Guest Post by Ira Glickstein
Solar “light” radiation in = Earth “heat” radiation to Space out! That’s old news to those of us who understand all energy is fungible (may be converted to different forms of energy) and energy/mass is conserved (cannot be created nor destroyed).
My Visualizing series [Physical Analogy, Atmospheric Windows, Emission Spectra, and Molecules/Photons] has garnered almost 2000 comments, mostly positive. I’ve learned a lot from WUWT readers who know more than I do. However, some commenters seem to have been taken in by scientific-sounding objections to the basic science behind the Atmospheric “Greenhouse Effect”. Their objections seemed to add more heat than light to the discussion. This posting is designed to get back to basics and perhaps transform our heated arguments into more enlightened understanding :^)
As I’ve mentioned before, during my long career as a system engineer I’ve worked with many talented mathematical analysts who always provided precise results, mostly correct, but some precisely wrong, usually due to mistaken assumptions. I got into the habit of doing a “back of the envelope” calculation of my own as a “sanity check” on their results. If their results matched within reasonable limits, I accepted them. If not, I investigated further. In those days my analysis was really done using a slide rule and scrap paper, but I now use spreadsheets.
The graphic above is based on an excellent spreadsheet from http://serc.carleton.edu/files/introgeo/models/mathematical/examples/XLPlanck.xls. It uses Planck’s Law to calculate the black body radiation spectrum from the Sun, as observed at the top of the Earth’s Atmosphere. It also may be used to calculate the radiation spectrum from the Earth System (Atmosphere and Surface, see below for explanation) at any assumed temperature. (I will refer to this spreadsheet as “Carleton” in this posting.)
I modified the Carleton spreadsheet to compute the mean Solar radiation per square meter absorbed by the Earth System, which turns out to be 240 Watts/m^2. I then used the spreadsheet to determine the effective mean temperature of the Earth System that would emit an equal amount of energy to Space, and that turned out to be 255 Kelvins (-18ºC which is 1ºF).
Since the mean temperature at the surface of the Earth is 288 Kelvins (+15ºC which is 59ºF), that leaves 33 Kelvins (33ºC which is 58ºF) to be accounted for. Guess how we acount for it?
The yellow curve (above left) shows that Solar radiation is in a tall, narrow “shortwave” range, from about 0.1μm (microns, or millionths of a meter) to about 4μm, which we call ultra-violet, visual, and near-infrared. The vertical axis is Intensity of the radiation, measured in Watts/m^2/μm, and the horizontal axis is Wavelength, measured in μm. If you divide the area under the yellow curve into vertical strips, and add up the total area, you get 240 Watts/m^2.
Since we humans sense the visual portion of this radiation as “light”, that is the name we give it, and that has led to the false assumption that it contains no “heat” (or “thermal”) energy.
The violet curve (above right) shows that, assuming a mean temperature of 255 K, Earth System radiation to Space is in a squat, wide “longwave” range, from about 5μm to beyond 40μm, which we call mid- and far-infrared. If you divide the area under the violet curve into vertical strips, and add up the total area, you get the same 240 Watts/m^2 as is under the yellow curve.
DETAILED EXPLANATION
The graph on the left shows the actual observed Solar radiation spectrum (in red) as measured at the top of the Atmosphere. It is superimposed on a black body model (in blue) showing very good correlation. Thus, while the Sun is not exactly a black body, it is OK to assume it is for this type of “sanity check” exercise.
If you calculate the area under the curve you get about 1366 Watts/m^2. That means that a square meter of perfect black body material, held perpendicular to the Sun, would absorb 1366 Watts.
However, the Earth is not a perfect black body, neither is it a flat surface perpendicular to the Sun! So, to plot the yellow curve at the top of this posting, I had to adjust that value accordingly. There are two adjustments:
- The Earth may be approximated as a sphere, with the Sun shining on only half of it at any given time. The adjustment factor for this correction is 0.25.
- The albedo (reflectiveness) of the Earth system, primarily clouds and light-colored areas on the Surface such as ice, causes some of the Solar radiation to be reflected back out to Space without contributing any energy to the Earth System. The adjustment factor for this correction is 0.7.
After applying these adjustments, the net Solar energy absorbed by the Earth System is 240 Watts/m^2.
