A guest post by Ken Coffman and Mikael Cronholm
In clicking around on the Internet, I found an outstanding paper called Thermodynamics of Furnace Tubes – Killing Popular Myths about Furnace Tube Temperature Measurement written by Mikael Cronholm. The paper was clever and wise…and made a lot of sense. Clearly Mikael knows a lot about infrared radiation and I’m a guy with questions. A match made in heaven?
We exchanged e-mails. I want to be clear about this…Mikael corrected some of my wrong ideas about IR. I’ll repeat that for the slow-witted. Some of my ideas about infrared radiation were wrong. I am considered a hard-headed, stubborn old guy and that’s completely true. However, I want to learn and I can be taught, but not by knuckleheads spewing nonsense and not by authoritarians who sit on thrones and toss out insults and edicts.
Ken Coffman (KLC) is the publisher of Stairway Press (www.stairwaypress.com) and the author of novels that include Hartz String Theory and Endangered Species.
Mikael Cronholm (MC) is an industry expert on infrared radiation, a licensed, level III Infrared Training Center Instructor and holds two Bachelor of Science degrees (Economics and Business Administration).
The following is a summary of our conversation.
KLC: Hello Mikael. I found your paper called Thermodynamics of Furnace Tubes and I found it very informative, practical and interesting. I hope you’ll bear with me while I ask a couple of dumb questions. I am an electrical engineer, so I have some knowledge about thermodynamics of conduction and convection, but not so much about IR radiation. In return for your time, I would be happy to make a donation to the charity of your choice.
If I take an inexpensive IR thermometer outside, point it at the sky and get a temperature reading of minus 25°C, what am I actually measuring? Is there anything valid about doing this?
MC: Just as a matter of curiosity, how did you find my paper? I checked your website and I guess this has to do with the Dragon, no? If you want to make a donation I would be happy to receive that book. If you can, my postal address is at the bottom. I don’t follow the debate more than casually, but I am a bit skeptical to all the research that is done on climate change…it seems that the models are continuously adjusted to fit the inputs, so that you get the wanted output…and they argue “so many scientists agree with this and that”…well, science is not a democracy…anyway…
About radiation, then. There is more to this than meets the eye. Literally!
Looking at the sky with an infrared radiometer you would read what is termed “apparent temperature” (if the instrument is set to emissivity 1 and the distance setting is zero, provided the instrument has any compensation). Your instrument is then receiving the same radiation as a blackbody would do if it had a temperature of -25°C, if that is what you measure. It is a quasi-temperature of sorts, because you don’t really measure on a particular object in any particular place, but a combination of radiation, where that from outer space is the lowest, close to absolute zero, and the immediate atmosphere closest to you is the warmest. (I have once measured -96°C on the sky at 0°C ground temperature.) What we have to realize though, is that temperature can never be directly measured. We measure the height of a liquid in a common thermometer, a voltage in a thermocouple, etc, and then it is calibrated using the zeroth law of thermodynamics and assuming equilibrium with the device and the reference.
KLC: Global warming (greenhouse gas) theory depends on atmospheric CO2 molecules absorbing IR radiation and “back radiating” this energy back toward the earth. If you look at the notorious Ternberth/Keihl energy balance schematic (as shown in Figure 1 of this paper: http://www.cgd.ucar.edu/cas/Trenberth/trenberth.papers/TFK_bams09.pdf ), you see the back radiation is determined to be very significant…more than 300W/m2. From your point of view as an IR expert, does this aspect of the global warming theory make any sense?
MC: The paper you sent me mentions Stefan-Boltzman’s law, but it does not talk about Planck’s law, which is necessary to understand what is happening spectrally. I suggest you read up on Planck and Stefan-Boltzman at Wikipedia or something. Wien’s law would be beneficial as well—they are all connected.
Planck’s law describes the distribution of radiated power from a blackbody over wavelength. You end up with a curve for each blackbody temperature. The sun is almost a blackbody, so it follows Planck quite well, and it has a peak at about 480nm, right in the middle of visual (Wien’s law determines that). The solar spectrum is slightly modified as it passes through the atmosphere, but still pretty close to Planckian. When the radiation hits the ground, the absorbed part heats it. The re-radiated power is going to have a different spectral distribution, with a peak around 10um (micrometer). Assuming blackbody radiation it would also follow Planck’s law.
S-B’s law is in principle the integral of Planck from zero to infinity wavelength. Instruments do not have equal response from zero to infinity, but they are calibrated against blackbodies, and whatever signal they output is considered to mean the temperature of the blackbody. And so on for a number of blackbodies until you have a calibration curve that can be fitted for conversion in the instrument.
