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
Discover more from Watts Up With That?
Subscribe to get the latest posts sent to your email.
Willis, I am glad to see that you are still here. I hope that I am not being a broken record, more like a 78 with a few cracks and chips. Since there is no material in a vacuum, how are you raising the temperature? There are no gasses to excite, no material to absorb and emit, no conduction or convection. How can you raise the temperature of nothing? If a steel shell surrounded a planet with no atmosphere, no pressure, complete vacuum- would not the space between them be zero?
Oh, I almost forgot- I laughed my @** off when I saw that pic of Foghorn Leghorn. I love that old bird. Good times…
par5 (23:00:31) :
There’s three ways that heat moves around – conduction (things touch), convection (a gas moves the heat), and radiation (electromagnetic waves). All of these, of course, have to act on something, so a vacuum has no heat and no temperature. There’s nothing there to heat up. The temperature of a vacuum is not zero, as many people think. It has no temperature at all.
In the situation in Fig. 1, there is no conduction and no convection between the planet and the shell. But the radiation can still move through the vacuum to heat something at the other side of the vacuum.
And Foghorn Leghorn? He’s my main man …
Thanks for the question.
Over 450 comments so far- Anthony should give you a free subscription or something. Great job Willis. Hat tip to the mods…
Willis you said :
.
“First let me say that you are correct. Second let me say that the differences between what you say and the results I show above are trivial, on the order of 0.2%. This is because for all practical purposes, there is no difference between the situation I describe above and two infinite parallel planes.
This is because the radii of the sphere and the shell are equal to within less than a tenth of a percent. As is common in analysing this particular situation, this tiny error is ignored, and they are treated as though they were infinite parallel planes. See e.g. the Kiehl/Trenberth diagram I show in Fig. 3.
I think that answers your questions, if not, let me know.”
.
First I am glad that you saw now what I have been meaning since the beginning 🙂
Second is that I had no questions but only a few comments .
Third is that I said from the very beginning that I shared the opinion of L.Motl , e.g there were no substantial errors with your toy system even if it had nothing to do with the atmospheric dynamics and couldn’t simulate it .
.
I also said from the very beginning that P.r²/R² ~ P when r~R .
So of course I agreed with your approximation and said so .
What I did NOT agree with and what is still wrong is the use of the word CONSERVATION of W/m² .
I hope you see the difference between a valid numerical approximation (e.g r~R) and a statement about a general law (e.g “W/m² are always conserved”) .
While the first is valid , the second is wrong .
.
My second contribution was meant to be more pedagogical in nature .
I have experience that many people consider the SB law as being a fundamental unique law what it is not .
When I was teaching , those people always had a very hard time to understand why a 2 faced plate could radiate 2 times “more” than what it absorbed on 1 face .
However the reality is that there is an infinity of SB laws and SB constants depending on the geometry .
So experience showed me that the right approach to explain this “paradox” was to say that the SB law applying to 2 faced objects had an SB constant which was TWO TIMES bigger than the SB constant of the law applied to a 1 faced (convex) object .
I believe that this explanation is short , correct and easy to understand .
.
Besides as the usual SB law formulation is derived from a plane surface radiating in a half space , people can then easily understand that as each differentiable surface can be approximated by tangent planes at each point , the (usual) SB law form can be generalised to any convex body like f.ex a sphere .
.
This insight avoids the HORRIBLE mistake to apply the (usual) SB law to a non convex cavity as a whole .
I had a nice exercice asking students to compute the SB constant of a T form cavity 🙂
Of course this has now nothing to do with your toy system because you did use only convex bodies .
But it is a caveat that what you did cannot be generalised to ANY form of bodies .
TomVonk (06:49:02) :
“However the reality is that there is an infinity of SB laws and SB constants depending on the geometry .”
