Guest Post by Reed Coray
On Dec. 6, 2011 12:12 am Lord Monckton posted a comment on a thread entitled Monckton on sensitivity training at Durban that appeared on this blog on Dec. 5, 2011. In that comment he wrote:
“First, it is not difficult to calculate that the Earth’s characteristic-emission temperature is 255 K. That is the temperature that would obtain at the surface in the absence of any greenhouse gases in the atmosphere. Since today’s surface temperature is 288 K, the presence as opposed to absence of all the greenhouse gases causes a warming of 33 K”.
Since I’m not sure what the definition of the “Earth’s characteristic-emission temperature” is, I can’t disagree with his claim that its value is 255 K. However, I can and do disagree with his claim that 255 K is “the temperature that would obtain at the surface in the absence of any greenhouse gases in the atmosphere”.
When computing the Earth’s surface temperature difference in “the presence as opposed to the absence of all greenhouse gases”, (i) two temperatures (A and B) must be measured/estimated and (ii) the difference in those temperatures computed. The first temperature, A, is the temperature of the Earth’s surface in the presence of an atmosphere that contains both greenhouse gases and non-greenhouse gases. The second temperature, B, is the temperature of the Earth’s surface in the presence of an atmosphere that contains non-greenhouse gases only–i.e., an atmosphere that contains non-greenhouse gases but is devoid of greenhouse gases.
For temperature A almost everyone uses a “measured average” of temperatures over the surface of the Earth. Although issues may exist regarding the algorithm used to compute a “measured average” Earth surface temperature, for the purposes of this discussion I’ll ignore all such issues and accept the value of 288 K as the value of temperature A (the temperature of the Earth’s surface in the presence of an atmosphere that contains both greenhouse gases and non-greenhouse gases).
Thus, we are left with coming up with a way to measure/estimate temperature B (the temperature of the Earth’s surface in the presence of an atmosphere that contains non-greenhouse gases only). We can’t directly measure B because we can’t remove greenhouse gases from the Earth’s atmosphere. This means we must use an algorithm (a model) to estimate B. I believe the algorithm most commonly used to compute the 255 K temperature estimate of B does NOT correspond to a model of “the temperature of the Earth’s surface in the presence of an atmosphere that contains non-greenhouse gases only”. As will be evident by my description (see below) of the commonly used algorithm, if anything that algorithm is more representative of a model of “the temperature of the Earth’s surface in the presence of an atmosphere that contains both greenhouse gases and non-greenhouse gases” than it is representative of a model of “the temperature of the Earth’s surface in the presence of an atmosphere that contains non-greenhouse gases only.”
If I am correct, then the use of 255 K in the computation of the Earth surface temperature difference with and without greenhouse gases is invalid.
Although there are many algorithms that can potentially lead to a 255 K temperature estimate of B, I now present the algorithm that I believe is most commonly used, and discuss why that algorithm does NOT represent “the temperature of the Earth’s surface in the presence of an atmosphere that is devoid of greenhouse gases”. I believe the algorithm described below represents the fundamental equation of radiative transfer for the Earth/Sun system assuming (a) an Earth absorption albedo of 0.3, and (b) an Earth emissivity of 1.
(1) The “effective temperature” of the Sun [i.e., the temperature of a sun-size spherical blackbody for which the radiated electromagnetic power (a) is representative of the total solar radiated power, and (b) has a power spectral density similar to the solar power spectral density] is approximately 5,778 K.
(2) For a spherical blackbody of radius 6.96×10^8 meters (the approximate radius of the sun) at a uniform surface temperature of 5,778 K, (a) the total radiated power is approximately 3.85×10^26 Watts, and (b) the radiated power density at a distance of 1.5×10^11 meters from the center of the blackbody (the approximate distance between the center of the Sun and the center of the Earth) is approximately 1,367 Watts per square meter.
(3) If the center of a sphere of radius 6.44×10^6 meters (the approximate radius of the Earth) is placed at a distance of 1.5×10^11 meters from the center of the Sun, to a good approximation the “effective absorbing area” of that sphere for blackbody radiation from the Sun is 1.3×10^14 square meters; and hence the solar power incident on the effective absorbing area of the sphere of radius 6.44×10^6 meters is approximately 1.78×10^17 Watts (1.3×10^14 square meters x 1,367 Watts per square meter).