That means that the instrument can only measure correctly on targets that are either blackbodies, or greybodies with a spectral distribution looking like a Planck curve, but at a known offset. That offset is emissivity, the epsilon in your S-B equation in that paper. It is defined as the ratio of the radiation from the greybody to that of the blackbody, both at the same temperature (and wave length, and angle…). Some targets will not be Planckian, but have a spectral distribution that is different. If you want to measure temperature of those, you need to measure the emissivity with the same instrument and at a temperature reasonably close to the one you will measure on the target later.
So, of course, the whole principle behind the greenhouse effect is that shorter wavelengths from the sun penetrates the atmosphere easily, whereas the re-radiated power—being at a longer wavelength—is reflected back at a higher degree. I have no dispute about that fact. It is reasonable. So I think the Figure 1 you refer to is correct in principle. My immediate question is raised regarding the numbers in there though. The remaining 0.9 W/m2 seems awfully close to what I would assume to be the inaccuracies in the numbers input to calculate it. You are balancing on a very thin knifes edge with such big numbers as inputs for reaching such a small one. An error of +/- 0.5% on each measurement would potentially throw it off quite a bit, in the worst case. But I don’t know what they use to measure this, only that all the instruments I use have much less accuracy than that. But with long integration times…well, maybe…but there may be an issue there.
KLC: I am interested in some rather expensive thermopile-based radiation detectors called pyrgeometers (an example is the KippZonen CGR 3 instrument http://www.kippzonen.com/?product/16132/CGR+3.aspx).
If a piece of equipment like this is pointed into the nighttime sky and reads something like 300W/m2 of downwelling IR radiation, what is it actually measuring? If I built a test rig from IR-emitting lightbulbs calibrated to emit 300W/m2 and placed this over the pyrgeometers, would I get the same reading?
MC: “What is it actually measuring?” Well, probably a voltage from those thermopiles…and that signal has to be calibrated to a bunch of blackbody reference sources to covert it either to temperature or blackbody equivalent radiation.
Your experiment will fail, though! If you want to do something like that, you have to look at a target emitting a blackbody equivalent spectrum, which is what the instrument should be calibrated to. IR light bulbs emitting 300W/m2 is simply impossible, because 300W/m2 corresponds to a very low temperature! Use S-B’s law and try it yourself. Like this: room temp, 20°C = 293K. The radiated power from that is 293K raised to the power of 4. Then multiply with sigma, the constant in S-B’s law, which is 5.67*10-8, and you get 419 W/m2 or something like that, it varies with how many decimals you use for absolute zero when you convert to Kelvin. For 300 W/m2 radiation I get -23.4°C at 300 W/m2 when I calculate it (yes, minus!). Pretty cool light bulb.
I don’t know what your point is with that experiment, but if it is to check their calibration you need a lot more sophisticated blackbody reference sources if you want to do it at that temperature. But you could do a test at room temperature though. Just build a spherical object with the inside painted with flat black paint, make a small hole in it, just big enough for your sensor, and measure the temperature inside that sphere with a thermocouple, on the surface. Keep it in a stable room temperature at a steady state as well as you can and convert the temperature to radiation using S-B’s law. You should get the same as the instrument. Any difference will be attributable to inaccuracy in the thermocouple you use and/or the tested instrument. Remember that raising to the power of 4 exaggerates errors in the input a lot!
I hope I have been able to clarify things a little bit, or at least caused some creative confusion. When I teach thermography I find that the more you learn the more confused you get, but on a higher level. Every question answered raises a few more, which grows the confusion exponentially. It makes the subject interesting, though.
Let me know if you need any more help with your project!
KLC: I found your paper because one of the FLIR divisions is local and I was searching their site for reference information about IR radiation. I know what a 100W IR lamp feels like because I have one in my bathroom. If someone tells me there is 300W/m2 of IR power coming from space, and I hold out my hand…I expect to feel it. What am I missing?
MC: Yeah, you put your hand in front of a 100W bulb, but how big is your hand…not a square meter, I’m sure. It is per area unit, that is one thing you are missing. The 100W of the bulb is the electrical power consumption, not the emitted power of the visual light from it. That’s why florescent energy-saving lamps as opposed to incandescent bulbs give much more visual light per electrical Watt, because they limit the radiation to the visual part of the spectrum and lose less in the IR, which we cannot see anyway. The body absorbs both IR and visual, but a little less visual.
And, here is the other clue. Your light bulb radiation in your bathroom is added to that of the room itself, which is 419 W/m2, if the room is 20°C. Your 300 W/m2 from space is only that. You will feel those 300 W/m2, sure. It will feel like -25°C radiating towards your hand. But you don’t feel that cold because your hand is in warmer air, receiving heat (or losing less) from there too.
