a blanket can make you warmer. Why? Because your body is generating heat, and that heat is escaping from your body into the environment. When you wrap yourself in a blanket, the loss of heat is reduced, some is retained at the surface of your body, and you warm up. You get warmer because the heat that your body is generating cannot escape as fast as before.

If you put the blanket on a tailors dummy, which does not generate heat, it will have no effect. The dummy will not spontaneously get warmer. That's obvious too!

Is using a blanket an accurate model for global warming by greenhouse gases? Certainly there are differences in how the heat is created and lost, and our body can produce varying amounts of heat, unlike the near-constant heat we receive from the sun. But as far as the second law of thermodynamics goes, where we are only talking about the flow of heat, the comparison is good. The second law says nothing about how the heat is produced, only about how it flows between things.

To summarise: Heat from the sun warms the Earth, as heat from your body keeps you warm. The Earth loses heat to space, and your body loses heat to the environment. Greenhouse gases slow down the rate of heat-loss from the surface of the Earth, like a blanket that slows down the rate at which your body loses heat. The result is the same in both cases, the surface of the Earth, or of your body, gets warmer.

So global warming does not violate the second law of thermodynamics. And if someone tells you otherwise, just remember that you're a warm human being, and certainly nobody's dummy.

Basic rebuttal written by Tony Wildish

**Update July 2015:**

Here is the relevant lecture-video from Denial101x - Making Sense of Climate Science Denial

**Update October 2017 :**

Here is a walk-through explanation of the Greenhouse Effect for bunnies, by none other than Eli, over at Rabbit Run.

Last updated on 7 October 2017 by skeptickev. View Archives

Fred Staplesat 21:16 PM on 7 April, 2011lesat 21:31 PM on 7 April, 2011Dikran Marsupialat 21:35 PM on 7 April, 2011Tom Curtisat 22:24 PM on 7 April, 2011Tom Curtisat 22:42 PM on 7 April, 2011only when we account for all energy flows within a boundary can we apply the 2nd law to the system within the boundary, and then it applies only to the entropy of the whole system, not to any arbitrary subpart of the system.As the worked example quoted above shows, when all energy flows are accounted for, the interchange of energy between Earth and atmosphere including back radiation involves a net gain in entropy, and hence no second law violation. And the second law applies only to the whole system, not to any arbitrary subpart - say, the backradiation alone. Plainly your accounting considers only one energy flow within the system, not all of them. Therefore it is an attempt to apply the second law to an arbitrary subpart of the system rather than to the system as a whole, which is as explained above, a misapplication of the second law.lesat 22:45 PM on 7 April, 2011L.J. Ryanat 04:31 AM on 9 April, 2011^{S}/_{4}* (1-A) +σ T_{A}^{4}=σ T_{S}^{4}For Atmosphere σ T_{S}^{4}=2 σ T_{A}^{4}(blackbody)Compare these equations to my Blackbody table. Notice the similarities.The highlighted row confirms Jin-Yi Yu conclusion. The other rows simply evaluate this rightful equation prior to and following equilibrium. Why examine beyond equilibrium...simply as a point of comparison. Before I school you further, indulge me with a couple self-serving quotes:

Do you favor your crow warm or cold? Back on point, the white and gray tables explore the same base equation save a change in emissivity. An atmosphere perfectly reflective to LW radiation will, according to GHG physics, get hotter faster as compared to ε= 1. The key question is, at what surface flux will SW emissions of 235 W/m

^{2}be therein...when added to albedo reflection achieves equilibrium. According to blackbody emission curves ~1200K nets ~235 W/m^{2}at 700 nm.Intuitively, these value make sense...(if you subscribe to GHG physics). With an atmosphere which reflects all LW radiation back to the surface, surface energy accumulation with respect to time is geometric. Lastly, a gray atmosphere with ε= .618 will, (according to GHG physics) increase equilibrium temperature. As KR said:

As stated before, given a known amount of outgoing radiation, the black body temperature is an absolute minimum on the temperature of an equivalently radiating graybody, due to the relationship of emissivity and temperature.So, what is actually borne out by the math. Assuming a gray body reflectivity = .388 the following table results:Look KR is right! A gray body emissivity does confer a higher temperature....

