Guest Post by Tom Vonk
In a recent post I considered the question in the title. You may see it here : http://wattsupwiththat.com/2010/08/05/co2-heats-the-atmosphere-a-counter-view/
The post generated great deal of interest and many comments.
Even if most of the posters understood the argument and I answered the comments of those who did not, I have been asked to sum up the discussion.
Before starting, I will repeat the statement that I wished to examine.
“Given a gas mixture of CO₂ and N₂ in Local Thermodynamic Equilibrium (LTE) and submitted to infrared radiation, does the CO₂ heat the N₂?”
To begin, we must be really sure that we understood not only what is contained in the question but especially what is NOT contained in it.
- The question contains no assumption about the radiation. Most importantly there is no assumption whether a radiative equilibrium does or does not exist. Therefore the answer will be independent from assumptions concerning radiative equilibrium. Similarly all questions and developments concerning radiative transfer are irrelevant to the question.
- The question contains no assumption about the size or the geometry of the mixture. It may be a cube with a volume of 1 mm³ or a column of 10 km height. As long as the mixture is in LTE, any size and any geometry works.
- The question contains no assumption about boundary conditions. Such assumptions would indeed be necessary if we asked much more ambitious questions like what happens at boundaries where no LTE exists and which may be constituted of solids or liquids. However we do not ask such ambitious questions.
Also it is necessary to be perfectly clear about what “X heats Y” means.
It means that there exists a mechanism transferring net (e.g non zero) energy unidirectionaly
from X to Y .
Perhaps as importantly, and some posters did not understand this point, the statement
“X heats Y” is equivalent to the statement “Y cannot cool X”.
The critical posts – and here we exclude posts developing questions of radiative transfer which are irrelevant as explained in 1) above – were of 2 types.
Type 1
The argument says “LTE never exists or alternatively LTE does not apply to a mixture of CO₂ and N₂.”
The answer to the first variant is that LTE exists and I repeat the definition from the original post : “A volume of gas is in LTE if for every point of this volume there exists a neighborhood in which the gas is in thermodynamic equilibrium (TE)”
2 remarks to this definition:
- It is not said and it is not important how large this neighborhood of every point is. It may be a cube of 1 mm³ or a cube of 10 m³ . The important part is that this neighborhood exists (almost) everywhere.
- LTE is necessary to define local temperature. Saying that LTE never exists is equivalent to saying that local temperatures never exist.
The second variant admits that LTE exists but suggests that a mixture of CO₂ and N₂ cannot be in LTE.
The LTE conditions are given when energy at every point is efficiently spread out among all available degrees of freedom (translation, rotation, vibration).
The most efficient tool for energy spreading are molecular collisions.
Without going in a mathematical development (see statistical thermodynamics for those interested), it is obvious that LTE will exist when there are many molecular collisions per volume unit.
This depends mostly on density – high density gases will be often in LTE while very low density gases will not.
For those not yet convinced, hold out a thermometer in your bedroom and it is probable that it will show a well defined temperature everywhere – your bedroom is in LTE .
We deal here with a mixture of CO₂ and N₂ in conditions of the troposphere which are precisely conditions where LTE exists too.
Type 2
The argument says “The mean time between collisions is much shorter than the mean decay time (e.g time necessary to emit a photon) and therefore all infrared energy absorbed by the CO₂ molecules is immediately and unidirectionaly transferred to the N₂ molecules.”
In simple words – the CO₂ never has time to emit any IR photons because it loses vibrational energy by collisions instead.
This statement is indeed equivalent to the statement “CO₂ heats N₂”.
Now let us examine the above figure.
The good understanding of this figure will do much better than only answering the original question. It will also make clear to everybody what is really happening in our gas mixture in LTE.
The figure shows the distribution of the kinetic energy (Ox axis) among the N₂ molecules (Oy axis).
This typical curve is called the Maxwell Boltzmann distribution, has been known for more than 100 years and experimentally confirmed with high accuracy.
We know that the temperature is defined by <E>, the energy average.
Hence it is the curve shown in the figure that defines the temperature of a gas.
Another way to say the same thing is to say that the curve depends only on temperature. If we wanted to have the distribution for another gas than N₂ , f.ex CO₂ or O₂, it would be given by an identical curve.
The blue curve gives the distribution of kinetic energy at 25°C while the red curve gives the distribution at 35°C.