(4) If the sphere of radius 6.44×10^6 meters absorbs electromagnetic energy with an “effective absorption albedo” of 0.3, then the solar power absorbed by the sphere is 1.25×10^17 Watts [1.78×10^17 Watts x (1 – 0.3)].
(5) A spherical blackbody (i.e., a spherical body whose surface radiates like a surface having an emissivity of 1) of radius 6.44×10^6 meters and at a temperature 254.87 K (hereafter rounded to 255 K) will radiate energy at the approximate rate of 1.25×10^17 Watts.
(6) If independent of the direction of energy incident on a sphere, the surface temperature of the sphere at any instant in time is everywhere the same, then the sphere possesses the property of perfect-thermal-conduction. Thus, for (a) an inert (no internal thermal energy source) perfect-thermal-conduction spherical body of radius 6.44×10^6 meters and uniform surface temperature 255 K whose center is placed at a distance of 1.5×10^11 meters from the center of an active (internal thermal energy source) spherical blackbody of radius 6.96×10^8 meters and uniform surface temperature 5,778 K, and (b) the inert perfect-thermal-conduction spherical body (i) absorbs electromagnetic energy with an effective absorption albedo of 0.3, and (ii) radiates electromagnetic energy with an emissivity of 1 then the perfect-thermal-conduction inert spherical body at temperature 255 K will be in radiation rate equilibrium with the active spherical blackbody at temperature 5,778 K. If the phrase “inert perfect-thermal-conduction spherical body of radius 6.44×10^6 meters” is replaced with the word “Earth,” and the phrase ” active spherical blackbody of radius 6.96×10^8 meters and uniform surface temperature 5,778 K” is replaced with the word “Sun”, it can be concluded that: If (a) an “Earth” at temperature 255 K is placed at a distance of 1.5×10^11 meters from the “Sun” and (b) the “Earth” (i) absorbs electromagnetic energy with an effective absorption albedo of 0.3, and (ii) radiates energy with an emissivity of 1, then the “Earth” will be in radiation rate equilibrium with the “Sun.” For the above conditions, the temperature of the “Earth” in radiation rate equilibrium with the “Sun” will be 255 K.
This completes the algorithm that I believe is commonly used to arrive at an “Earth’s characteristic-emission temperature” of 255 K, and hence is used to compute the 33 K temperature difference.
Even ignoring the facts that (1) it is incorrect to use the “average surface temperature” when computing radiation energy loss from a surface, and (2) in the presence of an atmosphere, (a) the blackbody radiation formula may not apply, and (b) blackbody radiation from the surface of the Earth is not the only mechanism for Earth energy loss to space (the atmosphere even without greenhouse gases will be heated by conduction from the Earth surface and both conduction and convection will cause that thermal energy to be distributed throughout the atmosphere, and the heated atmosphere will also radiate energy to space), the problem with using the 255 K temperature computed above to determine the difference between the Earth’s temperature with and without greenhouse gases is that the effective Earth absorption albedo of 0.3 used to generate the 255 K temperature is in part (mainly?) due to clouds in the atmosphere, and atmospheric clouds are created from water vapor, which is a greenhouse gas.
Thus an effective absorption albedo of 0.3 is based on the presence of a greenhouse gas–water vapor. It is illogical to compute a difference between two temperatures both of whose values are based on the presence of greenhouse gases and then claim that temperature difference represents the temperature difference with and without greenhouse gases. Without water vapor, there won’t be any clouds as we know them. Without clouds, the effective absorption albedo of the Earth will likely not be 0.3, and hence without the greenhouse gas water vapor, the Earth’s surface temperature in the absence of greenhouse gases is likely to be something other than 255 K. Thus, the 255 K “Earth characteristic-emission temperature” as computed using the algorithm above is NOT relevant to a discussion of the Earth surface temperature difference for an atmosphere that does and an atmosphere that does not contain greenhouse gases. Only if 0.3 is the effective absorption albedo of the Earth in the presence of an atmosphere devoid of all greenhouse gases is it fair to claim the presence of greenhouse gases increases the temperature of the Earth by 33 K.