The graph on the right shows the black body model for an Earth System at a mean temperature of 255 K, a temperature that results in the same 240 Watts/m^2 being emitted out to Space.
Of course, the Earth System is not a perfect black body, as shown by the graph in the upper panel of the illustration below, which plots actual observations from 20 km looking down. (Adapted from Grant Petty, A First Course in Atmospheric Radiation, Figure 8.2, http://www.sundogpublishing.com/AtmosRad/Excerpts/index.html.)
The actual measured radiation is the dark squiggly curve. Note that it jigs and jags up and down between the topmost dashed curve, which is the black body spectrum for a temperature of 270 K and a lower dashed curve which is the black body spectrum for 230 K. This data was taken over the Arctic, most likely during the daytime. The Petty book also has a graph looking down from over the Tropical Pacific which ranges from 300 K down to 210 K. Observations will vary by tens of degrees from day to night, summer to winter, and Tropical to Polar.
However, it is clear that my result, based on matching 240 Watts/m^2, is within a reasonable range of the true mean temperature of the Earth System as viewed from Space.
NOTE ABOUT THE ABOVE ILLUSTRATION
WUWT readers will notice some apparent inconsistencies in the graphs above. The top and bottom panels, from Petty, peak at 15μm to 20μm, while the purple, blue, and black curves in the middle panel, and the Earth System curves from the Carleton spreadsheet I used (see above) peak in the 9μm to 11μm range. Also, the Petty black body curves peak at a “Radiance” around 100 mW/m^2/sr cm^-1 while the black body curves from Carleton peak at an “Intensity” of around 14 W/m^2/μm. Furthermore, if you look closely at the Petty curves, the labels on the black body curves are mirror image! What is going on?
Well, I know some of the reasons, but not all. (I hope commenters who are more fluent in this than I am will confirm my explanations and provide more information about the differences between “Radiance” and “Intensity”.) I have Googled and Wikied the Internet and am still somewhat confused. Here is what I know:
- The horizontal axis in Petty’s plots are what he calls “Wavenumber”, increasing from left to right, which is the number of waves that fit into a cm (centimeter, one hundredth of a meter).
- This is proportional to the frequency of the radiation, and the frequency is the inverse of the wavelength. Thus, his plots are the mirror image of plots based on wavelength increasing from left to right.
- The spreadsheet I used, and my previous experience with visual, and near-, mid-, and far-IR as used in military systems, always uses wavelength increasing from left to right.
- So, when I constructed the above illustration, I reversed Petty’s curves, which explains why the labels on the black body curves are mirror image.
- Fortunately, Petty also included a wavelength legend, which I faithfully reproduced, in non-mirror image, at the top of each plot.
But, that still does not explain why the Petty black body curves peak at a longer wavelength than the Carleton spreadsheet and other graphics on the Internet. I tried to reproduce Petty’s blackbody curves by multiplying the Carleton values by the wavelength (μm) and that did not move the peak to the right enough. So, I multiplied by the wavelength again (μm^2) and, voila, the peaks agreed! (I hope some WUWT reader will explain why the Petty graphs have this perverse effect. advTHANKSance!)
ANSWERING THE OBJECTIONS TO BASIC ATMOSPHERIC “GREENHOUSE EFFECT” SCIENCE
First of all, let me be clear where I am coming from. I’m a Lukewarmer-Skeptic who accepts that H2O, CO2 and other so-called “greenhouse gases” in the Atmosphere do cause the mean temperature of the Earth Surface and Atmosphere to be higher than they would be if everything was the same (Solar radiation, Earth System Albedo, …) but the Atmosphere was pure nitrogen. The main scientific question for me, is how much does the increase in human-caused CO2 and human-caused albedo reduction increase the mean temperature above what it would be with natural cycles and processes? My answer is “not much”, because perhaps 0.1ºC to 0.2ºC of the supposed 0.8ºC increase since 1880 is due to human activities. The rest is due to natural cycles and processes over which we humans have no control. The main public policy question for me, is how much should we (society) do about it? Again, my answer is “not much”, because the effect is small and a limited increase in temperatures and CO2 may turn out to have a net benefit.
So, my motivation for this Visualizing series is not to add to the Alarmist “the sky is falling” panic, but rather to help my fellow Skeptics avoid the natural temptation to fall into an “equal and opposite” falsehood, which some of those on my side, who I call “Disbelievers”, do when they fail to acknowledge the basic facts of the role of H2O and CO2 and other gases in helping to keep temperatures in a livable range.