341K. Now, all we have to do is get those thermometers to fall in line with GHG physics.Response:[mc] Please limit width to 500 when posting images: <IMG SRC="" width=500>KRat 04:47 AM on 9 April, 2011P = SB const. * emissivity * T^4; hence T = [ P / (SB const. * emissivity) ] ^0.25 [ 240 / (5.6704*10^-8 * 0.618) ] ^0.25 =That's 287.675°K required to radiate that power, not 341°K, as you claim. I haven't bothered to track your math and logical errors down in this case - I don't consider it worth my time. You are quite simply287.675°K, or 14.53°Cin error, as both Tom and I have noted in the past.eat 05:07 AM on 9 April, 2011_{eq}as a parameter. At equilibrium, by definition, there is no longer any net accumulation of energy or temperature change, so your results are nonsensical as given.L.J. Ryanat 05:12 AM on 9 April, 2011L.J. Ryanat 05:21 AM on 9 April, 2011KRat 05:25 AM on 9 April, 2011effective emissivityis from the surface to space, as has been repeatedly stated. The actual radiation point (depending on the wavelength) may be from upper tropospheric CO2, lower tropospheric H2O, or directly from the surface; determining that on a per-wavelength basis is the realm of atmospheric modelling. Theeffective emissivity, however, is the sum effect on power as radiated from the surface, through a GHG atmosphere, to space, relative to a black body at surface temperature.Please cease to misinterpret clearly defined terms. Particularly if you are using those misinterpretations to claim I'm contradicting myself - which I am not. Can you describe on what basis you disagree with the Stefan-Boltzmann equation? There could be a Nobel prize involved if you can. Your computations, again, reflect your incorrect assumptions and calculations. You've shown no interest in correcting said mistakes - I don't consider it a good use of my time to beat that dead horse.eat 05:46 AM on 9 April, 2011describedwhat you are trying to show qualitatively but does not tell me how you derived your results mathematically. For example for emissivity of 1 and time of 3T_{eq}, you get a flux of 470 Wm^{-1}. Can you please show us themathematical equationyou used to output the number 470? It should somehow have time as an input and energy flux or temperature as an output. None of the equations you cited so far do this. Also you didn't comment on my note that, by definition, temperature does not increase past equilibrium time. The fact that it does so in your results shows you are making a serious error.L.J. Ryanat 07:30 AM on 9 April, 2011If you feel that integrated spectra are not measurements, well, then, I'll pass that on to the spectroscope manufacturers I work with.Don't forget, when chortling with your work chums,"Effective emissivity" is a useful summary of surface temperatures, atmospheric temperatures, band blocking, emission altitudes, and the lapse rate.Ah...what knee slapping fun. As I said, ee is calculated...calculated explicitly to obfuscate GHG physics. Also, if effective ε= .618 and 1- ε= reflectivity, then .382 of incident power is effectively reflected. Or maybe, working beyond the gray slab model, the atmosphere is partially transparent...like a crystal. A crystal that DOUBLES the light flux which transverses it's body. Now tell me KR, if an enterprising fellow was to manufacture crystal with REAL (not effective) ε= .618, will it also double light input? Or is energy doubling process contained to GHG physics alone?L.J. Ryanat 07:30 AM on 9 April, 2011eat 08:13 AM on 9 April, 2011you said: "temperature does not increase past equilibrium time." Yes I agree.Your table clearly shows temperature increasing past equilibrium time. Since temperature / flux does not increase past equilibrium time, the fact that you are including T_{eq}in your calculations suggests that you are doing something very very wrong.eat 09:05 AM on 9 April, 2011it's rather a simple ratio of energy accumulation based on ε= 0 energy calculations.I'm sorry but your ε= 0 energy calculations also do not make any sense. If you meant emissivity as intotal emissivityacross all wavelengths, then ε= 0 would mean that no energy is absorbed or emitted. Your temperature and flux would be 0 across the board into infinite time. If you meantspectral emissivityas in the emissivity in the LW is 0, while other wavelengths have ε>0, then that just means the surface will get hot enough until the power is emitted in wavelengths other than LW. It would certainly not accumulate infinitely as you have depicted. Again, without even posting your derivations it is obvious something is very very wrong with your math.Tom Curtisat 12:24 PM on 9 April, 2011so long as it is clearly understood that they do not represent any actual state in the universe. Attempts to calculate the surface temperature of the Earth are bound to be in error (and yours more in error than if the model was used correctly), and therefore are pointless except exposition of the relevant concepts. That is, they are not actual predictions of the real surface temperature. For a calculation of the Earth's average surface temperature using the simplest model of the GHE that is at the same time reasonably accurate to the actual physics, I refer you to equation six in my post 944 above. Having said that, if we wish to use the single slab model above for exposition, we need to carefully adhere to the assumptions of the model. Otherwise we just produce another example of "Garbage In - Garbage Out", and there are more than enough examples of that on the internet. Of course, the second assumption of the model is that it absorbsalloutgoing radiation. Therefore any attempt to include a reflecting atmosphere with this model constitued GIGO, and in particular, this is true of your second and third tables. Your first table is better, but does not include a column for (1-A)*S/4, an