The minimal energy is small but non-zero and there is no maximal energy.
A very important point on the Ox axis is the energy of the first vibrationally excited state of a CO₂ molecule.
You notice that at 25°C the majority of N₂ molecules has insufficient kinetic energy to excite this vibrational state.
Only the proportion of them given by the dark blue surface has enough energy to excite the vibrational state by collision.
When the temperature increases to 35°C, you notice that the proportion of N₂ molecules able to excite the vibrational CO₂ state by collision has significantly increased .
This proportion is given by the sum of the dark blue and light blue surface.
You also notice that as there exists no maximal energy, there will be a proportion of N₂ molecules able to excite the vibrational CO₂ state at any temperature.
Trivial so far? Well it will not get much more complicated.
First 2 technical points which play no role in the argument but which I would like to mention for the sake of completness.
- The figure shows the translational kinetic energy. Even if in some (popular) literature the temperature is defined as being an average of the translational kinetic energy, this is not strictly true.
The temperature is really defined as an average of all energy modes. So what about the vibrational and rotational energy?
At the low tropospheric temperatures we are considering, the distribution of the vibrational energy is extremely simple : about 5% or less of the molecules are in the first excited state and 95% or more are in the ground state.
As for the rotational energy, it can be computed classically without quantum corrections and the result is that it also follows a Maxwell Boltzmann distribution.
Therefore if we wished to plot the total energy (Etranslational + Evibrational + Erotational) we would rescale the Ox axis and obtain exactly the same curve as the one that is shown.
However as we are interested in studying the T/V interactions, it is the curve of the translational kinetic energy that interests us.
- We find the omnipresence of LTE again. This curve has been derived and experimentally confirmed only, and only if, the gas is in TE. Therefore the following 2 statements are equivalent :
“The gas is in LTE” , “The energy distribution at every point is given by the Maxwell Boltzmann distribution” .
If you feel that these statements are not equivalent, reread carefully what is above.
Now we can demonstrate why the Type2 argument is wrong.
Imagine that you mix cold N₂ represented by the blue curve in the Figure with highly vibrationally excited CO₂. The mixture would then not be in LTE and a transient would take place.
In the molecular process (1) CO₂* + N₂ → CO₂ + N₂⁺ which says that a vibrationally excited CO₂ molecule (CO₂*) collides with an N₂ molecule , decays to the ground state (CO₂) and increases the translational kinetic energy of N₂ (N₂⁺) , there would be a net energy transfer from CO₂* to N₂ .
As a result of this transfer the temperature of N₂ would increase and the blue curve would move to the red one.
However doing that, the number of molecules able to excite CO₂ vibrationally would increase (see the blue surfaces in the figure).
That means that during the increase of the temperature of N₂ , the rate of the opposite molecular process (2) CO₂ + N₂⁺ → CO₂* + N₂ where N₂ molecules (those from the blue surface in the figure) vibrationally excite CO2 molecules, will increase too.
Of course the transient net energy transfer from CO₂ to N₂ will not continue forever because else the mixture would transform into superheated plasma.
A local equilibrium will be established at each point and in this equilibrium the rate of the process (1) will be exactly equal to the rate of the process (2).
The curve of energy distribution will stop moving and the Maxwell Boltzmann distribution will describe this distribution at every point.
This is exactly the definition of LTE.
The transient will stop when the mixture reaches LTE and its characteristic feature is that there is no local net energy transfer from CO₂ to N₂.
This result demonstrates both that the Type2 argument is wrong and that the answer on the question we asked at the beginning is “No”.
In very simple words, if you take a small volume (for example 1 m³) of the CO₂ and N₂ mixture in LTE around any point , then there cannot be any net energy transfer from CO₂ to N₂ within this volume.
To establish the last step we will take the following statements.
- The result obtained for the CO₂ and N₂ mixture in LTE is equally true for a mixture containing 78% of N₂ , 21% of O₂ , x% of CO₂ and 1-x % H₂O in LTE.
- The mixture defined above approximates well the troposphere and the troposphere is indeed in LTE
- From the 2 statements above and the demonstrated result follows :
“The CO₂ does not heat the troposphere” what is the answer on the question asked in the title.
Caveat1
I have said it both in the initial post and in this one.
