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…
izen says:
February 14, 2011 at 12:53 pm
“But the first LoT kicks in. The energy is still around. The only way it can get of the planet is by radiating IR photons into space. All the hydrological cycle can do is move the location of that emission around.
——————————–
You make it sound like water just shuffles energy around horizontally.
This is claiming the taxi delivers you to your holiday hotel door, not the airplane.
frog, jim,
the details show in kt97 that the assumptions are 100% optical thickness for the two main cloud types and 50 or 60% thickness for the 6% coverage contributor. The 62% result they come up with is assuming 100% thickness with random overlap. It does seem to agree well with what little is known of the actual coverage, which can vary substantially over time, something like over +/- 5% as I recall. That also leads to an albedo variation of around 5% and a peak to peak difference in reflected incoming light power of about 10 w/m^2, far more than a mere co2 doubling.
Hans Erren says on February 13, 2011 at 4:57 pm
“Adding water vapour in the tropics gives the tropics a greenhouse effect, I remember very well when I left the airconditioned hotel in Dhaka Bangladesh, a wall of humid heat hit me and I had to wipe my glasses.
Every added particle to the atmosphere is a small downward radiator, adding more particles therefore adds more radiators. That’s in a nutshell the greenhouse effect.”
Hmmm. Then why is the maximum temperature there NEVER as hot as it is almost every day from June-Sept. in Phoenix, AZ?
“”””” Ian W says:
February 13, 2011 at 12:35 pm
When water vapor in the atmosphere condenses into liquid water and then changes state again and becomes ice, it gives off latent heat for both state changes.
Does that latent heat release follow Stefan-Boltzmann’s radiative equation ? “””””
Latent heat and anything to do with heat, is a property of physical matter; it has nothing to do with Electromagnetic Radiation; so it has nothing to do with the Stefan Boltzmann equation.
EM radiation can go anywhere it wants to; heat can’t.
Interesting article, and Al Tekhasski covers a good point too.
Detectors are tuned to work in a certain wavelength. We can make materials that have energy levels (or bands) that only allow transitions of certain energy. One common way of doing it is to create a ‘quantum well’ (look it up!) using sandwiched layers of different materials, but there are others.
The atmosphere is largely transparent in some wavelengths, it doesn’t absorb well here. This means that there aren’t many available transitions, so it can’t emit here well either. So if you point your scanner upwards and measure, and your scanner is tuned to this energy band then you won’t measure many photons coming down. This will mean a small electrical signal (the voltage output is related to the number of photons) and your device thinks ‘small voltage, so not many photons, so what I’m measuring is cold’.
This works because most objects have emissivity close to 1 at these ranges, whilst the atmosphere doesn’t. So you can measure the photons coming from far away objects and not the atmosphere; if the atmosphere had emissivity 1 for these wavelengths then it would absorb the light from the object before it could reach your scanner, and you would always measure the atmospheric temperature – not useful for a long range thermometer!
The strength of CO2 absorption can be tested empirically by spectral measurements from satellites (and changes can be measured from ground stations) for a variety of conditions. Look up some of Philipona’s papers or the Harries 2001 paper. They provide experimental confirmation of the physics.
physics schymisics,
real world
When CO2 levels were 4000 ppm – we had an ice age
when CO2 levels were 3000 ppm = we had an ice age
when CO2 levels were 2000 ppm – we had an ice age
and here we are wringing our hands and wetting the bed over 390 ppm……………….
real world says the only tipping point is when CO2 levels get in the thousands, we can have another ice age……………..
I will correct my answer to Slacko, taking into account Robert Clemenzi’s point.
Yes the IR at night is mostly due to H2O and CO2. During the day, the sun produces near-IR in the 1-2 micron range, while the atmosphere only emits at the wavelengths longer than about 4 microns. These two types of IR may be of comparable magnitude in the upper atmosphere, but much of the near-IR is absorbed before reaching the surface.