Because clouds reflect a significant amount of incoming solar power, without water vapor I believe the effective absorption albedo of the Earth will be less than 0.3. If true, then more of the Sun’s energy will be absorbed by an Earth whose atmosphere is devoid of greenhouse gases than by an Earth whose atmosphere contains clouds formed from the greenhouse gas water vapor. This implies a higher Earth surface temperature in the absence of water vapor than the “Earth’s characteristic-emission temperature of 255 K”.
For an effective absorption albedo of 0, the temperature of the Earth in radiation rate equilibrium with the Sun will be approximately 278.64 K (hereafter rounded to 279 K). If this value is used as the Earth temperature in the presence of an atmosphere devoid of greenhouse gases, then it can be argued that the presence of greenhouse gases introduces a warming of approximately 9 K (288 Kelvin minus 279 K).
In summation, using the simplified arguments that I believe are also used to arrive at the 33 K temperature difference (i.e., assumed perfect-thermal-conduction Earth, blackbody Earth emission, greybody Earth absorption with an effective absorption albedo between 0 and 0.3, and ignoring atmospheric radiation to space for an Earth atmosphere devoid of greenhouse gases), I conclude the presence of greenhouse gases in the Earth’s atmosphere increases the Earth’s temperature by somewhere between 9 K and 33 K. Thus, I believe the claim that the presence of atmospheric greenhouse gases increases the temperature of the Earth by 33 K is based on an argument that has little relevance to the Earth’s temperature in the presence of an atmosphere devoid of greenhouse gases; and hence at best is misleading and at worst incorrect.
Note: Upon first publication – the guest author Reed Coray was accidentally and unintentionally omitted.
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Yes your analysis is correct.
The albedo of the Earth surface is likely to be even less than the 0.12 value for the Moon.
Reason, the 70% of Earth surface covered by water.
Without the greenhouse gases the extra radiation reaching the surface will largely be short wave infra red(slightly longer than visible red down to 5um).
For this EM region the albedo must be getting near zero.
“…the sphere possesses the property of perfect-thermal-conduction….”
This assumption of perfect thermal conduction for the earth without greenhouse gasses is at least as bad as the albedo assumption. I just looked up the average surface temperature of the moon. Without any atmosphere at all the average lunar surface temperature found after a 2 minute search of the web (so take this value for what it’s worth) is -9 degrees Fahrenheit. The moon is about the same distance from the sun as the earth, so this value could be regarded as a ballpark estimate for what you want to know.
The original post is here …
http://wattsupwiththat.com/2011/12/05/monckton-on-sensitivity-training-at-durban/
Monckton’s comment is here …
http://www.scienceandpublicpolicy.org/
Who wrote this post?
There are many omission in the +33°K warming calculation.
The direct warming of the atmosphere from direct sunlight is ignored as well as the ocean warming from direct sunrise and the way how the seas get their average temperature, through enthalpy & radiation.
The snowball earth theory without CO2 makes no sense as it ignores the 1365 W/m2 direct solar radiation on a spherical surface in rotation not 1/4 of it on a hypothetical flat world.
Greenhouse is only a fraction of the 33° and no earth energy budget ignoring the oceans has any scientific value.
It is used as it is a useful tool for the CAGW scare.
There are strong arguments to suggest that the atmosphere if anything cools us rather than heats us. Certainly, if we did not have an atmosphere day time temperatures would be a lot hotter than we experience. Night time temperatures may be cooler but the extent to which this would be the case depends upon the heat capacity of the liquid storage reservoirs which cover approximately 3/4 of the surface area of the planet..
The average temperatrure of the globe may be out by several degrees, We do not accurately know its albedo nor its emissivity (which are not constants) and the oceans (which themselves are in effect a greenhouse gas albeit in liquid form) is a vast heat storage reservoir. The earth is not a blackbody and I consider that these fundamental problems make the entire BB assessment unreliable.
Further, I consider that it is misconceived to view the earth’s surface temperature as measured today as the base against which the comparison should be made. We are in the midle of an interglacial period and the earth’s surface is untypical warm. The oceans have an average temperature of about 4 degC and this low temperature will come back to bite. Some weighting should be given to take account that over geological time the average surface temperature of the earth is far less than we are privileged to enjoy today.
What if your calculation (or the accepted BB calculation) was performed 20,000 years ago? Many of the fundamentals would be the same and yet in this scenario, the cAGW crowd would have one believe that GHGs were raising the temperature by only about 23K (not 33K).