Actually, we cannot really feel temperature—that is a misconception. Our bodies feel heat flow rate and adjust the temperature accordingly. It is only the hypothalamus inside the brain that really has constant temperature. If you are standing nude in your bathroom, your body will radiate approximately 648 W/m2 and the room 419 W/m2, so you lose 229 W/m2. That is what you feel as being cooled by the room, from radiation only. Conduction and convection should be added of course. The earth works the same way—lose some, gain some. It is that balance that is being argued in the whole global warming debate.
KLC: I still feel like I’m missing something. IR heat lamps are pretty efficient, maybe 90%? Let’s pick a distance of 1 meter and I want to create a one-square meter flooded with an additional 300W/m2. It must be additional irradiation, doesn’t it? That’s going to take a good bunch of lamps and I would feel this heat. However, I go outside and hold out my hand. It’s cold. There’s no equivalent of 300W/m2 heater in addition to whatever has heated the ambient air.
Perhaps I’m puzzled by something that is more like a flux…something that just is as a side-effect of a temperature difference and not really something that is capable of doing any work or as a vehicle for transporting heat energy.
It’s a canard of climate science that increasing atmospheric CO2 from 390PPM to 780PPM will raise the earth’s surface temperature by about 1°C (expanded to 3°C by positive feedbacks). From my way of thinking, the only thing CO2 can do is increase coupling to space…it certainly can’t store or trap energy or increase the earth’s peak or 24-hour average temperature.
Any comments are welcome.
MC: Efficiency of a lamp depends on what you want, if heat is what want then they are 100% efficient, because all electrical energy will be converted to heat, the visible light as well, when it is absorbed by the surrounding room. If visible light is required, a light bulb loses a lot of heat compared to an energy saving lamp. Energy cannot be created or destroyed—first law of thermodynamics.
When you say W/m2 you ARE in fact talking about a flux (heat flow is what will be in W). If you have two objects radiating towards each other, the heat flow direction will be from the hotter one, radiating (emitting) more and absorbing less, to the cooler one, which radiates less and absorbs more (second law of thermodynamics). The amount of radiation emitted from each of them depends on two things ONLY, the temperature of the object and its emissivity. So radiation is not a side effect to temperature, it is THE EFFECT. Anything with a temperature will radiate according to it, and emissivity. (If something is hotter than 500°C we get incandescence, emission of visible light.) Assuming an emissivity of unity, which is what everyone seems to do in this debate, the radiation (flux. integrated from zero to infinity) will be equal to what can be calculated by Stefan-Boltzmann’s law, which is temperature in Kelvin, raised to the fourth power, multiplied by that constant sigma. It’s that simple!
With regard to your thought experiment, it is always easier to calculate what an object emits than what it absorbs, because emission will be spreading diffusely from an object, so exactly where it ends up is difficult to predict. I am not sure where you are aiming with that idea, but it does not seem to be an easy experiment to do in real life, at least not with limited resources.
CO2 is a pretty powerful absorber of radiated energy, that fact is well known. Water vapor is an even stronger absorber. In the climate debate it is also considered a reflector, which probably also true, because that is universal. Everything absorbs and reflects to a degree. So I guess that the feedback you mention has to do with the fact that increasing temperature increases the amount of water vapor, which increases absorption, and so on. But my knowledge is pretty much limited to what happens down here on earth, because that is what matters when we measure temperature using infrared radiation. However, it is important to remember, again, that we talk about different spectral bands, the influx is concentrated around a peak in the visual band and the outgoing flux is around 10 micrometer in the infrared band, and the absorption may not be the same.
With so many scientists arguing about the effects of CO2 I am not the one to think I have the answers. I really don’t know what the truth is. And the problem that all these scientists have is that they will never be able to test if their theories are correct, because the time spans are too long. For a theory to be scientifically proven, it has to be stipulated and tested, and the test must be repeatable and give the same results in successive tests for the theory to be proven.
If not, it is not science, it is guessing.
More like a horoscope…
George E. Smith says:
“So whatever your particles are emitting downwards, they must be emitting a like amount upwards; which escapes to space.”
Actually, the lower atmosphere is mostly IR opaque with a few windows. For most of the spectra were absorption occurs, 99% of the available energy is absorbed within 20 meters of the surface. In the “wings” it requires about 500 meters to absorb 99%. From this, I deduce that very little radiation from the lower troposphere ever gets to space. Based on my analysis of lapse rate plots, it is pretty obvious that, near the surface, the atmosphere is IR opaque and the net radiation is toward the surface. Since it is opaque, the radiation emitted toward space is simply reabsorbed within a few meters. At the tropopause, the atmosphere becomes IR transparent and a significant amount of energy is emitted into deep space.
cba says:
“For the sun to essentially emit a 5800k BB curve from the photosphere, one has many heavier elements present that are ionized and so are capable of emitting a continuum rather than merely a spectrum.”