“Sensor operator says:
February 14, 2011 at 11:52 am
Think of it like Children’s Hospital. The doctors there have one job: to put themselves out of a job. If they could solve all the medical issues they see, they would no longer be needed. And that is actually their goal! They don’t want kids to suffer. Do people really think climate change scientists want AGW to be true? If they are right, the end result is really bad.”
Sensor, I don’t have to THINK that is what some climate scientists think. Phil Jones actually stated it in his e-mails.
How does Nitrogen and Oxygen warm up and cool down in the atmosphere.
The way it is decribed by some here, the radiation theory assumption is that both of these gases are at absolute zero.
Infrared is Heat, or rather far and mid is heat. Near infrared is not felt as heat, think remote control. You’re not going to get warm by flicking your remote at yourself.
Visible light and UV do not feel hot, they are cool. UV doesn’t penetrate very far into the body, it tans, or burns, the skin – can do so even on cloudy days as Robert Clamenzi says (Feb 14, 11:23 am).
We receive IR here on earth from the Sun. Why is this excluded in AGW literature?
“Don V says:
February 14, 2011 at 12:00 pm
AC Osborne: re photon-photon collision”
Wikipedia has an interesting article on photon-photon collisions.
http://en.wikipedia.org/wiki/Two-photon_physics
Of course it points out that photons can’t collide since they don’t become photons, or quantized, until they interact with matter. This gets to the root of the issues with backradiation. Until the radiation from the CO2, or the earth for that matter, actually interacts with a particle they are best described through wave mechanics which was well mapped like here:
http://en.wikipedia.org/wiki/Phase_cancellation
http://www.mrelativity.net/Papers/4/Rykov.htm
Many people ask what would happen to the energy if 2 waves cancelled.
http://newsgroups.derkeiler.com/Archive/Rec/rec.radio.amateur.antenna/2005-12/msg00243.html
Sometimes the most simple concepts are missed. Actually it may not be simple reflection but scatter, although the scatter would seem to be when less than 100% cancellation occurs.
“”””” Bill Illis says:
February 14, 2011 at 4:11 pm
How does Nitrogen and Oxygen warm up and cool down in the atmosphere.
The way it is decribed by some here, the radiation theory assumption is that both of these gases are at absolute zero. “””””
Funny you should notice that too Bill. N2 and O2 at 288 or 300 K may not seem to be emitting any thermal radiation; but that is because we are used to feeling our “heat” at much shorter wavelengths, and much higher Temperatures.
Considering how difficult it is to observe and measure thermal radiation at 288 K; especially spectrally resolved; it is no wonder that ordinary humans are not even aware of its existence. We certainly can’t feel it on our skin.
But shift the wavelength down by ten, and raise the Temperature by ten, and the Total Radient emittance by 10,000 times, and the spectral peak emittance by a factor of 100,000 and humans finally can be made aware of it’s presence.
But I don’t see a whole lot of 100 Watt “heat lamps” turned up skywards, anywhere I’ve ever been !
Mikael Cronholm,
You set up a wonderful discussion here. Thanks.
It is a main event.
John
Turns out I was right about you on the other thread, Mr. Mosher. You have shown your true colors. Why do you feel you have to trap anybody? You are not a nice man.
REPLY: David, you don’t know jack, I know Mosher personally, you don’t. You’re simply wrong on this point. – Anthony
Bill Illis, nitrogen and oxygen heat and cool by conduction and convection only, not by radiation.
“”””” JAE says:
February 14, 2011 at 1:58 pm
Hans Erren says on February 13, 2011 at 4:57 pm
“Adding water vapour in the tropics gives the tropics a greenhouse effect, I remember very well when I left the airconditioned hotel in Dhaka Bangladesh, a wall of humid heat hit me and I had to wipe my glasses.
Every added particle to the atmosphere is a small downward radiator, adding more particles therefore adds more radiators. That’s in a nutshell the greenhouse effect.” “””””
So what is the magic of these particles that they know to only radiate downwards? Common sense would say that whatever in the atmosphere is emitting radiation of any kind, is doing so pretty much isotropically with no directional bias.