Objection #1: Visual and near-visual radiation is merely “light” which lacks the “quality” or “oomph” to impart warmth to objects upon which it happens to fall.
Answer #1: A NASA webpage targeted at children is sometimes cited because they say the near-IR beam from a TV remote control is not warm to the touch. Of course, that is not because it is near-visual radiation, but rather because it is very low power. All energy is fungible, and can be changed from one form to another. Thus, the 240 Watts/m^2 of visible and near-visible Solar energy that reaches and is absorbed by the Earth System, has the effect of warming the Earth System exactly as much as an equal number of Watts/m^2 of “thermal” mid- and far-IR radiation.
Objection #2: The Atmosphere, which is cooler than the Earth Surface, cannot warm the Earth Surface.
Answer #2: The Second law of Thermodynamics is often cited as the source of this falsehood. The correct interpretation is that the Second Law refers to net warming, which can only pass from the warmer to the cooler object. The back-radiation from the Atmosphere to the Earth Surface has been measured (see lower panel in the above illustration). All matter above absolute zero emits radiation and, once emitted, that radiation does not know if it is travelling from a warmer to a cooler surface or vice-versa. Once it arrives it will either be reflected or absorbed, according to its wavelength and the characteristics of the material it happens to impact.
Objection #3: The Atmospheric “Greenhouse Effect” is fictional. A glass greenhouse works mainly by preventing or reducing convection and the Atmosphere does not work that way at all.
Answer #3: I always try to put “scare quotes” around the word “greenhouse” unless referring to the glass variety because the term is misleading. Yes, a glass greenhouse works by restricting convection, and the fact that glass passes shortwave radiation and not longwave makes only a minor contribution. Thus, I agree it is unfortunate that the established term for the Atmospheric warming effect is a bit of a misnomer. However, we are stuck with it. But, enough of semantics. Notice that the Earth System mean temperature I had to use to provide 240 Watts/m^2 of radiation to Space to balance the input absorbed from by the Earth System from the Sun was 255 K. However, the actual mean temperature at the Surface is closer to 288 K. How to explain the extra 33 K (33ºC or 58ºF)? The only rational explanation is the back-radiation from the Atmosphere to the Surface.
grrr I should’nt use > for average. The correct thing is that
T = Taverage + ∆T
with the condition ∆Taverage =0
Mike McMillan, you seem to need something like these:
Here are a couple of energy budgets using KT97 and TFK09′s data for the overall averages of reflection (albedo) per the papers, irradiance at the surface, and measured TOA LW upwelling radiation. These are somewhat rough, a first stab at a cosine weighted across 24 hour view but they give you a closer idea of what actually occurs with Trenberth and Kiehls’s numbers. You will notice the totals all match what you see on their graphics.
http://i56.tinypic.com/avc5g.jpg : KT97
http://i53.tinypic.com/ir6lie.jpg : TFK09
They need some small adjustment I can see already:
• Need a better estimate of the actual evaporation rate at nighttime compared to evaporation in the daytime.
• The exact time that the minimum and maximum temperatures are reached, on the average globally, might need to be moved 30 minutes or one hour sooner.
• Window radiation is strictly in those spreads gauged by the surface temperature and I don’t know if this is strictly correct since the cloud layer over 62% of the earth needs to be considered, but, the average should be close.
• …
These are built upon what is actually happening in our atmosphere. You will notice the lack of the mention of back radiation for back radiation is a mirage. A figment of you imagination brought to life by the way the Stefan-Boltzmann equation is applied. In a layered gravity held atmosphere there is no real backward movement of net radiation or any other energy for that matter of fact. When the Stefan-Boltzmann equation is applied correctly to each thin layer you will find this to be true.
Also, radiation in the CO2 and H2O main bands which lie outside the radiative window frequencies cannot pass downward to any great distance to any great degree for the atmosphere is nearly completely “black” to these frequencies. Another way to put it is the mean free path (mean distance a radiation will travel before being absorbed) is simply to short in the lower troposphere to allow this imaginary backward propagation from high in the atmosphere to the surface to ever really occur. Both of these GHGs are simply fast and longer reaching conduction of parcels of energy just like conduction but conduction has to move molecule to molecule, this radiation can jump many meters. Above the tropopause these limits are no longer present enhancing the movement of radiation to space.
