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.
Robert Stevenson says:
May 24, 2011 at 2:57 pm
Another much hyped GHG is methane; there is only 1 ppm in the atmosphere!
More like 1.7ppmv, absorbs about 2W/m^2 at about 1200-1300 cm-1
Robert Stevenson says:
jae says:
Good God, are you guys blind? How many times do we have to debunk Postma’s nonsense?
http://wattsupwiththat.com/2011/05/07/visualizing-the-greenhouse-effect-light-and-heat/#comment-655686
http://wattsupwiththat.com/2011/05/07/visualizing-the-greenhouse-effect-light-and-heat/#comment-655880
In this thread http://wattsupwiththat.com/2011/05/14/life-is-like-a-black-box-of-chocolates , Postma even came by to try to defend his nonsense after Willis and I tore into it, although he didn’t stick around long.
Further to our discussion concerning IR thermometers, Omega was kind enough to reply and provide links to some of their technical documents. The document which I think most relevant to our discussion is the following one:
http://www.omega.com/temperature/Z/pdf/z063-066.pdf
which describes the design and operation of such as unit. There is also a summary of radiative physics and the equations upon which the technology rests.
I quote from the document as follows:
“A basic infrared thermometer (IRT) design, comprises a lens to collect the energy emitted by the target; a detector to convert the energy to an electrical signal; an emissivity adjustment to match the IRT calibration to the emitting characteristics of the object being measured; and an ambient temperature compensation circuit to ensure that temperature variations within the IRT, due to ambient changes, are not transferred to the final output.”
So when I measure the temperature of a block of ice while standing in the sun at 23C my IR Thermometer is collecting energy emitted from a colder object to a hotter object.
jae says:
(1) Just saying the words “lapse rate” and “ideal gas equation” doesn’t get you around having to explain why the surface temperature of the earth is not affected by having the atmosphere radiating at it.
(2) Just saying the words “lapse rate” and “ideal gas equation” doesn’t get you around having to explain how one could conserve energy if one considers the current surface temperature of the earth and imagines that one had an atmosphere transparent to IR radiation from the earth’s surface.
(3) Just saying the words “lapse rate” and “ideal gas equation” doesn’t get you around having to explain what sets the temperature somewhere in the earth’s atmosphere or at the surface. The lapse rate only determines the slope of the temperature vs. altitude line. You still need to set the value of the temperature at some point along that line in order to determine the temperature at the surface (or anywhere). [Also, the lapse rate only holds in part of the atmosphere…the troposphere…because the adiabatic lapse rate sets a stability limit on the temperature decrease with height. Where the heating of the atmosphere occurs is also relevant in determining what part of the atmosphere has a lapse rate approximately at that stability limit vs what parts have a lower lapse rate or even an increase in surface temperature with height.]
K&T’s numbers aren’t output of models. They are empirical values obtained mainly from satellite data. And, there is plenty of empirical data at every level: There is empirical data on the basic absorption lines of the various atmospheric constituents, there is a wealth of empirical data backing up the basic equations of radiative transfer that are applied in calculating the greenhouse effect in just the same way that engineers and scientists use these equations everyday in other calculations, there is empirical spectra looking both up from the surface of the earth and down from satellites.
(1) You don’t double the backradiation when you double the water vapor. You increase it by some amount but definitely don’t double it.
(2) I don’t know where you get your intuition as to what is a reasonable amount of radiation for the earth to receive at high noon in the tropics. Even with no back-radiation, the “bb temp.” calculated would be 82 C. However, I put “bb. temp.” in quotes because this temperature has very little real meaning: It would mean something in the limit of an earth with no heat capacity and no advection, convection, etc. For the real earth, with a significant heat capacity and significant atmospheric and ocean transport, the one summary number that has meaning is the average of T^4 over the surface of the earth…That is what is going to go into determination of the global surface radiative balance. The local balance will tend to be dominated by these other issues.
Phil says:
‘Where do you get your value of 0.2 emissivity from, ‘
Work it out for yourself Phil its not rocket science.
