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The greenhouse effect and the 2nd law of thermodynamics

What the science says...

Select a level... Basic Intermediate

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

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Further reading

  • Most textbooks on climate or atmospheric physics describe the greenhouse effect, and you can easily find these in a university library. Some examples include:
  • The Greenhouse Effect, part of a module on "Cycles of the Earth and Atmosphere" provided for teachers by the University Corporation for Atmospheric Research (UCAR).
  • What is the greenhouse effect?, part of a FAQ provided by the European Environment Agency.

References

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Comments 976 to 1000 out of 1393:

  1. As I posted my last contribution, I noticed that we are to be favoured by another post from SOD. I debated back radiation theories with someone I believed to be SOD in Eli Rabett’s blog, some time ago. Quoting from memory, SOD based his argument on the conservation of Energy (“where does the energy go”?) He freely admitted that, for him, entropy “never seemed real” and its formula deltaS = DeltaQ/delta T did not make sense (which, since that version is wildly wrong, is not surprising). No-one should comment on AGW (or anything else, in my opinion) without a reasonable grasp of the second law of thermodynamics. For any spontaneous transaction, anywhere in the known universe, the change in entropy must be positive. Now if a quantity of energy deltaQ leaves the atmosphere at a temperature Ta, the change in entropy is -deltaQ/Ta.If it is absorbed by the surface, at a temperature Ts, the change in entropy is +deltaQ/Ts. Now, since Ta is less than Ts (the lapse rate again), the combined change in entropy is negative, which is forbidden. If it happens on the microscopic level (vibrating molecules in a steel bar, heated at one end, or photons from a cold plate to a hot plate), it must be accompanied by a greater transfer in the conventional direction, hot to cold. It cannot be considered or analysed in isolation, which would take us immediately to a search for perpetual motion, as G and T point out at excessive length. By transferring from a hot source to a cold sink, energy becomes heat which can produce work. No net change is possible in the opposite direction without a heat engine. So to answer the question posed in the introduction, have climate scientists made an elementary mistake in explaining AGW? Of course not. Have most of the bloggers, politicians and journalists? Undoubtedly.
  2. 976 Staples SoD seems quite comfortable with Entropy.
  3. Fred@975 It seems to me that the source of the disagreement here is that you view backradiation as a separate theory, rather than as a component part of the "top of the atmosphere" explanation. AFAICS this is incorrect, it is backradiation that causes the surface to warm until equilibrium is re-established following an increase in the height of the emitting layer at the top of the atmosphere.
  4. Les @977, an excellent read, and profitable for anyone following this debate. I particularly liked propater's comment:
    "If you take ScienceOfDoom example 3 and reverse it to consider radiation emmited by the atmosphere and absorbed by the earth’s surface, you can see that entropy is lost when the radiation from the atmosphere is absorbed by the hotter surface : Atmosphere: δS1 = -(390-301)/270 = -89 / 270 = -0.33 J/K Surface: δS2 = (390-301)/288 = 89 / 288 = 0.31 J/K This is right, net entropy equals -0.02J/K (as you say, “radiation quality” has increased since that energy will be re-emitted at a shorter wavelength) Entropy is reduced. This looks wrong. And it is! Because in that calculation you forgot one term: the radiation from TOA to space : If we suppose radiation from space to be j* = sigma * T^4 = 5.67*10^-8 * 3^4 = 4.59*10^-6 W/m² Let’s consider the entropy budget for one square meter for a second : Atmosphere: δS1 = -(390-0.00000459)/270 ~= -390/ 270 = -1.44 J/K Space: δS2 = (390-0.00000459)/3 ~= 390/ 3= 130 J/K Net entropy production from TOA-Space exchanges : 128.56 J/K Net entropy production from atmosphere exchange with the earth’s surface and space : 128.56 J/K – 0.02 J/K = 128.54 J/K. When you take all terms into account, no entropy is lost (even though some of it gets moved away.) In fact you see a massive amount of entropy created and, as Nick Stokes points out, it is the place where the entropy production is the largest (Sun-earth ~ 10 J/K.s.m², earth-atmosphere ~ 10^-2 J/Ksm² , atmosphere-outer space ~ 10² J/Ksm² )"
    What more could we ask for - an actual worked example proving that back radiation does not violate the 2nd law of thermodynamics. Given this example, and the ease of the mathematics, I think it should be a minimum denier wanting to mount the tired old 2nd law argument that they provide such a worked example of their model of the greenhouse effect. It should make the arguments much shorter and to the point, and provided an easy way to keep score of who is correct, and who is simply full of it.
  5. Fred Staples @976, I believe Les has adequately rebutted your comments about SOD.
    For any spontaneous transaction, anywhere in the known universe, the change in entropy must be positive.
    This is incorrect. The Earth is infested with a spontaneous transaction which results in the formation of very low entropy products. It is called life. Of course, if you look at all energy flows involved in life, then the net entropy of the whole system increases, even though that in the living system decreases temporarily. So, only 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.
    "Now if a quantity of energy deltaQ leaves the atmosphere at a temperature Ta, ..."
    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.
  6. 979 Tom - "What more could we ask for..." I wait, with trepidation, to find out! To me, it all seems so straightforward. Still, just because this discussion has certainly not gone on long enough - nor has it been repeated nearly sufficiently enough times on nearly enough blogs - can I suggest a further place for doubt? It turns out that in many multiverse models it is not clear that conservation of energy holds for the whole universe nor, indeed, for every universe. As such, I'm guessing I admit, it should be possible to have infinitely many universes in which the 2nd law doesn't hold, or possibly doesn't hold only for those gases which don't transmit/radiate/absorb evenly at all wavelengths or something. see, always room for doubt.
  7. Tom Curtis 966 and KR Tom Curtis you said: “First, anybody who has read anything that I have written knows I am not given to knee jerk responses from my writing style alone, even if they do not understand the content.” Ok Tom, I have to admit, for the most part you are well considered;...though still wrong. So knee jerk was not appropriate, rather I should have said dismissive response. Regarding your arguments to my post 924. You set up hallow challenges to my equations. For example, your parsing of emissivity between SW and LW is irrelevant. Why irrelevant you ask...simple, I do not conflate the two. For example, lets compare the equations I used to the simplified models equations via slide 14 of Jin-Yi Yu's lecture. From Jin-Yi Yu's lecture: For Earth's surface S/4 * (1-A) +σ TA4=σ TS4 For Atmosphere σ TS4=2 σ TA4 (blackbody) Compare these equations to my Blackbody table. Notice the similarities. Blackbody
    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:
    Tom you said: a) You have not established the appropriate groundwork, and are instead working on a host of demonstrably false assumptions. b) Your tables which carry your argument have unclear symbols, and are derived by an unexplained method. In other words, they are simply bare assertions. KR you said: Please, L.J.Ryan - read some of the references you have equation-mined. Learn a bit more about the science. You're approaching the issue with a lot of erroneous preconceptions, and those are leading you to incorrect conclusions. Do some reading, L.J. Ryan, including the sources you yourself have linked to.