Actually, the line spectrum is temperature and pressure broadened to produce the continuous spectrum. Heavier elements are not required.
@ur momisugly Matt. That does not sound right to me, as someone who deals mainly with terrestrial stuff. There is the sun, there is a very long bit of near vacuum, there is the atmosphere, there is the earth, and radiation goes from left to right in that sentence. Assuming the radiation from the sun is constant, the near vacuum will be near perfectly transmitting, (leaving the atmosphere out for a while), we come to the earth that will not transmit at all, so it can only reflect and absorb (and re-emit).
Now, are you saying that the transmissivity of the atmosphere will *increase* when there is more CO2 in it? You will have to explain to me how, then!
The way I understand the theory is that the CO2 supposedly lets the shorter wavelength solar radiation through, while blocking the longer wavelengths the earth emits from radiating back into space, and that any possible decrease in the influx is counteracted by an even greater decrease in the outflux. And so the balance is disrupted, according to thetheory. Or have I misunderstood something?
When you say “you should believe them” you begin to sound a bit like a missionary, and that makes me suspicious. Before I believe something I need to understand it.
@ur momisugly Matt. Another thing: “Think for a bit. Is there scientific *proof* that Miami’s climate next winter is going to be warmer than Minneapolis’s?” Well, yes, I think there is, and the point there is that the proof for the theory is repeated cyclically every year. The cycles that are discussed in the climate debate are much longer, if I am not mistaken, which means that there are no chances of repeating the “CO2 test” and see if the same result occurs again.
Cal says:
February 14, 2011 at 8:27 am
I accept that I should have said that the sun radiates at all wavelengths from UV to far infrared, but that the peak is in the visible region. Moreover a lot of the incoming energy in the infra red region is absorbed by CO2 and H2O in the atmosphere and is then re-radiated downwards (and upwards ultimately to space) in exactly the same way as the energy radiated to the surface. UV is also absorbed (particularly by ozone) but the peak energy radiated at the surface is still in the visible region of the spectrum. So at night all the infrared radiation is from CO2 and H2O and during the day the majority is. However the main point I was making was about infra red radiation at night.
But, even if true, peak energy of visible light is not heat. Heat is IR (far and mid).
Take a plant, it absorbs blue and red light from the visible spectrum. It is taking in energy for photosynthesis, it is not taking in heat in these colours, they don’t have any heat. Visible light is not hot.
Both day and night the infrared radiation is from everything to some extent, from the earth, from plants, and depending on how hot these are will be the different IR radiating.
What heat has been absorbed during the day by the earth is radiating out at night, and what is being generated by life is being radiated out at night, as heat in the far and mid IR.
http://science.hq.nasa.gov/kids/imagers/ems/infrared.html
So, it’s not true that all the IR radiated out at night is from H2O and CO2, no more is that true that the majority of IR is from these during the day.
P Wilson says:
February 14, 2011 at 8:35 am
actually, watervapour above this level acts as an absorber of radiation. the so called greenhouse effect”
Water vapour is lighter than air, it absorbs heat radiation and transports it up and away from earth, where it cools and comes down as rain. The greenhouse effect is keeping the earth cool enough for life overall. Compare deserts with hot areas with available water.
Our whole atmosphere is a greenhouse, both cooling and warming via greenhouse gases, predominantly water vapour.
Without an atmosphere the earth would bake during the day and freeze at night, the average surface temperature would be -18°C (0°F); the average temperature of our earth is 15°C (59°F).
With our atmosphere but with little or no water the temperature of the earth would be 67°C.
Sorry, I’m being horribly distracted. Missed a close italics before “Water vapour is lighter than air”.
@Myrrh. Oh yes, visible light will cause a temperature rise, hence heat an object, as it is absorbed. Google “Herschel experiment” and I am sure you will find food for thought!
Another proof for that is that objects with different colors will absorb differently in the visual, white the least, black the most, and other colors in between. In the IR, the absorptivity (or emissivity) will NOT depend on color. And this is definitely my area of expertise, I assure you!
On the blackbody ranking of 0 to 1 (with 1 being a perfect blackbody), what are the rankings for Nitrogen and Oxygen in the IR spectrum that Earth produces.
If they are not Zero in the IR spec, what does that say about the radiation theory (considering that there is 35 times more N2 and O2 in the atmosphere than H20 and CO2).
Jim D and the radiation theory seems to assume they are Zero.