Unfortunately, I know that it can’t be avoided and that some readers will still be confused about the result established here and start considering radiative transfers or radiative equilibriums.
That’s why I stress again that LTE and the result established here is totally independent of radiative equilibriums and radiative transfer properties.
However it does falsify one misconception concerning radiative properties of CO₂ that has also figured in the comments and that is that “CO₂ does not radiate at 15µ because it “heats” N₂ instead”.
It is also to be noted that we consider only the T/V process because it is only the vibrational modes that interact with IR radiation.
There are also rotational/translational and rotational/vibrational transfers.
The same argument used for T/V applies also for the R/T and R/V processes in LTE – e.g there is no net energy transfer between these modes in LTE even if for example the R/T process has a much higher probability than a T/V process.
For the sake of clarity we don’t mention specifically the R/T and R/V processes.
Caveat2
The result established here is a statistical thermodynamics bulk property.
This property is of course not sufficient to establish the whole dynamics of a system at all time and space scales.
If that was our ambition – and it is not – then we would have to consider boundary conditions and macroscopic mass, momentum and energy transfers, e.g convection, conduction, phase changes, lapse rates etc.
More specifically this result doesn’t contradict the trivial observation that if one changes the parameters of the system, for example composition, pressure, radiation intensity and spectrum, etc, then the dynamics of the system change too.
Yet it contradicts the notion that once these parameter are fixed there is a net transfer of energy from CO₂ to the troposphere. There is not.
Caveat3
It will probably appear obvious to most of you but it has also to be repeated.
This result says little about comparisons between the dynamics of 2 very different systems such as, for example, an Earth without oceans and atmosphere, and an Earth with oceans and atmosphere. Clearly the dynamics will be very different but it stays that in the case of the real Earth with an atmosphere in LTE, there will be no net energy transfer from the CO₂ to the atmosphere.
kfg
“Some physicists leave the tower and go to work as engineers.”
An undergraduate degree in physics is something of a general purpose kind of thing. It’ll get your foot in the door at entry level a lot of different engineering specialties. A graduate degree in physics won’t help much. As a general rule years of experience and years in graduate school are equitable. An established salary history in the desired occupation trumps everything else though and you don’t get that salary history as a full time student.
@scott
Condensation plays a huge role when the sensible temperature reaches the dewpoint.
Dave Springer says:
“That’s what makes this so damn frustrating. This is knowledge from experimental physics 150 friggin’ years ago!”
———–
It’s astonishing to me that people with serious degrees in Physics can be arguing about this, too.
However, the discussion is carried on in such truncated fashion that it’s not often clear what exactly is being disputed.
It’s hard to reconcile statements that emission is irrelevant at 1 bar with the assertion that IR is radiated back to the surface. Then , the twice-spent dollar crops up again when a photon that has gone translational is also able to speed back to Earth, warm up the ground, then zip back for a logarithmically reduced second shot.
Being mathematically naive, I am limited to simple arithmetic and a lot of what is said doesn’t add up.
Your link to Tyndall’s book was good enough to hint at the contents but I didn’t see much of the actual content. I’ll try Amazon or some such.
According to Anthony:
It’s more than a bit hard to believe at this point.
Nylo says:
August 31, 2010 at 8:49 pm
The easiest way to see it is the opposite reasoning. Take a bottle of CO2+N2, put it next to a source of 15 microns radiation. It gets some temperature, and when the energy comming in equals energy getting out it is in LTE. Now, chemically or mechanically, whatever, take out the CO2 content alone, replace it for N2 at the same temperature that the mix had before. After the transient, the temperature of the new gas will be lower, because again energy out = energy in, and suddenly there is no energy in, so there must be no energy out, so the temperature must be 0 K.
—
I thought it was accepted that N2 does not radiate or absorb at normal temperatures so how is your example every to cool since you put it in without the CO2 at the same temperature? Does not absorbability = emissivity? (Or are you allowing conduction through the jars walls which makes little sense in your pure example?)
To me the N2’s temperature would never change, not absorbing or emitting.
“Is any of these points wrong?”
No.
“In case not, doesn’t 3 (increasing CO2 content of the atmosphere) inequivocally lead to 9 (temperature gets higher)?”