important term without which the equilibrium state cannot be determined. Equilibrium is reached when (and only when) (1-A)*S/4 = σTa^4, ie, AtmU in your first table. The failure to include a term for insolation probably explains why your terms do not stabilize over a certain time period (unless I am misinterpreting your first column).Tom Curtisat 12:26 PM on 9 April, 2011Tom Curtisat 13:24 PM on 9 April, 2011L.J. Ryanat 13:52 PM on 11 April, 2011^{o}delta is do to non-radiative energy input. My calculations set out to disprove back radiation by using the very tenets of GHG physics. That is, by using proponents equations to confound GHG physics conclusions, I will prove my supposition. Start with a solar input 240 W/m^{2}SW. Due to my confusion of KR take on albedo and earths surface emissivity/absorptivity I applied .98 absorption a second time...netting 235 W/m^{2}SW absorbed by the surface. Nevertheless, the number workout nearly the same. For consistency within this explanation, 235 W/m^{2}SW input will be used. All of the following presumes GHG physics: With the understanding a white atmosphere will reflect all terrestrial LW, the white atmosphere will “force” the surface to accumulate energy faster then any other emissivities. When the input flux 235 W/m^{2}SW to the surface equals surface output flux 235 W/m^{2}LW the earth SURFACE is at equilibrium (not the entire system). It is this instant which is designated T_{eq}. Since the subsequent instance receives more energy then the prior instance, the accumulation happens faster, followed by an ever faster instance...ect. That said; I= solar input 235 W/m^{2}SW E= earths radiated flux A=atmosphere flux LW A_{U}= atmosphere up A_{D}= atmosphere down A=A_{D}+A_{U}With ε= 0 T_{eq}when E=I=A 1.5T_{eq}E'= I+A and E'=A' 1.75T_{eq}E''= I +A' and E''=A'' 1.875T_{eq}E'''=I + A'' and E'''=A''' Assuming energy accumulation is linear, the time to reach 1.5T_{eq}is half as long as time to reach T_{eq}...1.75T_{eq}is half as long as time to reach 1.5T_{eq}fromT_{eq}...etc. These calculation were continued some 20 times, as a geometric series 2T_{eq}will never be reached. At the same time, surface flux will accumulate infinitum. Now, of course the visible spectrum will be emitted prior to runaway thereby achieving TOA equilibrium. When ε= 1 the equation are: 2T_{eq}E= I+A_{D}and E=A 3T_{eq}E'= I +A_{D}' and E'=A' The time it takes E=A=I, is twice as long as when ε= 0. When ε= .612 reflectivity equals .388 . This introduces another value: A_{R}= .388 A = atmosphere reflected A_{D}=A_{U}A_{D}= .5 (.612A)= .306 A E= I+A_{R}+A_{D}and E=A E'= I+A_{R}'+A_{D}'and E'=A' The T_{eq}is found by dividing A for white atmosphere of a particular surface radiation by A_{R}+A_{D}of the same E (gray), then multiplying by the T_{eq}for that white E interval. For example; for the 470 W/m^{2}interval: A (white) =470, A_{R}+A_{D}=182+144....470/( 182+144)*1.5T_{eq}=2.16T_{eq}. The T_{eq}provides only a non critical time component of surface equilibrium. In hindsight, I should not have included this information, it only confused my conclusions.L.J. Ryanat 14:29 PM on 11 April, 2011_{eq}is, as I explained to e in the previous post, a time interval to mark surface equilibrium. That is, when SW flux to the surface equals LW flux out of the surface. Regarding your 995, I think you have double counted atmosphere radiation. Specifically: The entropy of the back radiation (AtmD) = 240 J/255K = 0.94 J/K The entropy of the radiation to space (AtmU) = 240 J/255 K = 0.94 J/K. Total atmospheric radiation 480J...confers 303K which is obviously wrong and/or it unbalances your conservation of energy equations.Tom Curtisat 02:23 AM on 12 April, 2011eat 08:57 AM on 12 April, 2011When the input flux 235 W/m2 SW to the surface equals surface output flux 235 W/m2 LW the earth SURFACE is at equilibrium (not the entire system)This is incorrect. Equilibrium only exists when there are no longer any temperature changes. The fact that the surface is emitting the same flux as the overall system input is not particularly relevant in this example, since equilibrium is not reached at that point.Nothingin the system will be in equilibrium until thenetoutput of the system matches thenetinput of the system. That should be intuitively easy to understand, if input and output are different, then energy will either be accumulating or depleting. Nothing is changing if input and output are the same. Using your terms, you are looking for the point in time when A_{U}= I. The other problem is that you are mixing up the effective emissivity model with a simple grey slab model as depicted in Prof Yu's lecture. These two models are describing the system in very different ways, and cannot be mixed and matched in the simple fashion you are attempting. The grey slab model assumes that each component in the system is a blackbody (among other assumptions) and describes the flow of energy between individual components in the system. The effective emissivity model completely abstracts away the internal exchanges of energy that are illustrated in the grey slab model. When someone says that the effective emissivity of the earth is .612, the value .612 already captures all the internal behaviors of the system. It applies to the system as a whole, it would not make any sense to insert this back into the grey slab model as the emissivity of the atmosphere. I would suggest starting by clarifying exactly which model you are using to draw your conclusions. If you're using effective emissivity, then it is simply a matter of plugging the emissivity into the Stefan-Boltzmann law and computing the temperature that results. If you are using the grey slab model, then you would calculate the result as given by the equations in professor Yu's lecture slides.Fred Staplesat 22:05 PM on 14 April, 2011