I think this is a horrific way to teach the concept. There is one SB constant; by definition it does not change with geometry. However, if you want to know how two blackbodies exchange radiation with each other, then geometry is taken into account by using the view factor (or shape factor).
http://en.wikipedia.org/wiki/View_factor
P Wilson (18:17:03) :
“Final comment prior to censorship: What happens to heat from a red hot poker from a hot furnace and exposed to normal temperature air? Does the heat disippate/disappear or does it have to be integrated into some other entity, or is it just a case that when objects cool and emit radiation they thermalise to the temperature set by the air, as the electrons and atoms lose their excited state?”
Through the processes of radiation and convection, the thermal energy is transferred to the surrounding air and objects. The surrounding air and objects will end up with a higher temperature than what they began with. Energy does not disappear.
I have a great deal of difficulty understand what you’re saying in most of your comments, but it seems to me that while you know the mathematical form of the SB law, you are having trouble applying it within an energy balance. Since you are not correctly writing the energy balance, you turn back and think the mistake is in the SB law itself.
carrot eater (08:17:40)
No problem applying the constant and the results thereof.. nothing more to say on the matter as its upsetting the balance of the discussion. Out of respect of Willis
addendum *for respect of Willis’s request*
This has been the most thought provoking post I have seen. A steel greenhouse also illustrates how low level particulate pollution (especially black soot) has a greenhouse effect. The greenhouse affect of low level particulate pollution has been greatly under rated.
You can argue cause and effect, but the warmest temperatures in the US occurred during the dust bowl years.
Smudge pots prevent orchards from freezing.
Over the last century, farmers have been tilling more and more land causing more soil erosion and low level particulate pollution. This has an affect.
Over the last couple of decades farmers have been switching to no plow farming. This lessens erosion and low level particulate pollution. This has an affect.
Climate models should pay more attention to these affects and a lot less attention to carbon dioxide.
Willis Eschenbach (20:41:08)
“I can’t get to whatever is the nature of your complaint,…”
Sorry if I was too vague. TomVonk says it much more clearly and concisely, than I was able to:
“I hope you see the difference between a valid numerical approximation (e.g r~R) and a statement about a general law (e.g “W/m² are always conserved”) .
While the first is valid , the second is wrong .”
He also added that “your toy system … [has] nothing to do with the atmospheric dynamics and couldn’t simulate it .” (I hadn’t even gotten that far yet.)
Finally, the temperature (in deg K) wouldn’t double, because the heat output is dependent on the fourth power of the temps, so, if I’m correct, it would at most increase by a factor of “2^(1/4)”, not “2” (for a sphere radius only slightly greater than that of the source sphere). The same for the temp of the penultimate sphere, with every additional sphere, which would consequently increase the heat source sphere’s temp, though not in a linear way, etc. After doing the calculation, I saw that there would be a substantial increase, but as TomVank said, it has “nothing to do with the atmospheric dynamics.”
Anyway, thanks for getting me to remember a little bit of some course material from about 30 years ago.
Let’s try this another way. Let’s say that we find a rogue planet with the same conditions as you descibe- no atmosphere, total vaccuum and in the middle of nowhere. Now let’s say we don’t have any steel with us. Willis and I go down to the planet to do a survey. I’ve got the flashlight, Willis has the thermometer. Willis measures 235W/m2. Since we have no steel, I decide to take the outside skin of the planet, one meter thick, and raise it above our heads by three meters. So now we are standing on the new surface, in our pressure suits, in a vaccuum, with a one meter thick old surface over our heads. Willis measures the new surface we are standing on, and it is 236W/m2. And so is the roof three meters above. And the outside of the shell is still 235W/m2. Now we know the transfer rate of the planets surface material. It doesn’t matter how high we put the shell, three meters or three km, the outside of the shell will still be 235, and the inside will be 236. Let’s do another. We take the next one meter under our feet, and raise it three meters just like the last. So now we have two shells above the planet. Willis takes a new measurment of the new surface and gets 237W/m2. He then takes a measurment of the roof and gets the same.