So, that is why the familiar 390 & 396 W/m^2 huge red arrows of upward LW radiation and the 324 & 333 W/m^2 of downward IR radiation are missing. They do not belong there. However, they are real and can be measured by radiometers at the surface but this radiation that is being measured there is in the bottom few hundred meters or so except within the window frequencies. It is local, very local as to compared to vastness of the atmosphere. They deserve to be mentioned on those famous budget diagrams but would have more honestly portrayed as a footnote.
Mike, fell free to use these to make you contention a bit stronger.
Alcheson;
Since the earth’s core is molten (very hot), there must also be convection of heat from the core to the surface. How much of the extra 33K heat can be attributed to earth’s own heat source? Is it none?… a little?…. or perhaps a significant amount of the 33K?>>>
When I first became interested in climate science that was one of my questions. After much reading and research and my own back of the envelope calculations, the answer came out “insignificant”. Same for radioactive decay of elements in the earth, and I even calculated the energy released by burning of fossil fuels… in the end, all insignificant. Solar rules.
wayne says:May 7, 2011 at 11:56 pm . . .
Thanks, Wayne. Seems to me the atmosphere is more of a blanket than greenhouse, just slowing the outward movement of energy. The temperature rises down here to the point where the increased outward movement once again roughly equals the insolation, not that they’re ever exactly equal.
“The only rational explanation is the back-radiation from the Atmosphere to the Surface.”
Has anyone actually calculated how much energy can be absorbed by the GHG molecules in the air? Radiation is not the only heat transfer mechanism.
I have difficulty rationalizing how 2% of the molecules in the air (H2O and CO2) can heat themselves, the other 98% of the air molecules, and the earth surface, by 33K only through back radiation of IR. The GHGs absorb and emit IR but the other air molecules can’t so the heat transfer has to be through collisions. What is the probability of a non-GHG molecule to collide with an excited GHG molecule compared to the probability of colliding with the earth surface and have direct energy transfer from there? (Equally reasonable, we could have “back-collisions” as well, when an air molecule with higher energy than the earth surface, transfer its excess energy back when colliding). If the energy can be inhibited from instantly going out into space by being absorbed by 2% air molecules then it must also be inhibited by being absorbed by e.g. oceans, and released at a later moment. (Some of the energy is also stored in vegetation through the photosynthesis which uses energy from the sun to convert H2O and CO2 into oxygen and cellulose, but that is a longer time perspective.) As has been pointed out in previous postings, we can’t gain more energy than we receive from the sun, but we can delay its disappearance so the 33K energy that is missing has to originate from the sun anyway. Earth is not at instant equilibrium (never at equilibrium at all), and there are many variables that can affect the energy budget at any instant.
“However, the actual mean temperature at the Surface is closer to 288 K. How to explain the extra 33 K (33ºC or 58ºF)? The only rational explanation is the back-radiation from the Atmosphere to the Surface.”
Suppose you cover all the ocean of the world with plywood floating “floor”.
You can paint the plywood any color- a nice blue, perhaps.
Would this cause any significant change of global temperature?
And would average global temperature increase or decrease?
This “plywood ocean” could be hotter [though depends on type of paint- not necessarily on simply what color pigment it has] during the day. Since we are not sealing the ocean just covering it, the ocean water would still have some evaporation, but still the nite should be cooler.
Of course such thing would cause a massive extinct event and have all kinds wild effects.
But the point is, would it warm the planet or cool it?
And my answer is it would cool it, significantly.
It could increase global average daytime temperatures- especially if you included the area of the plywood oceans and have somewhat lower global average nite time temperatures. But the northern hemisphere [where most earth land mass is] would get colder- mostly nite time temperature. And other land masses would get colder, and the ocean [beneath the plywood] would get colder.
And looking from space and measuring earth’s radiation would indicate that earth has lost a significant part of it’s “greenhouse affect”
[And if painted the “plywood ocean” black, or put asphalt on it, the earth would still cool.]