In posing that question I am aware that ice extent will have changed the albedo, but may be it was less cloudy elsewhere which would tend to offset the change in albedo made by ice.
I would not be surprised if in the real world GHGs add only a few degrees. It is the presence of water and its properties that keep us warm.
From the last few sentences one has therefore to conclude that the climate sensitivity to an additional greenhouse gas must be less than “previously thought”. Or in other words, CO2 adds only 9 K instead of 33 K. Am I right?
OK, that seems like a pretty silly argument. Take a look at Mars. I think Mars would be pretty close to earth without ghgs. No oceans, no vegetation, no ice no animals, just lots of red dirt. But Mars does not have an albedo of 0, it has an albedo of 0.25 – not too disimilar to that of Earth, and certainly a value which will get you closer to 255K than 279K.
As for the assumption of a perfectly thermally conducting Earth, if you abandon this, you will lower the average temperature. Because of the T^4 dependancy of outgoing radiation, having one part of earth at (say) 245 and another of equal area at 265 will result in more outgoing radiation than having it all at 255. Therefore to achieve the same amount of outgoing radiation, the average temperature needs to be lower than 255.
Anyway, well done for writing this, but make a bit more effort to think things through.
This is yet another example of a set of calculations that simply ignores the role of sunlight into the oceans
Since most of the Greenhouse Gases are in fact water vapour the whole exercise is meaningless unless one removes the oceans and all surface water too.
Some warming proponents have suggested that without non condensing GHGs (all those other than water but especially CO2) then in that situation all the water on Earth would freeze. That is their way of attributing ultimate importance to the non condensing GHGs.
However I don’t see that as likely. Sunlight would still penetrate the oceans and water would still evaporate. The water cycle might be slower but it would still be present would it not ?
To my mind this issue goes to the heart of AGW theory which assumes that the Earth is about 33C warmer than it ‘should’ be as a result of GHGs.
What I would like to know is how they feel able to exclude the oceans in setting that temperature. The liquid in the oceans is a vastly more effective system for slowing down the exit of solar energy from the Earth system than are all the GHGs present in the air even including water vapour.
As far as I know Arrhenius et al only considered the atmosphere. They did not consider the role of the oceans so in order to maintain AGW theory it is essential to propose that it is only the non condensing GHGs that keep the oceans liquid.
But that is not the case. It is solar shortwave penetrating the ocean surface that keeps the oceans liquid, creating water vapour in the process and thus making the Earth’s atmosphere warmer than it otherwise would have been.
That solar shortwave into the oceans and the retention of much of that energy for a lengthy period of time is the primary mechanism for slowing down the loss of that energy back to space. Vastly greater than the slowing down caused by GHGs.
It is the atmospheric pressure that dictates the baseline energy content that the oceans can achieve because that pressure determines the energy cost of evaporation by fixing the energy value of the latent heat of evaporation. Namely the difference between the amount of energy required to break the bonds between water molecules relative to the energy required by the breaking process. At current atmospheric pressure the latter is about 5 times the former which is why evaporation is a very powerful net cooling process.
GHGs do NOT affect that baseline energy content for the oceans because they do not significantly affect atmospheric pressure and therefore they can only affect the rate of energy flow from surface to space. Therefore GHGs can only result in a faster water cycle and NOT a change to the baseline energy content for the system as a whole.
The rate of energy flow from surface to space at any given moment is reflected in the relative sizes, intensities and latitudinal positions of the permanent climate zones. It is the shifting of those zones that we perceive as climate change and NOT any significant change in total system energy content.
Surely if you want to determine the effects of the sun on a body at the distance of the earth with no atmosphere, just look to the moon.
I’ve mentioned before that I thought this computation was missing from the literature. I believe you’ve made a great start. However, there is more to the question than just albedo.
The surface of a bare Earth would still have some albedo, but even more important, an atmosphere of O2 and N2 would still absorb and radiate some energy. There would also be some conduction from the surface and 100% of the surface area would be land. Since this kind of atmosphere would NOT radiate very effectively due to its low emissivity, it’s important to determine how much heat would get trapped and what the equilibrium temperature would be.
This is painful to read. You seem ignorant of many things. You even start by saying you don’t know what is meant by a characteristic emission temperature. I mean, that’s not particularly clear language but anyone with a basic physics education knows what is meant by that. You wrongly claim that you can’t use an average temperature to calculate radiation losses. And you have used nearly 2000 words simply to say that you don’t agree with the albedo commonly used in this simplified calculation.