We have not added any exotic materials- everything is the same. The transfer rate, planetary suface, vaccuum- same. We can continue to do this, towards the core, until the material changes. This implies three things:
1) The outside shell will always emit 235W/m2
2) An outer shell will always be colder than an inner shell
3) No matter what layer you are in, the roof and the floor will always measure the same
Willis, if you agree to this, I can build your steel green house. However, there will be one more implication.
Cheers!
Hi Willis,
Thanks for the interesting post. I think your clever double membrane model makes the greenhouse effect a lot easier to understand.
The other reason for my comment: I think I see an asymmetry in Figure 4. The figure has a term (53 W/m2) that represents the upwelling longwave window from the surface through the troposphere membrane to the stratosphere membrane and beyond. However, the figure does not have a term for the downwelling window from the stratosphere to the surface. I would have thought that the clarity of the downwelling window would be the about the same as the upwelling window (ignoring temperature differences).
If so, then this would mean the size of the downwelling term would be about 20 W/m2. I’m not a Windows person (pun intended) so I was not able to run your Excel model and check how adding this term would affect it.
And I am sorry if this issue has already been addressed, but there are a lot of comments to this post and it would be easy for me to miss the issue.
George Crews (22:37:46) :
You are correct. It is a term I left out of the drawing, although it is included in the model. However, if it is very large, the surface won’t warm, so …
The model was written in Excel on my Mac, but it should run in a Windows version of Excel, I’d be surprised if it didn’t.
w.
Everyone, read
Smith (2007), Proof of the Atmospheric Greenhouse Effect [arXiv:0802.4324v1]
It’s only nine pages, and it has actual physics in it.
Sorry Willis, your physics fails.
stunned (09:54:49), out of curiosity, where does Smith’s paper disagree with Willis?
stunned (09:54:49) :
The article you referenced – http://arxiv.org/abs/0802.4324 – was published as a response to this “falsification” of atmospheric greenhouse theory – http://arxiv.org/abs/0707.1161
I see no attempt to falsify anything with Willis’ work here, refine maybe but not falsify
My question: How would a thermometer read in a stevenson screen on the moon. Obviously there’s no need for the louvers, but how would a thermometer register in the shade of a box. My guess is that it would end up taking on about the same temperature as whatever the roof temperature is.
I still think the greenhouse metaphor is broken. The air in our atmosphere tempers the earth’s temperatures in both directions (judging by the highs and lows of surface temperature seen on the moon). How does the tired out old greenhouse that everyone kicks around illustrate that point?
Also of interest:
http://www.sciencedaily.com/releases/2009/09/090917191609.htm
Paragraphs 9 and 10 indicate that there is something like regional climate on the moon based on the nature of the surface materials. I suppose that would be things like heat capacity, conductivity, depth of material, and blackness. This seems suggestive of the difficulties of modeling climate. One needs to account for various kinds of surface material. I also want to think that somewhere one needs to account for the length of a day in any good model since we know that the extreme heat of a day does not coincide with high noon.
I like the steel greenhouse, because everyone knows that a steel shed gets hot inside. Still can’t tell if the radiative greenhouse is real or falsified.
Can someone please explain why it is that climate scientists measure “energy” in W/m2? If any of us tried that in high school physics, we would flunk. All the math/physics that follows falls over because conservation of energy is robust but conservation of radiative flux does not exist.
I note that the Kiehl/Trenberth Global Energy Budget does not seem to make any account of earth-generated thermal pollution. Neither does IPCC. Hence climate scientists have trouble understanding Nordell’s analysis of energy, which claims to account for 74% of observed warming. I think there are some missing assumptions in Nordell and Gervet’s work but not sure of their magnitude – ie how much energy is stored in industrially generated materials and not turned to heat.
Paper via
B Louis (13:32:04), thanks for your note. you say:
While radiative flux is not conserved, W/m2 is. If a body in space is receiving X W/m2, if it is in equilibrium it is also radiating X W/m2.
I was interested by your description of “earth-generated thermal pollution”. What is that? Do you have a reference?