Ira,
Very detailed explanation. Hard to argue with the fact that when one breaks the data down into that much detail, one gets only one possible conclusion. Either that or all the physics we know that engineers use every day to make planes fly, internal combustion engines that work, air conditioners that cool, furnaces that heat, are wrong. Since the planes they design do fly, the engines work as expected, the air conditioners cool exactly as designed and furnaces heat right to specification, either the physics is right or a few billion calculations based on the physics got it right by pure coincidence. I’ll put my money on the physics!
That said, you can break this down to a much simpler calculation. The sun’s radiance doesn’t get absorbed 100% by earth surface. Some of it is absorbed by the atmosphere. Similarly, the earth’s radiance is not released 100% from the earth’s surface, some of it is released by the atmosphere itself. So the earth surface might be 288K, but as seen from space, any given outward bound photon could have come from the earth’s surface or from the atmosphere. Seen from space, earth is not a “surface” but a sort of fuzzy/murky/semi-transparent sphere. Take any “area” that you can “see” from space, and measure the w/m2 being emitted, and you get a total that is mostly originating from the surface, and some that originated from the atmosphere.
If we assume that your 240 w/m2 is correct, we can use Stefan-Boltzman Law to calculate the “effective” black body temperature of the earth, which if you accept “average” as an argument, would occur not at earth surface, but somewhere between earth surface and the top of the atmosphere.
SB Law being P=5.67*10^-8*T^4
If we use P of 240 watts/m2 we get T of 255.07 degrees K or -17.93 degrees C.
The same logic still applies. Since we can measure what’s going into the system at TOA (top of atmosphere) as well as what is coming out, and the various satellite measurements and experiments like ERBE and others all arrive at a MEASURED value of arounf 240 w/m2, then there is only one possible conclusion. Earth surface is 288K or +15C on average because the atmosphere retains heat that keeps the surface warmer than it otherwise would be. Call it “less cooling” or “back radiation” or “invisible pajamas” but the measurements and the math arrive at the same conclusion. 240 watts/m2 going in =240 watts/m2 going out = -18C. Since the earth surface is +15C, it can only be 33 degrees warmer because the invisible pajamas are making it warmer.
Cherry Pick says:
May 7, 2011 at 8:38 pm
“Energy is not just radiation. Earth System is more complex than that. For example, absorbed radiation energy may result in increased cloudiness that has impact on who much radiation reaches the earth. Ignoring major part of the system does not help in believing your conclusions.”
What you have just used as an example is one that would cause a cooling effect. I’m not sure if that was your intent, while trying to educate Dr. Glickstein.
Good nickname, by the way.
Thanks David. Seems the radioactive decay of elements is estimated to be ~30TW (thought to be ~80% of the earth’s internal heating based on Wikipedia) which would result in about 0.1K of the 33K difference. Wonder how well that value is actually known? Off hand, 30TW doesn’t seem like a lot to keep the earth’s core molten. I might have to spend some more time investigating how that 30TW was derived, and see if it makes sense given the thermal conductivity and temperature gradient of the earth to convince myself as well that it is totally insignificant.
Energy in = Energy out
Only if integrated over period spanning centuries !
Ira
You should read the article below.
It will answer all of your misconceptions about the “greenhouse theory” which in fact is pure fiction.
1. It is true that the radiative equilibrium temperature of the Earth is -18C.
2. This figure is confirmed by Satellite measurement from space.
3. It is NOT true that all the radiation emitted from the Earth surface.
4. The emission is mostly from the cloud level.
5. If we pick say around 5km as the average emission height to space we will be near enough.
6. Now use thermodynamics expressed in the form of the lapse rate.
7. An average moist lapse rate is -6.6K/km.
8. The lapse rate has nothing to do with back radiation as even scienceofdoom admits.
9. Work back down from -18C at 5km to the Earth surface and we get 5×6.6 = 33K.
So now you have an Earth surface temperature of 15C without any “greenhouse effect”.
http://www.ilovemycarbondioxide.com/pdf/Understanding_the_Atmosphere_Effect.pdf
Alcheson says:
May 7, 2011 at 11:54 pm
Since the earth’s core is molten (very hot), there must also be convection of heat from the core to the surface. How much of the extra 33K heat can be attributed to earth’s own heat source? Is it none?… a little?…. or perhaps a significant amount of the 33K? Just wondering.
=====
From memory … about a tenth of a watt per meter squared. Big enough that it should probably be explicitly counted in really serious models, but probably not enough to really affect anything. As I recall, the number is in Wikipedia somewhere.