    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/m2 be therein...when added to albedo reflection achieves equilibrium. According to blackbody emission curves ~1200K nets ~235 W/m2 at 700 nm. White
    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:Gray
    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>
  8. L.J. Ryan A gray body with a LW emissivity of 0.618 (as you postulated above), radiating 240 W/m^2, will have a temperature calculated by the Stefan-Bolzmann equation of: P = SB const. * emissivity * T^4; hence T = [ P / (SB const. * emissivity) ] ^0.25 [ 240 / (5.6704*10^-8 * 0.618) ] ^0.25 = 287.675°K, or 14.53°C 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 simply in error, as both Tom and I have noted in the past.
  9. LJ, You have again neglected to explain the derivation of your tables, none of the equations you cited list Teq 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.
  10. KR 984 You are calculating the temperature of the atmosphere not the earths surface. Do you agree with my table showing blackbody atmosphere? Do you agree with the equation via Jin-Yi Yu's lecture...the exact equation, you suggested I consider? You also contradict your own words;"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."
  11. e 984 Teq is defined as the time it will take the earth to accumulate 235 W/m^2 and therefore radiate 235 W/m^2. The white atmosphere will reach Teg first...thus defining the interval.
  12. L.J. Ryan - The effective emissivity is 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. The effective 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.
  13. LJ, Your response described what 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 3Teq, you get a flux of 470 Wm-1. Can you please show us the mathematical equation you 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.
  14. KR 964 you said: If 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?
  15. e 988 I will post the equations this evening...I'm pressed for time at the moment. But quickly, it's rather a simple ratio of energy accumulation based on ε= 0 energy calculations. you said: "temperature does not increase past equilibrium time." Yes I agree. My rather simple spread sheet did not include comparative functions.
  16. LJ, BTW please do not post every single equation for every single value in that spreadsheet. Just show us the general equations you are using to derive it. >you 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 Teq in your calculations suggests that you are doing something very very wrong.
  17. LJ >it'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 in total emissivity across 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 meant spectral emissivity as 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.
  18. LJRyan @982, thank you for the greater clarity about the source of your calculations. From your comments, I understand you to be basing your calculations on the following model: Science of Doom discusses just such a model in a recent post, and has this to say:
    So in this first model, which is very common in introductory books on atmospheric physics, three things are assumed - and none of them are true: * the atmosphere is isothermal – a slab of atmosphere all at the same temperature * the atmosphere is completely transparent to solar radiation * the atmosphere is completely opaque to terrestrial radiation
    The reason such simple but factually false models are used in introductory courses is the same reason frictionless surfaces are used in simple models of dynamics - it allows the introduction of important concepts without unnecessary complications. Because they do allow the introduction of important concepts, I am happy to work with such models so 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 absorbs all outgoing 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).
  19. Where I said second assumption, I should have said third. Sorry for any confusion.
  20. With regard to the slab model illustrated @993, and with (1-A)*S/4 = 240 w/m^2, then: The entropy of the incoming solar radiation per meter squared per second = 240 J/6000 K = 0.04 J/K. The entropy of the surface radiation = 480 J/303 K = 1.58 J/K 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. The challenge for the deniers is to find any partition of the system such that conservation of energy is maintained for that partition, and such that the Entropy decreases for that partition. That is, the partition must show an energy flow from E1 to E2 such that E1 = E2, but such that the Entropy of E1 is greater than that of E2. As an example, we have: 1) Insolation + Back radiation => surface radiation which in terms of energy (per meter squared per second) is 1') 240 J + 240 J = 480 J (so we have conservation of energy); but in terms of entropy we have: 1") 0.04 J/K + 0.94 J/K < 1.58 J/K so there is no violation of the 2nd law in this partition. There is in fact no partition satisfying these conditions in which the energy in has a higher entropy than the energy out. Therefore, the 2nd law of thermodynamics is not violated by this model.