The calculation neglects the greenhouse effect; it doesn’t neglect the reflective properties of the atmosphere. This is because we wish to estimate the magnitude of the greenhouse effect, so we only take that out.
And not forgetting that our EARTH HAS NO LID.
Then, the question arises: Where is heat saved?
Alternatives:
1) Atmosphere: Air: Volumetric heat capacity: 0.00192 joules /cu-cm.
2) Oceans: Water: Volumetric heat capacity:4.186 joules/cu-cm, i.e., 3227 times than that of Air.
3) Soil: Ground: volumetric heat capacity: About 2.0 joules /cu-cm.
Green House Effect = Confined Heat Effect
No confinement = No effect.
Remember: How soon atmosphere cools down during an eclipse.
Stephen Wilde says:
December 26, 2011 at 4:23 am
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I am with you on much of what you have said.
On several occassions I have had arguments with Willis regarding the oceans freezing. He considers that without GHGs the oceans would freeze. I say to him, he is basing that claim on average conditions and one needs to look not at the average condition but rather to look at the considtions prevailing over the tropics. In the tropics, there is enough sloar energy to ensure that the oceans do not freeze at the tropics. The heat in the tropical ocean is then circulated poleward. May be the circulation pattern and extent would be different to that observed today but the material point is that the tropical ocean would not freeze and some of its heat would be distributed elsewhere.
The key to Earth’s climate is the oceans, the properties of water and the water cycle.
Stephen Wilde nails it when he says:
“Since most of the Greenhouse Gases are in fact water vapour the whole exercise is meaningless unless one removes the oceans and all surface water too.”
Further, I don’t think the 255K number is supposed to represent the surface temperature of the earth without greenhouse gases. It is supposed to represent the temperature of the earth without the greenhouse EFFECT. If that is the case, then the logic presented to arrive at 255K seems quite reasonable.
(I’d rather eat bark than admit that the IPCC et al got anything right, but in this case, I think they did)
Try this for simplicity:
incoming total radiation 1364 W/m^2 on a disk with a radius the same as earth.
Since a sphere has a four times larger area than a disk divide by 4.
30% albedo: 0,7 x 1364 / 4 = 239 W/m^2 average incoming radiation.
Stefan Boltzmann (SB) then gives the infamous 255K
More realistic seems to divide the incoming radiation over half the sphere, as on the real earth.
The fourth power in SB then has an interesting effect:
0,7 x 1364 / 2 = 477 W/m^2
SB now gives 303K !!!!
So half the earth receives on average enough energie to reach 303K (30 C)
Since the average temp is lower 288K ? (15 C), this surplus is stored and released at the nightside of the earth. Seems very plausible that this results in an average temp of 288K.
No greenhouse effect needed.
One must also understand that there is in inherent temperature due to gravity. Gravity pulls a non greenhouse atmosphere down and this produces a pressure. The ideal gas law can be used to approximate this inherent temperature. This effect occurs on any gravitating body with any type atmosphere and is already neglected in the calculations.
who is the author of this? the article only says “news staff” and it isn’t signed either?
It surprises me when people believe 5x10ee18kg of cold, rarefied atmosphere can heat 1350x10ee18kg of water. You can heat a material with a large thermal mass (like water) with a material with a small thermal mass (like a gas), but never if the gas is colder than the liquid. This is an intelligence test.
Imagine a gallon of water on a table in front of you. There are lots of methods that can be used to increase its temperature by 10% (33C). However, which of these methods are at work in the atmosphere?
A thought experiment. If the earth had an albedo of 0 and you could put whatever imaginary layer you wanted around it (with physical properties, i.e. you can’t use one-way glass that doesn’t exist in the real world). What is the maximum temperature?
Could you get the earth to have a higher temperature than a theoretical perfect black-body? If so, does that seem logical?
John Brookes (December 26, 2011 at 4:18 am)
(245+265) / 2 = 255
sqrt( sqrt( ((245^4)+(265^4)) / 2 ) ) = 255.5864
Not what I was expecting.