I’d like to run a little thought experiment here, and ask a question. I’m not pushing an agenda. It’s a very long time since I had the maths to follow the physics above, and I’m too idle to re-learn.
Given a warm radiating surface, and cooler atmosphere of, say, 1 molecule. The surface radiates 10 units of energy out. Nine escape, one unit is trapped by the molecule, and re-radiated back to the surface. Net result is a loss of nine radiation units from the surface. The 2nd law of thermodynamics is satisfied: the cool body is not heating the warm body, merely slowing the cooling rate. Effectively “insulating” the surface (but not in the sense of a “blanket” preventing convection).
Isn’t this the effect of “back-radiation”?
At a quick flick through I don’t think there was much discussion on the fact that the earth is a rotating sphere.
The solar constant for Earth is commonly given at 1366 W/m^2, and to make a simple average of what is absorbed over the surface and atmosphere, and absorption rates etc, (and the outgoing emissions, spread over the entire sphere), when the wattage is proportional to the fourth power of T, introduces a few complications. In short, might I suggest that the effective radiative T of Earth as is popularly touted is uhm questionable.
Gotcha Henry!
He thinks CO2 is like a toxin in the atmosphere!
An amazing toxin which is NOT TOXIC at 380ppm but suddenly BECOMES TOXIC at 400ppm. There in a single post is the fuzzy thinking of the Warmists! Fantastic.
Textbook example of ‘Carbonaphobia’!
Oh and by the way Henry, do bear in mind that every time you breath out you are breathing out that toxic gas…be careful not to inhale too quick after you breath out!!
Intellectual…ha…like shooting fish in a barrel.
JT says:
I am well aware that spontaneous emission is equally likely to be in any direction so that approximately 1/2 will head downward to the surface. However, there is another kind of emission – stimulated emission – and it is massively biased in the same direction of travel as the direction of travel of the stimulating photon.
This is I believe wrong in that you are assigning some kind of kinetic impact of the photon even though you call it stimulated emission. The photon has no mass, so when it is absorbed it is the electromagnetic properties of the photon that is acting on the molecular structure. In an instant the electromagnetic fields of its components are both attracted and repelled depending on the phase polarity of the photon at the time of absorption.
I find wave theory a better example of the action. At rest the molecule is in balance with its competing fields. The magnetic wave is an oscillation of both positive and negative fields. Dependant on it’s phase, the approaching wave will attract all components of the molecule that are the opposite polarity and repel all those the are the same polarity. As the wave progresses, the polarity is reversed and the components that were repelled are now attracted and those that were attracted are now repelled.
The energy of the wave has now been absorbed and for a brief time the molecule has kinetic energy as, notionally the competing fields struggle to reach equilibrium. In doing so the molecules components that are now vibrating disturb the electromagnetic balance of all the surrounding molecules electromagnetic fields which due to their state of equilibrium with the wider body of gas offer resistance but must acquiescent to the force of the field by propagating the energy though out medium.
The above thought bubble only applies to an ideal gas.
The average global surface albedo is about 0.121 to 0.124. The .3 figure includes the cloud contribution. The AR4 model albedos averaged about 0.14 per Roesch (2006), which amounts to about 3W/m^2 globally and annually averaged. This 3W/m^2 of absorbed radiation that the models currently under represent, chiefly from earlier observed spring snow melts while they allegedly “match” the 20th century climate, will be added as the models project the next 100 years, catching up with the snow melt, an addition of heat comparable to the projected CO2 forcing.
Another way to think of the significance of the CO2 concentration is that it is about 2.66 grams per square inch, that is about the mass of a penny. Imagine that penny flattened into a square inch plate, and contiguous over the surface of the earth with all the other pennies. Instead of having the optical properties of copper, this contigous shell is opague to certain characteristic frequencies of CO2. Obviously this is thick enough for several absorption of photons in that range and to allow practically no chance of these photons escaping directly from the surface to space. I don’t see how one can just hand wave this away with no deeper analysis than a dismissive mention of the concentration.
You state that the back radiation has been measured. Well it might, or at least some long wave IR that is more probably short wave IR that has lost energy traveling through the atmosphere. It would be impossible to tell the difference from measurements at the surface. It is also true that if a molecule of any so called ‘greenhouse gas’ adsorbs energy it must, by 2nd law rules, share this energy with the surrounding non ‘greenhouse’ gasses, ie it loses heat through conduction and radiation.