Another much hyped GHG is methane; there is only 1 ppm in the atmosphere!
More like 1.7ppmv, absorbs about 2W/m^2 at about 1200-1300 cm-1
A new definition of clutching at straws.
jae says:
I don’t think that you radiation freaks truly understand Local Thermodynamic Equilibrium. You seem to simply ignore heat capacity and thermalization–and prefer to look at radiation, in vacuuo. That is simplistic, to say the least.
yah, and they’re still saying it and they won’t stop saying it. it’s the climate catechism.
they won’t say what the phase change of water looks like on IR, tho…lol
50,000 times more heat moved by phase change effects of 1% h20 gas compared to 500ppm co2 in any volume of atmosphere.
and it’s well known that any improvement in heat capacity improves coupling from heat source to sink, not the opposite.
and they wish to imagine that every co2 molecule never touches a neighbor – no kinetic transfer, so they must imagine that the co2 gets hotter than the surroundings?
otherwise, it would be the same temperature as the surroundings and contribute its tiny tiny share of thermal capacity to the transfer of heat from source to sink – which is virtually an infinite sink.
lol – there’s a reason why houses are not insulated with nothing but mirrors, eh?
Phil says:
… over what range of wavelengths does your value of 0.2 apply?
13 to 17 microns.
All this talk about “radiation and back-radiation” has confused my cat. But she is Siamese and may not fully understand the language. Having said that, she is – still pretty good at mathematics.
Now then, – she has worked out, – mathematically of course – that once x units of energy have been transported away from the surface, the surface must cool. – And, furthermore – that if then those x units of energy are transported (by radiation) to atmospheric GHGs that eagerly consume or absorb them, those GHGs must warm. – Well, so far, so good and I agree with her there.
However then her mathematical model shows that if those same x units of energy are radiated back to the “Surface” the surface must warm. – Yes, of course it does, her model shows – but not to a higher temperature than what it was before it lost the energy in the first place.
Shall I give my cat a “Cat-treat” today?
Joel Shore says:
May 25, 2011 at 10:36 am
(2) I don’t know where you get your intuition as to what is a reasonable amount of radiation for the earth to receive at high noon in the tropics.
My heat transfer book shows 1061 W/m^2 for sun over head. SOD has an example from a book showing 700 W/m^2 in Saskatchewan (sp) in July at the surface. That is a measured 700 which would indicate a T of 60 C, but actual T reading was 40 C ,I think. Gases dissipate heat.
http://scienceofdoom.files.wordpress.com/2011/02/hartmann-ch4-extract.png
So day time solar W/m^2 are quite large even at mid latitudes. I though you knew this stuff.
Oh, Joel, where to begin:
“(3) Just saying the words “lapse rate” and “ideal gas equation” doesn’t get you around having to explain what sets the temperature somewhere in the earth’s atmosphere or at the surface. The lapse rate only determines the slope of the temperature vs. altitude line. You still need to set the value of the temperature at some point along that line in order to determine the temperature at the surface (or anywhere). ”
That’s WHY I and some others here are trying to get you to actually read the links we posted that explain the “ghe” without using radiative balances (have you even bothered to LOOK at them, let alone tell us why they are incorrect???). As they explain so much better than I can, the temperature of the Earth’s surface is “set” by the radiation signal at about 5 km, where it is about -18 C (i.e., the so-called bb temp. of Earth). One then “works out” the surface temperature using lapse rate, and lo and behold it is about 15 C–without having to do radiation balances.
Please note that I DO believe the radiation exists; I just don’t think it is the CAUSE of much; but rather an EFFECT.
“(1) You don’t double the backradiation when you double the water vapor. You increase it by some amount but definitely don’t double it.”
Yes, I know. Probably some logarithmic function, eh? But this gets very funny to me when I hear “climate scientists” suggest that there will be all this “positive water vapor feedback” from a putative increase of 3 wm-2 from doubling OCO! You can’t have it both ways, bud.