Mikael Cronholm says:
February 14, 2011 at 5:24 pm
it may depend on what definition you use for heat, but the one I stick to is that heat is the total kinetic energy of the particles in a substance (mass times velocity squared over two). Then infrared is not heat, but is caused by, and can cause, heat. Emission creates radiation energy by converting heat energy to electromagnetic waves (or phtons, if you prefer) and absorption converts in the opposite way.
Mikael, this, I think, is the crux of the matter, definition.
By common science, IR is heat. We feel it as heat, it is deeply penetrating, it warms things up. This is how it is commonly used in science, for example: http://www.reynardcorp.com/cold_mirror.php
“The cold mirror is designed to separate heat from light. The visible spectrum is reflected (), while the infrared wavelengths are transmitted ().”
Heat is classically defined in science as a form of energy associated with the motion of atoms or molecules and capable of being transmitted through solid and fluid media by conduction, through fluid media by convection, and through empty space by radiation. & The transfer of energy from one body to another as a result of a difference in temperature or a change in phase. (answers.com)
The last is important, heat always travel from hot to cold.
Several posters remind us here and elsewhere that what we’re talking about is heat transfer, people who actually do understand thermodynamics. They’re usually ignored, eyes seem to glaze over when convection is mentioned.. AGW has actually excluded that from its calculations by concentrating on radiation, not all of which is heat, and this pulls in many skeptics who, istm, have also lost the plot, in arguing about radiation with AGW’s as ‘total kinetic energy etc.’ and how much of it ‘back radiates etc.’, it takes on a far greater importance than it is due – and the subject is Global Warming after all. It’s been mentioned many times in this argument, that it is impossible for a cold body to warm up a warmer one, because heat travels from warm to cold. How much does it matter then when CO2 is radiating from a colder atmosphere even if the photons reach the earth? IR will always be travelling from warmer to colder, heating up the earth in the day and cooling the earth at night. There isn’t anything that ‘traps’ it in the atmosphere globally capable of changing the climate by overriding the normal warming and cooling on a day to day basis, let alone from ice age to interglacials. (Water doesn’t, it also cools the earth.)
AGW arguments take sound bites from science and create an unreal world from these, ignoring the IR as heat penetration is such a sleight of hand by saying that it’s irrelevant in the energy balance received on earth and only kicks in when the heated earth radiates it back. We don’t live in a test tube in a lab, we live in a real atmosphere which has weight and volume and is subject to gravity, (and molecules don’t travel at superfast speeds through that to mix thoroughly in the atmosphere as AGW has it); we do not live in a vacuum.
Normally, the shorter the wavelength the better the penetration. If you consider human skin, UV, visible and IR penetrate in that order, UV the most, visible a little bit, and IR hardly at all.
Not as I know it. UV barely penetrates the skin, around 1mm, IR is deeply penetrating in the far infrared. It is what we feel as heat and it is what heats up the earth, the ground.
The reason a remote control does not feel hot is merely a question of magnitude. A couple of AA batteries don’t have a lot of energy to give off.
Again, not as I know it. This is in the near infrared and doesn’t carry heat of any significance. A million (I’m not a scientist, this could be exaggeration) remote controls flicking at you will not warm you up.
Both Far Infrared and Near are used in medical therapies. Far infrared warms up your insides, saunas are built using IR as the heat source. The closer to visible light the cooler the IR, UV can’t be felt at all. Microwaves and radio waves are perhaps not in concentrated enough form in the atmosphere to heat us up.
jae said:
“But, I would like to posit another “world” for the warmistas. Suppose the atmosphere of Earth consisted of ONLY N2 and O2. Now, these gases could not cool by IR emissions, but they would warm by conduction from the surface. And they could not radiate to space. So would they continually warm for millions of years?? Would they melt the planet? WTF, folks?”
Heat would be transferred to the atmosphere by conduction and convection, increasing the energy in non-radiative transfers (perhaps it is kinetic energy or there is no change in angular momentum of the energy states so that the transition probability integrates to zero).
The atmosphere would warm, but as soon as it became as warm as the surface, it would stop absorbing heat (2nd law of thermodynamics – atmosphere can’t get hotter than the surface!). The near surface atmosphere would approach the surface temperature, and the upper atmosphere would be cooler as defined by the dry adiabatic lapse rate (easily derived from PV = nRT plus hydrostatic balance).
So no, it wouldn’t warm forever. At equilibrium, it would have no net effect on the heat flows so the surface temperature would be the simple surface radiative balance (and much cooler than today, assuming ice is still allowed)
So the effect of conduction and convection eventually heads to an equilibrium value. Adding objects with an emissivity means that they absorb energy from the surface but they can also ‘bypass’ this absorption by increasing the emissivity higher up. However, since the surface is warmer than the higher level, it emits more. For an atmosphere where emissivity is not higher at higher altitudes then the effect is always warming.