I do not see in your calculations any adiabatic effect which is important. Taking this into account raises the temperature at the surface. Also, when any gas is heated it will convect. Not an easy effect to measure in the laboratory compared to the real atmosphere. Convecting air will cool.
Jupiter, with an atmosphere of mainly hydrogen, has warming at the 1 atmosphere level similar to Earth. So there it must be an adiabatic temperature rise not ‘greenhouse effect’. Solar input out there is much lower than Earth’s.
True, glass does not transmit IR energy but a greenhouse, transmits visible light and so doing allows energy losses inside, warming the interior by visible light. These losses mean that the lower wave light converts to IR which cannot be transmitted out through the glass. This, together with lack of mixing with cooler external air raises the internal temperature. At night the temperature falls rapidly. 2nd law dictates that the higher the temperature difference of two bodies the greater the heat flow between the two bodies. Entropy must increase.
Taking into account the adiabatic effect, which is real and measurable, there is no need for the GHG effect.
The greenhouse effect also has one fatal flaw. The warming in the mid troposphere in the tropics shown by all models has yet to be found in real observable data. I do not care what your calculations show (in reality another model), if this warm area shown by models is not there then the theory falls flat on its face!
Martin Lewitt says:
……Another way to think of the significance of the CO2 concentration is that it is about 2.66 grams per square inch………
Its an odd unit, but it seems much too high;
Ive worked it out in g/cubic inch as 0.000074!
Bob_FJ says:
The solar constant for Earth is commonly given at 1366 W/m^2, and to make a simple average of what is absorbed over the surface and atmosphere, and absorption rates etc, (and the outgoing emissions, spread over the entire sphere), when the wattage is proportional to the fourth power of T, introduces a few complications. In short, might I suggest that the effective radiative T of Earth as is popularly touted is uhm questionable.>>>>
One of… OK, my BIGGEST pet peave. There is no such thing as an average temperature or w/m2 that is all that meaningfull. The rough estimate commonly used in climate circles is 1366 x .5 (for day/night) = 683. 683 x .5 (cuz its a sphere) = 360. 360 x .7 (for albedo) = 250, pretty close to Ira’s 240.
But that of course would be around 500 w/m2 at high noon at the equator, and pretty much 0 for six months at a time at the poles. So despite being relatively cold, the poles actually beam way more energy into space than they recieve from the sun, and the equator, despite getting very high insolation year round (in daytime at least) is a net absorber (the excess being moved away by wind and water currents).
So trying to come up with an average insolation value and an average temperature value is meaningful in terms of the total of each. But since insolation varies with w/m2 and w/m2 varies with T^4… trying to track “warming” by following “temperature” is silly. 1 watt of extra insolation results in a much larger temperature anomaly at the south pole than it does at the equator. So when the IPCC says that the “average” temperature of the earth has increased about 1 degree “on average” since 1880, that is highly misleading.
But if they said about 0.3 degrees at the equator and about 1.5 degrees at the poles, it would sound a lot less alarming. And before anyone starts screaming about the ice caps melting, let’s keep in mind that T^4 applies by latitude as well as season. So to get an even more reasonable expression of that “one degree average” they would need to say something like:
Equator +0.3 degrees all year round.
Poles +0.5 degrees in summer
Poles +2 degrees in winter
In other words, that scary “one degree” is mostly confined to the dead of winter in high latitudes. Sorry “the ice caps are melting we’re all going to drown” crowd, but if the poles go from -40 in the dead of winter to -38, the extra melting would be…. between almost zero and zero. And neither the polar bears nor the seals will notice. Or mind.
I’m surprised that a “skeptical” site such as WUWT would print such nonsense:-
“Answer #2: The Second law of Thermodynamics is often cited as the source of this falsehood. The correct interpretation is that the Second Law refers to net warming, which can only pass from the warmer to the cooler object. The back-radiation from the Atmosphere to the Earth Surface has been measured (see lower panel in the above illustration). All matter above absolute zero emits radiation and, once emitted, that radiation does not know if it is travelling from a warmer to a cooler surface or vice-versa. Once it arrives it will either be reflected or absorbed, according to its wavelength and the characteristics of the material it happens to impact.”