Not only is the Earth not a black body but since oceans cover about 7/10th of the surface, a much smaller area than half of a sphere centered directly under the sun can even have the effect of absorption of incoming radiation as clearly shown in this simple experiment in relation to the specific critical angle:
I can find no reference of this in “climate science” and would appreciate any links someone might have, where this is taken into account or, whether this is calculated within the albedo ~0.30 figure used within such calculations used in this post.
“… and the heated atmosphere will also radiate energy to space)”
No, it won’t. When you start with the assumption that there are no greenhouse gases in the atmosphere, then the remaining gases (oxygen and nitrogen in this case) will neither absorb nor radiate in the relevant micrometer range, and hence can’t radiate energy to space. They are not a blackbody.
I am puzzled as to what exactly will happen in such an atmosphere: we will get an adiabatic temperature profile, we will get convection and redistribution of heat through circulation (“wind”), but all heat removal from earth must come from radiative emission from the earth itself, not from its atmosphere. The albedo will be determined solely from the earth’s surface, i.e. its various soils, as the atmosphere will not be radiating.
At the end, this heat redistribution on the globe will possibly result in a justification of the assumption of a single, global temperature which determines the radiative heat loss of the earth.
Or not?
steveta_uk says:
December 26, 2011 at 4:24 am
Surely if you want to determine the effects of the sun on a body at the distance of the earth with no atmosphere, just look to the moon.
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Two problems. First, the length of the lunar day. Eg., what would be the average temperature of the moon if a lunar day was only one hour rather than about 27;5 days? This becomes very important when considering the second problem. Second, the moon is a rocky world without the heat storage reservoirs found on Earth (ie., it does not have the equivalent thermal ocean reservoirs).
On Earth, the oceans absorb solar irradiance during the day and then slowly release this at night. The Earth’s rotation of 24 hours playing an important role in this. There is long enough daylight for the oceans to respond to the warmth of the sun, but not too much darkness to permit them to fully convcet/conduct and radiate away the heat so absorbed. If the Earth rotated on its axis once every 100 years, the conditions on Earth would be radically different even though we would still be receiving the same amount of solar energy input.
The Earth is a water world and cannot be likened to a rocky world. As far as I understand matters BB calculations do not work properly for the moon nor for any of the planets in the solar system and the Earth (due to its watery nature) is far more complicated.
I consider the entire rational behind the BB calculations for the Earth to be misconceived . .
I think were on the right track here. First, we need to consider the Earth with no oceans and an atmosphere just like the current one without GHGs. Then, we need to add back in the oceans but still keep the GHGs out of the atmosphere. Yes, I realize this is unphysical, but I think in doing those calculations we could learn a lot.
I think it’s very possible the first situation would have a surface temperature greater than what we currently have and what it would be in the second situation. I suspect, as others have indicated, that water cools the planet.
“If I am correct, then the use of 255 K in the computation of the Earth surface temperature difference with and without greenhouse gases is invalid.”
Yes of course it is. It’s the earth without an atmosphere or ocean and an albedo of zero. The average temperature of the moon is 250K. It has an albedo of 0.15.
Ask the fraudsters sometime to take the ocean out of their climate models and see what happens. The global ocean is where most of the greenhouse effect happens. Water is quite transparent to visible light and quite opaque to infrared. It’s more opaque than water vapor and far more opaque than CO2. Transparency to visible light and opacity to infrared are the properties which distinguish greenhouse gases from non-greenhouse gases. Since gases are technically fluids and liquid water has the requisite properties for greenhouse warming one can then quickly realize that greenhouse heating of the ocean is what does most of the surface warming above blackbody temperature. Sunlight penetrates the ocean to some 100 meters where it absorbed along the way by impurities. All the energy in the sunlight is transferred to the water at the speed of light. However because water is opaque at all far infrared frequencies, unlike CO2 which is opaque only in narrow bands, none of that energy absorbed at depth can escape radiatively – it must be mechanically transported to the surface. That delay in transport, whether it’s water or CO2 doing the delaying, is what causes the so-called greenhouse effect. The ocean, you see, has many times the mass of the atmosphere and thus many times the greenhouse effect. Adding insult to injury the ocean doesn’t absorb any significant energy from downwelling infrared from greenhouse gases because the LWIR energy is completely absorbed in a skin layer just a few micrometers deep which doesn’t mix downwards but rather evaporates and carries the LWIR energy immediately away as latent heat of vaporization.