We live in this world, hence the greenhouse effect warming us up.
So the Law of Conservation of Energy has been repealed? (***) Therefore Over Unity and Perpetual Motion devices are not only possible but practical? Electric cars that run off the alternator cannot be far behind. As another commenter pointed out, a vacuum would allow maximum transfer of EMR from the sun. Any amount of matter in any form can only decrease the total transfer.
Now Matt, you nailed yourself to the wall right there for all the world to see. Seriously, right this minute you *should* be dwelling on the fact that the foundation you built all your conclusions on (and perhaps your religion) is non-existant. An epiphany lies in your immediate future if you are a wise person.
However if you are otherwise, well, you will probably continue your journey on the AGW bandwagon. If so, I can recommend exciting offers that I find in my spam filter concerning ethanol, miracle fuel pellets and super duper windmills.
(***) For some reason Joe Pesci in My Cousin Vinny fits right in there!
Konrad says:
February 14, 2011 at 11:47
I have read through this thread and have not yet seen a reasonable answer to the question raised in the comment –
richard verey says:
February 13, 2011 at 4:11 pm
Because the reasonable agree with him.
The unreasonable AGW argument is that CO2 traps IR in a blanket, it can do this they say because CO2 stays up in the atmosphere for hundreds and even thousands of years accumulating. The unreasonable have created a supermolecule wearing its pants on the outside of its trousers. It can do many impossible things before breakfast, defy gravity, rise up through lighter molecules even though heavier, stay up accumulating forming a blanket trapping IR and so able to create runaway heating of the earth by continually bouncing back IR from earth to blanket and back again, and re Richard Verney’s post, it can do this because AGWCO2 doesn’t know it’s an average, it thinks it’s well mixed throughout the atmosphere at 390 ppm or whatever and with its superpowers it has spread that average everywhere..
………………………………………………………..
Mikael Cronholm says:
Oh yes, visible light will cause a temperature rise, hence heat an object, as it is absorbed. Google “Herschel experiment” and I am sure you will find food for thought!
http://coolcosmos.ipac.caltech.edu/cosmic_classroom/classroom_activities/herschel_example.html
“Herschel was interested in measuring the amount of heat in each color. To do this he used thermometers with blackened bulbs and measure the temperature of the different colors of the spectrum. He noticed that the temperature increased from the blue to the red part of the spectrum. Then he placed a thermometer just past the red part of the spectrum in a region where there was no visible light and found that the temperature there was even higher. Herschel realized that there must be another type of light which we cannot see in this region. This light is now called infrared.”
My emphasis in italics. The earth and all on it are not painted black.
Re: penetration: http://www.ncbi.nlm.nih.gov/pubmed/9784938
“Wavelengths in the near-infrared range have much better penetrance in organic substances than visible light. ”
This subject is AGW claims that global warming is caused by increasing CO2 molecules in the atmosphere. It is a trace gas. It has even less capacity to hold heat than oxygen or nitrogen, and all these far below water. (We fill our radiators with water, not with any of these other gases.) However, there is more oxygen and nitrogen in the atmosphere than water and carbon dioxide.
Eadler says
“Calibration error doesn’t affect the temperature anomaly if the error remains constant. In the case of equipment changes, so that the temperature trend becomes discontinuous, the change if detected is corrected for by the use of adjacent location data that is consistent. These adjustments are normally done by using computer programs. Calibration of the equipment is not really necessary.”.
First I’ll agree this is somewhat off the thread, but temperature sensors include IR, and their results form the bases of these data sets, so much discussed. And much of what is discussed here relates to temperature and how accurate it is measured.
However if the sensors are in error, and that error is not accounted for, the resultant anomaly is subject to error. The reason for the Metrology Dept., is to correct for that slow insidious “drift” that takes place in instrumentation, which is why most reputable manufactures recommend periodic calibration. An earlier, and very well written item is:
http://wattsupwiththat.com/2011/01/22/the-metrology-of-thermometers/
As far as using adjacent stations to correct for others, it may be the “blind leading the blind”, as noted by the quality of stations presented by the following site.
http://www.surfacestations.org/
As far as “adjustments” using computer programs, the old saying “garbage in, garbage out” comes to mind”.
Sensor operator says:
February 14, 2011 at 11:52 am
“Do people really think climate change scientists want AGW to be true?”
Yes. Full stop. As stated previously, many of the best known climate scientists have said as much. How else to justify your life’s work? Why would M. Mann not switch to using mussel shells and such for proxies, instead of continuing with dodgy tree ring reconstructions? Because (I think) he’s not interested in actually doing an accurate historical temp. reconstruction but instead wishes to find proof of something he really, really wants to find exists. For most warmongers I fear, AGW theory HAS to be true . . . too much invested to now say, “maybe not” in spite of what observation tells them.