[b]No![/b] The earths surface is already emitting radiation in the Infra Red. It’s atoms and molecules are ALREADY excited sufficiently to be emitting IR. Therefore that IR scattering back to the surface CANNOT warm the surface any further because the atoms and molecules of the ground are already kinetically excited to this level. The IR will merely be scattered back to the atmosphere again- and eventually espace to space. The radiation from a body can ONLY transfer heat to a cooler body. The second law applies also to radiation.
This sentence is true however:-
” Once it arrives it will either be reflected or absorbed, according to its wavelength and the characteristics of the material it happens to impact.”
Yes, and temperature of that surface is one of those characteristics. The radiation, does not need to “know” anything! If the surface is ALREADY emitting radiation that is “hotter” , i.e. more energetic then the “back radiation ” will in effect just be reflected again.
Heat always passes from hot to cold, never the other way round, THERE ARE NO EXCEPTIONS!
Ira Glickstein does not seem to understand how matter interacts with radiation.
“I modified the Carleton spreadsheet to compute the mean Solar radiation per square meter absorbed by the Earth System, which turns out to be 240 Watts/m^2. I then used the spreadsheet to determine the effective mean temperature of the Earth System that would emit an equal amount of energy to Space, and that turned out to be 255 Kelvins (-18ºC which is 1ºF).
Since the mean temperature at the surface of the Earth is 288 Kelvins (+15ºC which is 59ºF), that leaves 33 Kelvins (33ºC which is 58ºF) to be accounted for. Guess how we acount for it?”
We can account for the 33C difference without any need for an imaginary greenhouse effect. I urge everyone at WUWT to read this brilliant paper by physicist Joe Postma ;
http://www.globalwarmingskeptics.info/forums/attachment.php?aid=327
In this paper, Joe accounts for the 33C. The truth is that the radiative temperature of the Earth is measured from space IS -18C. And it is true that the average air temperature near the surface of the earth is 15C. However, if one includes the whole of the atmosphere right up to the edge of space and the surface of the earth in calculating the average radiative temperature of the Earths surface [i]and[/i] atmosphere, then we get to the -18C figure.
There is no discrepancy, there is no need for an imaginary “greenhouse effect”.
Surely the Phlogiston of our times!
….and just for the “cold things can’t heat warm things” crowd, here is a link to Denmark’s DMI which publishes average temperature in the arctic going back to 1958. The green line is the average and the red line is “this year”:
http://ocean.dmi.dk/arctic/meant80n.uk.php
Now take a close look at days 1 to about 60. That’s roughly January 1 to the end of February. The average is flat at about 247 K or -26 degrees C. Think about that for a moment. ITS DARK THE WHOLE TIME! NO SUNSHINE! NO INSOLATION!
Despite which it drops down to around -26C by December… and then just stays there. Why doesn’t it keep cooling off? After all, the atmosphere above the arctic is even colder than the arctic itself… Heck, at 25,000 feet its colder than that even at the equator! So… with 0 watts/m2 from the sun, the arctic nonetheless sticks right around -26 degrees. Hmmmmm…..wonder what keeps it from cooling off more….
Alcheson;
Off hand, 30TW doesn’t seem like a lot to keep the earth’s core molten.>>>
You’re thinking about it wrong way round. The 30 TW (actually I think it is 44 TW but no matter) is what ESCAPES from the molten core to the surface. OK correction, the outer core is molten, the inner core is actually solid. Fact is that dirt and rocks are pretty good insulators. In high latitudes where winter temps might be as low as -40, houses have to be constructed on a foundation that goes below the “frost line”. That line is a few feet below the surface and in even the harshest of winters, the dirt only reaches the freezing temperature by no more than an inch or so deeper than average. Get about 10 feet down and the temperature varies by almost nothing all year round. So figure that molten outer core is insulated by kilometers of dirt and rock…the heat gets out…but slowly.
There’s a decent summary at wikipedia (not the best source for climate info but good enough in this case) http://en.wikipedia.org/wiki/Earth's_energy_budget
Geothermal about 0.08 w/m2
Tidal about 0.006 w/m2
Fossil Fuel consumption about 0.025 w/m2.
Compared to 250 w/m2 from solar… pretty minor.