“(2) I don’t know where you get your intuition as to what is a reasonable amount of radiation for the earth to receive at high noon in the tropics. Even with no back-radiation, the “bb temp.” calculated would be 82 C. However, I put “bb. temp.” in quotes because this temperature has very little real meaning: It would mean something in the limit of an earth with no heat capacity and no advection, convection, etc. For the real earth, with a significant heat capacity and significant atmospheric and ocean transport, the one summary number that has meaning is the average of T^4 over the surface of the earth…That is what is going to go into determination of the global surface radiative balance. The local balance will tend to be dominated by these other issues.”
This makes absolutely no sense to me. My “intuition” on solar radiation comes from here:
http://rredc.nrel.gov/solar/old_data/nsrdb/redbook/sum2/state.html
What would be your guess, looking at this data. 900 wm-2 is probably quite conservative for high noon!
You folks are telling us that you can ADD wattages to arrive at a total, as in the K&T diagram. Now you are saying I cannot add the power from the sun to the power from backradiation to get a new total–and then back-calculate the temperature of a blackbody which would irradiate at that temperature. Again, sir, you cannot have it both ways.
“K&T’s numbers aren’t output of models. They are empirical values obtained mainly from satellite data. And, there is plenty of empirical data at every level: There is empirical data on the basic absorption lines of the various atmospheric constituents, there is a wealth of empirical data backing up the basic equations of radiative transfer that are applied in calculating the greenhouse effect in just the same way that engineers and scientists use these equations everyday in other calculations, there is empirical spectra looking both up from the surface of the earth and down from satellites.”
Baloney. K&T’s numbers are guestimates. You cannot get anywhere near their values using Modtran (averages of various conditions or any single conditions). Is MODTRAN empirical?
The rest of the data you mention are, indeed, empirical, but they do NOT demonstrate the magnitude of any greenhouse effect. They demonstrate that IR-active substances emit IR in very predictable amounts. This information, indeed, does allow many calculations, but not one that proves, from first-principles, that there is a purely radiative greenhouse effect that makes the surface 33 C warmer than it “should be.”
“(2) Just saying the words “lapse rate” and “ideal gas equation” doesn’t get you around having to explain how one could conserve energy if one considers the current surface temperature of the earth and imagines that one had an atmosphere transparent to IR radiation from the earth’s surface.”
If we had an atmosphere completely transparent to IR, we would not likely have LTE, either, so I don’t know what would happen–and it doesn’t matter for the purposes of this discussion. Again, you guys seem to be ignoring all other heat transfer mechanisms other than radiation in your “ghe” explanations. It think that is incorrect.
O.H. Dahlsveen said:
Shall I give my cat a “Cat-treat” today?
Alleyne replies:
Lucky for her your name is not Schroedinger 🙂 The way I understand it, she deserves a Cat-treat. The surface won’t warm up to where it was before, but also the heat loss slows as a result of only the net difference between the energy radiated from the surface and the energy “returned” from the atmosphere being lost from the surface.
Alleyne:
“The surface won’t warm up to where it was before, but also the heat loss slows as a result of only the net difference between the energy radiated from the surface and the energy “returned” from the atmosphere being lost from the surface.”
This is true. However, picture this scenario: If you use a flame to heat the end of a metal rod, the heat will at first travel toward the other end of the rod at a very fast rate. As equilibrium conditions are approached, however, the heat will flow towards the other end at a slower and slower rate. But we don’t call this effect a “greenhouse metal effect.” It is the “delta T effect” present in all equations of thermal change. Of course, in the air heat flows slower from the surface to the layer above as that above layer becomes warmer. But not just due to radiation, but due to “back -conduction” (i.e., kinetic energy transfer between molecules), also. This total focus on radiation is keeping folks from looking at other heat transfer mechanisms.
885 responses to this article, – and we still cannot agree. (or maybe we do agree to disagree) I cannot agree with the GHgT (Greenhouse gas theory) simply because it defies logic.