Nordell and Gerbett – see 2009 paper at bottom of page + some responses and replies by authors: (Sorry must have forgotten to paste the link.)
http://www.ltu.se/shb/2.1492/1.5035?l=en
W/m2 is radiative flux (also J/m2/s). Energy is Joules or kWh.
(Laws of physics include conservation of mass, energy, momentum, charge.)
So a body in space in thermal equilibrium must not retain energy. If the body was a steel sphere in the sun, the energy flux it receives is spread over its lit surface area. The heat it radiates if it conducts is over its whole surface with a different area and a different flux. Physics of a rotating earth is a bit more complex, since it never really reaches equilibrium it is always heating daytime and cooling at night. The response to the CO2 greenhouse falsification contained complex physics of that. (http://arxiv.org/abs/0802.4324). not sure how to tell who is right.
B Louis, here’s why I find Nordell’s explanation, that the warming is from human use of energy, totally unconvincing.
In the BP Statistical Review of World Energy June 2004, I find the following world energy consumption including all sources (fossil fuels, hydroelectric, nuclear) in Tonnes of Oil Equivalent (TOE)
9.7E+09: TOE for world, 2003
From the same source, the energy content of a TOE is:
4.2E+10: joules/TOE
Multiplying these gives us a total of:
4.1E+20: joules/yr from human fuel use.
Since a watt is a joule/second, dividing joules/year by the number of seconds in a year gives us:
1.3E+13: watts
And then we have the area of the earth:
5.1E+14: sq. metres planetary surface
Dividing one by the other, we get:
0.03: W/m2 from “thermal pollution”
Given that the total downwelling radiation at the surface (short and long wave) is on the order of half a kilowatt per square metre, this is a trivially small amount from “thermal pollution”, far too small to make any discernible difference.
Please check my numbers, but I believe that they are correct.
w.
Sorry Willis, the numbers may be correct but the assumptions have nothing to do with the reality.
The issue is: how does one convert energy to temperature? It has nothing to do with W/m2. There is a robust physical relationship between energy, specific heat and mass. The equations used by Nordell appear reasonably robust. One can compute energy from W/m2 time m2. What is the relevant specific heat? How does one account for thermal conduction? Nordells work accounts for all of that in a more complicated but apparently robust and verifiable manner. He has checked his numbers.
Climate scientists on the other hand defy the laws of physics. The magical formula to convert radiative flux to temperature change is (ΔTs): ΔTs = λRF, where λ is the climate sensitivity parameter
None of that formula has any robust basis in fundamental physics. Mass is unaccounted. Specific heat is unaccounted. Lambda is extremely wooly and is derived by observation of flawed temperature data and with a range that varies by an order of magnitude. Talk about junk science. W/m2 is not a measurement of energy. m2 from a sphere is not constant when the radius of interest changes. In fact m2 has nothing to do with the planetary surface, since in the radiative balance is represents area irradiated by the sun (assuming a flat earth), hopefully correcting for the earths curve and atmospheric effects and rotation. Climate math is a lot easier if the earth is flat! Conservation of radiative flux is great for perpetual energy.
If we assume the earth is in long-term equilibrium for a moment. (Gerlich and Tscheuschner argue that a mean temperature for an earth is nonsense.) Any particular patch of earth warms in the daytime and cools at night. The net energy gain is positive positive in daytime, negative at nighttime, positive in spring/summer and negative in autumn/winter more or less. So for a whole year on average, despite the high radiative flux that hits the earth on the sunny side, net energy gain is near zero because of losses on the dark side. So large radiative flux numbers on average have not a lot to do with long-term warming.
The fact that heat is generated by burning fossil fuels is absolutely proven. The magnitude of this energy gain over the mass of the earth has been calculated. One could argue that this will be lost as radiation or convection, but that doesn’t help the flux warmist cause since a warmer earth resists further heating. (Unless one doesn’t believe in the second law of thermodynamics.)
I read somewhere that liquids don’t emit blackbody radiation, so that puts a big dent in the radiative flux theory.