” If they are right, the end result is really bad.”
And that statement is based on what? Historically speaking; how much better things were when it was colder than when it got warmer? And please spare me the horrors of some areas (now considered a shit-hole that can’t sustain life) COULD be worse.
-Barn
If I hold my last utility bill in one hand and my wallet in the other, they will hit the floor at the same time if I drop them at the same time. Well established physics with centuries of consensus. Only what happens . . .?
@Myrrh.
IR is electromagnetic radiation. Or photons. The radiation is emitted when excited electrons shift from one shell to another, roughly speaking. The hotter an object is, the more excited the molecules and atoms are, and the more radiation will be emitted. When the radiation hits another object, the reverse happens. If that object happens to be you, you will feel the absorbed radiation as heat on the surface of your skin, and it will spread from there by conduction and convection (blood flow). If you roast a chicken, does it heat up evenly all the way through at once? Of course not! It will absorb the radiation on the surface and then it will conduct (no convection, no blood flow, dead chicken, hehe) throughout the meat. A microwave oven heats from the “inside” though, because it excites water molecules directly, as they are dipoles.
Visible light will definitely heat things. So will parts of UV and all of IR. I don’t just know it, I could prove it live to you, but not on a blog. So that ends my discussing of that issue.
Around the type of IR I work with, roughly 2-24 micron with the exception of the atmospheric absorption band at 6-8um, human skin will absorb 98%. The rest is reflected. But my knowledge stops there, so maybe at even longer wavelengths there will be penetration, I don’t know. In shorter wavelengths I would expect penetration to increase.
A penetration depth of UV of 1 mm is something I would consider deep penetration in skin. At shorter wavelengths still, X-ray and gamma, we don’t just get penetration, we have complete transmission through, except the very small portion that is absorbed.
If you ask Max Planck he will tell you that the energy of a photon increases with shorter wavelengths.
I don’t know what kind of sauna you use, but is sounds more like a toaster. My sauna is a wet sauna, with a wood fired stove with rocks on it where I throw water to get steam. The heat transfer to my body is actually mostly through condensation, since my body is the coolest thing in there. So the latent heat from the steam is given off to my body as it condenses and mixes with sweat. There is radiation and convection taking place in there too, of course. But the real heat shock comes when I throw a good splash of water on the rocks!
Common sense does not always make scientific sense.
@Myrrh. Good, you found old Herschel! Now, did you notice what that radiation went through before it hit his thermometers? A glass prism. Spectral things will happen to white light as it goes through glass. And I can guarantee you that above approximately 2.3 micron there would be no more radiation to heat his bulbs.
Now go and find a Planck curve for the solar spectrum and see all that energy right there in the visual part of the spectrum. What do you think happens to that ENERGY when it hits a surface? Three things could happen, absorption, transmission, reflection. And no, you are right the earth and us humans are not painted black, BUT WE ARE NOT SHINY POLISHED MIRRORS EITHER. Sorry, don’t know how to emphasize that in a better way. So all objects will be absorbing visual light, more or less, but for sure never zero. Not even close to zero, unless it is very shiny metal.
Most energy that reaches earth from the sun is in the visual spectrum. It will be absorbed and be converted to heat.
@- Myrrh says:
February 15, 2011 at 4:39 am
“The last is important, heat always travel from hot to cold.
…It’s been mentioned many times in this argument, that it is impossible for a cold body to warm up a warmer one, because heat travels from warm to cold.”
Heat or energy in the form of photons or radiation travels in ALL directions including from cold to hot, otherwise it would be impossible to see yourself in a mirror which is colder than you are.
It is just that MORE heat travels from the warm region to the cold than travels from the cool to the warm so the NET flow of heat is from warm to cold. But the amount flowing back from cold to warm modifies the NET flow.
A thought experiment.
Lets us construct a vessel in order to produce a simulated blackbody, this will essentially be a spherical chamber with an opening and a black interior. We shall construct it so that produces 1000w of continuous blackbody spectrum IR from the opening when at its desired target temperature.
Let us place it somewhere very cold, perhaps 2.7 Kelvin degrees and airless. (In orbit on the dark side of a dark moon perhaps?)
If we now heat our chamber to 193.18456 Kelvin, the IR leaving the opening should have its peak emission with a (continuous blackbody spectrum) wavelength at 15 µm.
We are used to dealing with CO2 in its gaseous form, from Wiki I found…
“At 1 atmosphere the gas deposits directly to a solid at temperatures below −78 °C (−108 °F; 195.1 K) and the solid sublimes directly to a gas above −78 °C. In its solid state, carbon dioxide is commonly called dry ice.”