Once upon a time, when I was the chief engineer on a ship that had the misfortune of suffering “a breakage of a camshaft in one of it’s engines” I reported to the shipping company that no further usage of that engine was possible until ’a (new) renewed camshaft had been installed – I was, promptly treated to a visit from “The Company’s Doctor of Mechanical Engineering” who informed me that my report showed that no fractures of “External Engine Panels” had taken place. He therefore drew the conclusions that none of the parts that formerly made up the composition of the cam-shaft had left the engine and that it would be a (maybe tedious but not impossible) task for my crew to search out, reassemble and reinstall the said camshaft.
Well, I never attained a doctorate, so I could not fix it. – nor can I – to this date, fix it.
Alleyne says on May 25, 2011 at 2:01 pm:
“O.H. Dahlsveen said:
Shall I give my cat a “Cat-treat” today?
Alleyne replies:
Lucky for her your name is not Schroedinger :-)”
What! – You mean Schroedinger had a cat as well?
Alleyne says on May 25, 2011 at 2:01 pm:
“Lucky for her your name is not Schroedinger 🙂 The way I understand it, she deserves a Cat-treat. The surface won’t warm up to where it was before, but also the heat loss slows as a result of only the net difference between the energy radiated from the surface and the energy “returned” from the atmosphere being lost from the surface.”
So Alleyne does that mean that “as far as you can understand it” the surface temperature and my cat’s explanation thereof coincide?
Well, the lucky little beastie is in for two treats tonight
“lol – there’s a reason why houses are not insulated with nothing but mirrors, eh?”
That is a good observation. Why do we not see someone selling highly IR reflective paint to coat houses inside and out if back radiation is so significant? We should be able to focus this with IR mirrors and heat our houses at night, given that it is calculated to be a significant percentage of solar radiation.
So, why do we spend so much to heat our cities at night? Why not just focus the IR back radiation from the atmosphere and boil water to produce steam/electricity, as we do with solar radiation during the day.
We could place mirrors in farms just outside the cities and boil water at night using back radiation and use this steam to produce electricity and heat our cities without any need for batteries. How come engineers are so stupid that they cannot figure out how to do this? Why haven’t climate scientists shown then how it is done? Why are we trying instead to generate power with windmills when there is all this back radiation?
The part of your posting that I highlighted above has generated an example that may convince you that there is, indeed, what you call “back-conduction” equivalent to what we have been calling “back-radiation”.
In my time as a hobbyist and my early years as an electronics engineer, I did a good deal of soldering. When you turn a soldering iron on, an electrical heating element provides heat energy to one end of the soldering iron and the heat, as you describe, flows to the tip, at first very fast and then, as the temperature of the iron stabilizes and the tip reaches a temperature close to that of the heated end, very slowly.
OK, let us do an experiment. Take two identical soldering irons. Touch the tip of the first to a copper penny that is maintained at 15ºC. Touch the tip of the second to a similar penny that is maintained at 0ºC. Now, turn both soldering irons on and make a record of the temperature a half-inch from their tips. The first starts at around 15ºC, heats up fairly rapidly and stabilizes at some temperature “A”. The second starts at around 0ºC, heats up more slowly, and stabilizes at some temperature “B” which is lower than “A”. Right?
OK, so, since both soldering irons have exactly the same energy coming in at the heated end, and are identical in their heat conduction and capacity and so on, how to explain the difference in temperature 1/2 inch from their tips? Why is the first iron temperature higher than the second? I say it is the higher “back-conduction” from the warmer penny as compared to the lesser “back-conduction” from the cooler penny.
What is your explanation?
By the way, I accept that, in the case of both soldering irons, the direction of net heat transfer is from the very hot heating element to the less hot iron to the still less hot penny. But, if the energy flow is only going one way, how does the first tip get the “message” back that it has to be warmer and the second tip get the “message” it has to be cooler?
O H Dahlsveen said on May 25, 2011 at 2:58 pm:
“What! – You mean Schroedinger had a cat as well?”