Only if nothing else changes. The problem is other things change at the same time and these other things (all related to water in all its phases – albedo of snow, clouds, and liquid water – and phase changes which shuffle around enormous amounts of latent heat) and ) can easily drown (pun intended) the CO2 contribution. For instance when increased CO2 tries to throw the system further from equilibrium temperature it causes surface water to evaporate faster. Water vapor is lighter than air so it rises and condenses into a cloud and in the process carries an enormous amount of latent heat of vaporization straight through the densest layer of CO2 and releases it high in the sky where the underlying greenhouse gases then serve to insulate the warmer cloud from the cooler air below it. Latent heat of vaporization and to a lesser extent latent heat of fusion is the fatal flaw in the highly exagerated “climate sensitivity”. If there’s no positive feedback from CO2 heating that amplifies its effect by a factor of three, and there clearly ain’t no such thing indicated by any empirical observation, then catastrophic global warming becomes beneficial global warming.
Nylo says:
August 31, 2010 at 9:34 pm
“Is any of these points wrong?”
No.
“In case not, doesn’t 3 (increasing CO2 content of the atmosphere) inequivocally lead to 9 (temperature gets higher)?”
Only if nothing else changes. The problem is other things change at the same time and these other things (all related to water in all its phases – albedo of snow, clouds, and liquid water – and phase changes which shuffle around enormous amounts of latent heat) can easily drown (pun intended) the CO2 contribution. For instance when increased CO2 tries to throw the system further from equilibrium temperature it causes surface water to evaporate faster. Water vapor is lighter than air so it rises and condenses into a cloud and in the process carries an enormous amount of latent heat of vaporization straight through the densest layer of CO2 and releases it high in the sky where the underlying greenhouse gases then serve to insulate the warmer cloud from the cooler air below it. Latent heat of vaporization and to a lesser extent latent heat of fusion is the fatal flaw in the highly exagerated “climate sensitivity”. If there’s no positive feedback from CO2 heating that amplifies its effect by a factor of three, and there clearly ain’t no such thing indicated by any empirical observation, then catastrophic global warming becomes beneficial global warming.
Everything radiates at any temperature above absolute zero.
Everything absorbs unless its surroundings are at absolute zero.
http://en.wikipedia.org/wiki/Absolute_zero
OK, I will chime in though my thermodynamics course is fifty years behind me.
What Tom is saying is that, at normal pressures ( not vacuum) once in local thermodynamic equilibrium, the type of constituent molecule is irrelevant to the temperature. All types should be at the same temperature by the way temperature comes out in statistical mechanics. Since they are at the same temperature, one type cannot heat the other by tautology. Radiation,incoming or outgoing is within this LTE again by construction. In this thermodynamic formalism the modifying effect of a new constituent molecule appears in changes in the heat capacity of the whole volume of gas under observation.
H2O is very good at that, as Dave has shown, doubling the heat capacity of air for high humidity values.
Is it then correct to say that H2O heats N2 an O2? I would say it delays the radiative cooling of the total gas volume, which is a different concept than heating, though connected to heat transfer. I would also accept that it changes the gray body constants of the gas volume. Again this cannot be called heating.
Again I will repeat that in my opinion the introduction of absorption and radiation lines is an unnecessary double counting when using the formalism of classical thermodynamics and statistical mechanics.
The whole book is there. You can thumb back and forth a page at a time at the top left. The link I gave is the optical character recognition version so it has an occasional character that is wrong. You can also see the original pages as they were scanned and download the whole book in a single PDF (it’s a big file).
@Oliver Ramsay
The link below takes you to the table of contents for Tyndal’s book with hotlinks to the chapters and sections. Google doesn’t make it easy to find the ToC.
http://books.google.com/books?id=m-TUAAAAMAAJ&pg=PR16&dq=heat+a+mode+of+motion&output=text
George E. Smith says:
August 31, 2010 at 4:24 pm
Well CO2 most certainly does not absorb all of the LW radiation available to it. The surface emission spectrum; at least from ana average Temeprature surface can be expected to contain significant energy between 5.0 microns and 80.0 microns (about 98% of it.
“available to it” Has a specific meaning. It does not imply that CO2 absorbs all the LW radiation emitted from the black body. It means that CO2 absorbs all the energy in the wavebands that it can absorb, the ones outside those wavelengths are not available to it because of the atomic structure of the molecule.