I have not found a reference to the temperature that CO2 deposits to at zero atmospheres, (or close to) I did find reference to the fact that dry ice has been detected on Mars and in comets. So for now I will assume that it can exist in a near vacuum.
If we now introduce a chunk of CO2 (say 1kilo at 3 Kelvin) in front of the opening to absorb and get heated by this 15 µm peak IR, how warm could it get?
NoIdea
@Matt…
What’s up with all the boxes of cats?
The simplest way I know to understand the greenhouse effect is to realize that when you’re standing on the ground in daytime the Sun is shining on you in visible light, and the atmosphere is shining on you in infrared. If you change the atmosphere so it shines more (e.g. adding more CO2 and other greenhouse gases) you will be warmer on the surface because there is more electromagnetic radiation coming down on you than otherwise. That’s it. More CO2 = more radiation on you.
The cats is an illustration about what actually happens through the heat transfer from the Sun to the ground and from the ground to the air, as well as the impossibility that a colder system warms up to a warmer system. A scientist must adhere to observed phenomena and to well tested theories.
The simplest way you know is not the correct mechanism how it actually happens, though popular; however, popularity is not a step of the scientific method.
The atmosphere doesn’t shine. The Sun does shine.
@Blade;
There is no breaking of conservation of energy; to everyone who doesn’t believe in the greenhouse effect you have a lot more reading to do: I recommend Guenault’s statistical physics textbook as an excellent (and concise!) starter. Atmospheric Physics by Ambaum is also concise and deals with radiatve transfer in later chapters. Most of your claims are dealt with explicitly there, if you can follow the maths.
The emission ability of a substance is related to its absorption (if it has energy levels allowing emission, it has the same levels allowing absorption). This value depends on wavelength for most materials – you can be transparent to visible light but absorb a lot in the infrared for example.
The emissivity/absorptivity of the atmosphere in the wavelengths sent by the Sun is relatively small (with notable exceptions like ozone absorbing UV). CO2/H2O doesn’t absorb here so adding it doesn’t reduce the amount of energy hitting Earth.
But it does absorb in the lower wavelengths. Since it absorbs there, it must be able to emit there too. The atoms move quickly and interact with each other, which allows CO2 to dump heat into the O2/N2 and vice versa.
Adding CO2 doesn’t prevent sunlight hitting Earth, but it does absorb light going up from Earth. Hence the warming.
barn E. rubble says:
“If I hold my last utility bill in one hand and my wallet in the other, they will hit the floor at the same time if I drop them at the same time.”
Not true! Your utility bill will absorb your wallet before it hits the ground! 😉 GK
Mikael,
Did you see Bill Illis’ excellent question, and do you know the answer?
Bill Illis says:
February 15, 2011 at 4:38 am
Also Mikael you state at:
Mikael Cronholm says:
February 15, 2011 at 6:35 am
Quote: “Around the type of IR I work with, roughly 2-24 micron with the exception of the atmospheric absorption band at 6-8um”
Is 6-8 µm the absorption band of O2 and/or N2? Because it isn’t CO2 is it? CO2 absorption is 15 µm as we all know.
So when you refer to “the atmospheric absorption band at 6-8 µm” which part of the atmosphere is absorbing at 6-8 µm please?
Right, there is no Tyndall and Co. because they bios are on Wiki. Check. From the can’t- see-it-from-my-house-school-of -myopia. Check.
http://geosci.uchicago.edu/~rtp1/papers/PhysTodayRT2011.pdf
Learn.
Oliver Ramsay says:
February 14, 2011 at 9:58 am
A C Osborn says:
February 14, 2011 at 9:40 am
Oliver Ramsay says:
February 14, 2011 at 8:51 am “In fact the warmer the surface becomes the more the energy will be radiated to space without bouncing around in the atmosphere.”
WUWT?
So the surface has a direct connection to space then?
—————
Yes
Wow, I didn’t realise there were any Mountains that high.
The Earth’s Surface is in DIRECT contact with space?
Not going through any atmosphere for the photons to “bounce around” in then?
“A C Osborn says:
February 15, 2011 at 10:03 am
Wow, I didn’t realise there were any Mountains that high.
The Earth’s Surface is in DIRECT contact with space?
Not going through any atmosphere for the photons to “bounce around” in then?”
AC,
you should try a job as a comedienne, or not. You apparently need to read about atmospheric windows.
http://csep10.phys.utk.edu/astr162/lect/light/windows.html
I’ll leave it to others to decide whether particles can reach states in the atmosphere where they can no longer absorb in bands where they typicall are considered to absorb.