Alleyne replies:
Yes he did, or does, we can’t be sure 🙂 BTW, I am all too familiar with the opinions people with no field experience come up with, to the point where in the fabrication company I used to work for we had to developed a new material, ‘unatanium’, in order to implement some of the wonderful ideas and designs these people came up with 🙂
It seems to me that so far, having determined that radiative energy can slow the heat loss from the surface, the questions which remain unresolved are:
To what extent does CO2 account for warming? Now I am not denying that it does, but how much warming is the result of the lapse rate and adiabatic heating resulting from the depth of the atmosphere and how much from CO2 and other GHGs “trapping” heat?
Is increased water vapour a positive or negative feedback? More water vapour = more clouds = more albedo seems to be the dominant effect, and therefore a negative feedback.
Is the climate as sensitive as the AGW theory theorizes? It most likely isn’t based on the paleoclimatic record (as I understand it).
Is it worth spending trillions of dollars “decarbonizing” our society based on a very unlikely probability that AGW will have a significant impact on the climate, moreover one that is adverse to our interests? Obviously not given the significant risks such a policy poses.
jae says:
Not only have I read them but I have actually critically analyzed them, which is clearly more than you have.
Yes and yes http://wattsupwiththat.com/2011/05/07/visualizing-the-greenhouse-effect-light-and-heat/#comment-667513 …but it doesn’t seem to matter because you just ignore the explanations and pretend that I haven’t given them.
Why do you repeat this again and again when I have again and again told you what is wrong with it. The reason that the temperature is set by the the radiation from about 5 km is because of the current level of IR-absorbing matter in the atmosphere. If there were more greenhouse gases, the level would be lower and the ground temperature lower; if there were less greenhouse gases, the level would be higher and the ground temperature would be higher. Which part of this is so difficult for you to understand?
What the heck does that mean? Does it mean that we can just ignore conservation of energy because you find it inconvenient to what you want to believe?
There’s no “both ways”. There is nothing inconsistent in this picture.
My point about intuition was not applying to that. It was from saying (to paraphrase), “Oh, my god, the amount of radiation hitting the surface sounds so high. It corresponds to a blackbody temperature of X.” My point was that you have no good intuition as to what constitutes an excessively large amount of radiation hitting the surface. The blackbody temperature isn’t particularly relevant at a single point at the surface because there are lots of different heat transport mechanisms that affect the local surface energy balance and there’s lots of thermal inertia at the surface, particularly the oceans.
Is this really all so difficult for you to understand? I am not saying you can’t add powers. I am just saying that this “blackbody temperature” that you derive from it doesn’t mean a whole lot because it is at one particular point on the surface. At another point on the opposite side of the planet, there is only the radiation from the atmosphere and the “blackbody temperature” you’d calculate is much lower. What is actually relevant in figuring out how much the earth is going to absorb is averaging the powers over the entire planet and what is relevant in figuring out how much it is going to radiate is averaging the temperature over the entire surface (actually T^4…or, most technically, the emissivity*T^4 but the emissivity in the mid- and far-IR is very close to 1 for most surfaces).
No…MODTRAN is a model…which you are likely misapplying if you are getting nowhere close. The K&T numbers are derived from empirical data, mainly satellite data. Yes, there is data analysis involved and there are uncertainties…but if you want to call them “guestimates” for this reason, you’d have to call basically every measurement in the universe a guestimate.
Yes…The 33 C number is in fact derived from empirical values: One compares the actual surface temperature to the surface temperature that would be required to satisfy conservation of energy in the absence of absorption of emitted surface radiation.
We’re not ignoring all other heat transfer mechanisms. We are just saying that radiation alone from the surface would violate energy balance in the absence of IR-absorbing elements in the atmosphere; all that convection and conduction could do is make the violation worse. Between the earth system and space, radiation is all there is. You could try to argue that convection and conduction from the atmosphere to the surface adds enough heat to the surface close the energy balance…but claiming that convection and conduction lead to heat transfer from the colder atmosphere to the warmer surface would REALLY violate the 2nd Law.