And in the lower atmospehre the mean lifetime of the CO2 excited states is longer than the mean time between collisions; so that energy becomes thermalized before a spontaneous photon emission from the CO2 can occur. In the higher less dense regions in the stratospehre the time between collisins is great enough for spontaneous decay to occur.
I don’t dispute what you are saying but that is not what
R Stevenson asked.
The answer I gave is correct retention time of the photon doesn’t increase with the density of the gas.
Co2 does not heat the troposphere – correct in the sense that co2 is not a source of energy like the sun.
Co2 is however defacto heating the troposphere (in some amount) in the sense that it transforms energy that otherwise would pass through the troposphere and converts it to kinetic energy in the molecular level. I don’t suppose anyone seriously contends that?
There are a bunch of problems in this post that need to be addressed – being a circular argument, the best remedy might be just removing it as I can’t see how you could fix it if the basic logic is faulty. I do however appreciate the post, good read and made me think.
Some thermodynamisists state that the atmosphere can never be in LTE so temperature measurement will produce a meaningless number. The above argument also ignores the fact the warm air will convect and cool adiabatically a fact that the theory of GHG’s also ignores. I cannot see why an energetic CO2 molecule does not transfer this vibrational energy to other gasses. To say that this energy is transferred back only states that equilibrium will be established when all the gas in that system is at the same temperature, ignoring boundary exchanges. Unfortunately the atmospheric system does have boundaries and there are many energy exchanges.
The argument that temperature measurement in a room will produce a constant temperature shows that this gentlemen has not tried this experiment. Place the thermometer on the floor and the temperature will be fairly low and take the temperature at the ceiling and it will be higher. I have done this and the data shows this to be true so my room was not in LTE.
It is true that the troposphere does get warm and cools so must be by energy transfer from solar radiation and gas/gas interactions, convection and adiabatic cooling and warming due to vertical movement of gasses through the atmospheric column. What the alarmists state about backradiation and energy increasing does violate the laws of thermodynamics.
I think the point is that there is a lot more nitrogen in the atmosphere then there is co2 in the atmosphere ,it is going to be very difficult for co2 to raise the average energy in the troposhere by absorbing ir radiation but it would be very easy for nitrogen to raise the average energy in the troposphere by absorbing ir radiation.
It is good that we are now discussing the thermodynamics ( a well established body of knowledge) of the gas N2 that makes up nealy 80% of the atmosphere.
The idea that CO2 traps heat is a return of the phlogiston theory that heat is a substance. We now know heat is the result of motion of every substance so cannot be trapped.
I believe one the problem with the exposition of the main article is that Local Thermodynamic Equilibrium is a condition that can only exist if the given local region is absolutely *not* being heated or cooled by any other external region. In practice, I think LTE never really applies in our atmosphere, even taken as a whole, because it is always interacting with the universe at large.
What I believe we do have is a dynamic Energy Exchange Equilibrium, where a given local region rises to a temperature that allows it to expel the same average amount of energy that it is receiving from the surrounding external environment.
I believe the logarithmic effect of added CO2 is the most important reason for discounting the AGW alarm. According to the MODTRAN online radiation calculator utility, the CO2 increase over the past 10 years should only have had a raw effect of about 0.06 degrees C.
The so-called ‘hotspot’ refers to the tropical troposphere, and increased heating there is not an artefact of GHG heating, but of any heating. If the sun gets hotter, or albedo somehow reduced – the anticipated signature of global warming is a hotspot in the tropical troposphere. It has no direct relation to CO2.
Therefore, if there is a missing ‘hotspot’, then there is a problem with the understanding of heat transport in the atmosphere, not the theory of global warming from increasing GHGs.
However, it appears that the hotspot may previously been obscured by poor data.
Wow. I have been in the mountains for days and got a look at this for the first time when I was getting ready to go black again. I typed up some quick but long responses expecting to have to make extensive revisions to my comments to respond to what I expected to be further additions of (what I believe to be) incorrect statements about atmospheric heating in the follow-on comments, but am exceptionally pleased (and frankly shouldn’t be!) At the extremely sober comments on Tom’s article. The vast majority of Anthony’s visitors really get the basic issue with this contention:
Yes, of course, a system in thermodynamic equilibrium (local or otherwise) cannot possibly have net energy flow – that violates the initial statement. When used in the physical sciences, equilibrium is defined as that state of a system in which an external force is required to change the state of the system. Posit BOTH equilibrium AND no external forcing (remember – no radiative coupling by Tom’s stipulation) and the argument is rock solid. By definition.