  21. e 992 Understand e, it is my contention energy, radiative or otherwise, can NOT increase itself due to reflection, re-radiation or insulation. So when the input flux to the surface equals surface output flux the earth is at equilibrium...255K. The 33o 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/m2 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/m2 SW absorbed by the surface. Nevertheless, the number workout nearly the same. For consistency within this explanation, 235 W/m2 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/m2 SW to the surface equals surface output flux 235 W/m2 LW the earth SURFACE is at equilibrium (not the entire system). It is this instant which is designated Teq. 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/m2 SW E= earths radiated flux A=atmosphere flux LW AU = atmosphere up AD = atmosphere down A=AD+AU With ε= 0 Teq when E=I=A 1.5Teq E'= I+A and E'=A' 1.75Teq E''= I +A' and E''=A'' 1.875Teq E'''=I + A'' and E'''=A''' Assuming energy accumulation is linear, the time to reach 1.5Teqis half as long as time to reach Teq...1.75Teqis half as long as time to reach 1.5TeqfromTeq...etc. These calculation were continued some 20 times, as a geometric series 2Teq 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: 2Teq E= I+AD and E=A 3Teq E'= I +AD' 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: AR = .388 A = atmosphere reflected AD=AU AD= .5 (.612A)= .306 A E= I+AR+AD and E=A E'= I+AR'+AD'and E'=A' The Teq is found by dividing A for white atmosphere of a particular surface radiation by AR+AD of the same E (gray), then multiplying by the Teq for that white E interval. For example; for the 470 W/m2 interval: A (white) =470, AR+AD=182+144....470/( 182+144)*1.5Teq=2.16Teq. The Teq provides only a non critical time component of surface equilibrium. In hindsight, I should not have included this information, it only confused my conclusions.
  22. Tom Curtis 993 995 I read the SoD article and in fact was going to submit comments until the conclusions you list. That is, the model is wrong...but kinda works. I'll wait to see where SoD takes this simplified model stuff, before commenting. You said: "Your first table is better, but does not include a column for , 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 constant input (1-A)*S/4 is assumed...but I get your point. Teq 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.
  23. LJRyan @997, the atmosphere has an upper side and a lower side. Because thermal radiation is the same in all directions, if the atmosphere were 303 K, then it would radiate 480 w/m^2 up, and the same down. That would violate conservation of energy. However, with an atmosphere at 255 K, it will radiate 240 w/m^2 to space and 240 w/m^2 towards the surface. It follows that there is no violation of conservationof energy, and I did not double count.
  24. LJ, Thank you for taking the time to detail and formalize your thinking. There are several things wrong with your logic (none of which follow from standard GHG physics): One big issue is you are incorrectly defining the time to equilibrium. You wrote here: When 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. Nothing in the system will be in equilibrium until the net output of the system matches the net input 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 AU = 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.
  25. I agree, Les,977,a vast improvement but a much later, and much up-dated, post than the one I quoted. Though much has gone, much remains from the original – “where does the back-radiation energy go”. It is obvious that there is no second law entropy (quality) problem if we consider only the transfers sun to earth, earth to space, and define the whole solar system as our closed system. Sadly, this ignores completely the transfers we are interested in - earth to atmosphere and back again. The most fundamental point is that you cannot consider the out and back long wave energy transfers in isolation. It is the net transfer (heat transfer) that counts. Petty, page 6, is correct and the version of Trenberth quoted here is grossly misleading. It is not just a matter of using the difference, as someone posted. We can follow G and T, and use heat engines to make the point. It is always possible in principle to convert an energy flow from a warmer source to a colder sink into work (first law). In the process the energy degrades (second law). It is never possible to do the same with an energy flow from a colder sink to a warmer source. If it were possible we would have perpetual motion. It is the net flow that counts (out minus back). The back radiation is the negative term in the Stefan Bolzmann equation. As to the “higher is colder” mechanism, 978,it has nothing to do with back-radiation. Atmospheric emission must balance incoming solar energy, and will be at an appropriate “goldilocks” temperature and elevation to make this happen. If increasing CO2 concentration elevates the emission point (for the sake of the argument) outgoing radiation will be reduced. Incoming radiation will remain the same, so the whole system warms up. The lapse rate moves to the right, and the surface warms.

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