Ira:
“The part of your posting that I highlighted above has generated an example that may convince you that there is, indeed, what you call “back-conduction” equivalent to what we have been calling “back-radiation”.”
No argument at all! My point is only that the air also “back-conducts,” not just “back-radiates.” Water is another example where the reasons for a “delta T” involves conduction and no radiation. IMHO, there is way too much emphasis on radiation, to the virtual exclusion of convection and conduction and the simple storage of heat by matter. And this “obsession” with radiative-only explanations tends to focus emphasis just on GHGs and even tiny additions of them (CO2) which cannot possibly have any significant effect, to the exclusion of the immense importance of LTE (thermalization) and lapse rates.
BTW, you are just as quiet as Joel Shore on those articles which explain the 15 C average surface temperature without considering GHGs at all. Do you have no opinion? If these articles are correct, we don’t need a ghe to explain what is going on, so they are worth refuting, if it can be done.
And, Ira:
If you look closely at the K&T diagram and the whole ghg theory, it is not just saying that the ghgs are “slowing the cooling,” which indeed they are. THE THEORY is SAYING that the Sun is providing only 342 wm-2 into the planet, only 168 wm-2 of which is being absorbed by the surface, but SOMEHOW, through backradiation, the surface is capable of emitting 390 wm-2 and increasing in temperature to 15 C. I simply CANNOT accept this, even after about 5 years staring at that diagram. I still think it is nonsense. And I’m not exactly alone.
I admit to being basicly a B student, so maybe I simply cannot think and understand like the pros. But I can still keep asking “the pros” to explain it at a level that I can understand. And to provide some type of empirical demonstration for their theory (no empirical evidence, NO SCIENCE!). They have not done that, IMHO.
Ira:
…
“By the way, I accept that, in the case of both soldering irons, the direction of net heat transfer is from the very hot heating element to the less hot iron to the still less hot penny. But, if the energy flow is only going one way, how does the first tip get the “message” back that it has to be warmer and the second tip get the “message” it has to be cooler?”
Through your new “back-conduction”? LOL! ROTF! I could feel you moving that way all along. /sarc
Seriously, the “message” is through delta T, that simple, just as jae clearly explained, as I and others all up this thread have explained to you. Open a good physics book.
Your example is lacking. If you are “maintaining” the pennies’ temperatures; that could be a fixed energy sinks on the pennies, or, a variable ones. Specify. If fixed, B’s sink must be greater that A’s sink to keep B at 0 ºC and A at 15º Cif the room is warm. See, your example needs no “message” at all. If you turn off the cooling that “maintains” the lower temperatures then they will equalize in the end to the SAME temperature, all calculated by delta Ts.
This is the same problem you are having when viewing the atmosphere. Toss the “back-conduction”, toss the “back-radiation” and look at the problem frontwards from the standpoint of what the energy “sees”, always warm to cooler, always from higher energy density to lower energy density, if Ts are the same, all energy transfer ceases numerically and really, literally. It will do your mind nothing but some good.
It is (emphasized) because it is TRUE.
You have obviously never sat out on a car hood till the late hours (with a member of the opposite sex) have you? And witnessed TRUE radiative cooling of the car’s hood, roof, until they eventually ‘dew up’ … the reason why FOG FORMS (surface radiative cooling).
All your posts re: ‘radiative cooling’ show this deficiency.
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wayne says:
Really? So let’s put the atmosphere and the earth at the same temperature. I agree there will be no net heat flow from the earth to the atmosphere. Will there be no temperature change of the earth? If so, what happens to the energy that the earth is receiving from the sun? Does it just magically vanish?
jae says:
Every time you say this I am going to link to the most recent post I have that discusses it…or links back to a previous post discussing it: http://wattsupwiththat.com/2011/05/07/visualizing-the-greenhouse-effect-light-and-heat/#comment-667782
(Eventually people will have to jump back several posts to get to the actual substance, but it will serve to show how you are just saying this despite the fact that we have already refuted it…multiple times.)