The problem is, Tom isn’t saying that CO2 can’t heat N2 when the system is in LTE. If he stopped there he’d be on solid ground. The problem is that once having established that he then claims that therefore CO2 cannot heat the atmosphere at all and backs it up with very watery caveats about how the atmosphere is almost in LTE, etc. This is simply untrue.
I am so pleased that WUWT folks have so clearly understood this that I promise not to post my extremely lengthy replies unless I’m refuted again and am forced to.
And – by the way – contrary to what Tom says, all that V/R, R/T, and V/T transfer that Tom claims to be mere details IS the whole warp and weft of radiative heating in the atmosphere. Those modes, by and large, CANNOT get by energy coupling directly to the electromagnetic field (more details later if required) so the only way for energy to get into those modes is via the indirect path of collisional energy transfer after the electromagnetic field has energized the CO2 vibration. As long as the electromagnetic field is out of equilibrium with the atmosphere there is an external forcing. Period. If there is more capacity in the EM field then energy flows from the EM field into the CO2 vibration and energy moves from the CO2 vibration into all of the vibrations, rotations, and translations in all the atmospheric constituents until all of those constituents are in equilibrium with the EM field. This is called heating. The CO2 was the key constituent in the radiative heating process (water contributes in a similar fashion, but we’re discussing CO2 here). The higher the concentration of CO2, the more rapidly CO2 can couple EM energy into the atmosphere*. Period. So, yes Virginia, there is a “Greenhouse Effect” – even if it’s inappropriately named, as has been pointed out on many occasions. And, yes, there is ALSO a logarithmic effect, so at some point additional CO2 has a much smaller effect than the CO2 already there. We are well into that logarithmic rollover and IPCC CO2 doubling nightmare scenarios are simply rubbish.
*from above as marked, when the capacity of the EM field is less than that of the atmosphere (like when the earth cools after sunset) the CO2 also acts as an efficient mechanism for couplng the energy back out of the atmosphere. More photons go out of CO2 vibrations than are repopulated by the external forcing of the EM field, and energy flows from the V/R/T modes of the relatively warm atmospheric constitutents into CO2 vibrations to reestablish equilibrium and both of these energy flows continue until equilibrium is reestablished or the balance reverses and a warming earth under the risen sun starts pumping energy through CO2 (and water) back into the atmosphere.
Tom – just a suggestion: pull up a plot (perhaps accuweather.com?) of the hourly temperature records for yesterday for any place you’d like. Then tell me where it is so I can look at the same data and you tell me which parts of yesterday’s time record, according to the temperature record, were a good example of the atmosphere experiencing LTE conditions.
RE: George E. Smith: (August 31, 2010 at 4:24 pm)
“And in the lower atmospehre the mean lifetime of the CO2 excited states is longer than the mean time between collisions; so that energy becomes thermalized before a spontaneous photon emission from the CO2 can occur. In the higher less dense regions in the stratospehre the time between collisins is great enough for spontaneous decay to occur.”
This is an interesting comment. I have not considered that something like this might also be a factor in establishing the level of the tropopause before. I wonder at what altitude the mean time between collisions is equal to the typical CO2 or H2O excited state lifetime.
This article sounds like Xeno’s Paradox and Tom Volk has just proved that arrows can’t possibly fly.
Oops Zeno not Xeno
[Reply: that’s OK, you’re just a teenager.]
Your Maxwell Boltzmann distribution is mis-labeled. You figure shows velocity, not energy. Energy is velocity squared, the population distribution against E is typically a plot of logE.
Moreover, you are applying classical equilibrium thermodynamics, local thermal equilibrium, to a steady state system. This is a no-no. Imagine applying equilibrium thermodynamics to a juggler, it is not possible.
Tom are you absolutely sure of this?
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The figure shows the translational kinetic energy. Even if in some (popular) literature the temperature is defined as being an average of the translational kinetic energy, this is not strictly true.
————
Yes he is absolutely sure of it and he is also absolutely wrong.
This claim is also wrong!
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Of course the transient net energy transfer from CO₂ to N₂ will not continue forever because else the mixture would transform into superheated plasma.