So whatever your particles are emitting downwards, they must be emitting a like amount upwards; whcih escapes to space. So whatever the originals ource of the energy that your particles are radaiting; be it direct incoming sunlight or surface emitted LWIR radiation, it seems that half of it si going to escape to space, and not reach the ground. Particularly when the sun is the source of that energy your particles are radiating downwards; that is an amount of sunlight that will never reach the ground; so it will get less hot than if your particles did not intercept that solar raiation.
No matter how you try to skin the cat; anything in the atmosphere that absorbs any incoming solar energy or even widely scatters it, such as the blue skylight due to Raleigh scattering, must result in less solar energy reaching the surface of the earth (ocean) and getting stored in earth’s thermal sink.
izen says:
February 14, 2011 at 12:53 pm
Ian W says:
February 14, 2011 at 10:06 am
“As confirmed in response to my question at the top of this thread, the temperature of the cloud has no bearing on the amount of latent heat being released. The heat being released is linked to the amount of water changing state and the latent heat of that state change.
The missing heat is almost certainly in the misunderstood and underestimated hydrologic cycle that is ignored in favor of a simple Stefan-Boltzmann radiation equation.”
The ‘Science of Doom’ site linked to from here (top right of page) has a good discussion of some of these issues.
The amount of energy transported by latent heat changes is complex, there is one simple way to quantify it however. The total global yearly rainfall is a direct measure of how much energy was moved by evaporation and condensation in the hydrological cycle.
But the first LoT kicks in. The energy is still around. The only way it can get of the planet is by radiating IR photons into space. All the hydrological cycle can do is move the location of that emission around. That evens out the temperature so that water does not boil at the equator and can melt at the poles, but the same amount of energy has to leave the planet. Otherwise it will warm until the S-B T^4 relationship increase the emissions enough to compensate.
So when the heat is released at 30,000 feet (*) in the atmosphere as liquid water turns to ice, well past the dense part of the atmosphere it does not radiate to space?
How does it know that it can only radiate down? 😉
It certainly appears that significant IR is being emitted from these clouds and weather systems independent of their temperature
http://www.ssd.noaa.gov/goes/east/natl/flash-rb.html
(*) The updrafts in the ITCZ as aircraft have found out to their cost can be in the order of 100KnH or more and liquid water can get that high before freezing – hence my quote of the Nortwest Airlines Airbus.
Bill Illis
N2 and O2 are at the same temperature in a small parcel of air, just like the co2. The co2 absorbs some energy and the average time it takes to radiate it away is more than enough time for it to be transferred away by collisions, most likely n2 followed by o2. just because a co2 molecule excited by a photon is likely to have its energy reduced by a collision, so too is the likelihood that a co2 molecule will be excited by a collision and capable of emitting a photon or capable of being ‘defused’ by yet another collision. net result is that a certain fraction of these molecules will emit a photon of a particular energy (wavelength) based upon the temperature and upon the proclivity of the co2 molecule to absorb or emit that energy. absorption doesn’t really depend much on temperature but the emission is highly dependent. The blackbody curve for a give temperature is actually a portrayal of the fraction of molecules at particular energies that are capable of emitting photons. The bb curve must be a solid or liquid – or an optically thick enough gas to be thick at all wavelengths. 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.
Myrrh, 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 it is caused by, and can cause, heat. Emission creates radiation energy by converting heat energy to electromagnetic waves (or photons, if you prefer) and absorption converts in the opposite way.
Thermal radiation, if you define it as radiation that can heat or cool a surface by exchange of radiation, will be wavelengths from somewhere in the UV throughout visible and IR. Most of the heat the earth receives from the sun is in fact in the visible band. In the shorter wavelengths, X-ray and gamma, most objects will transmit most of it, and hence no transfer of energy. Longer wavelengths, microwave, and radio wave, do not significantly heat things, I am not sure why, but if they did heat us we would be cooking with all the radio waves around. Microwaves can only heat by directly agitating water molecules, because they are dipoles, so it is a different process than we normally think of as heat transfer by radiation. But it is a matter of definition, largely.
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.
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. By the way, you can use a normal CCD camera to check if your remote works, a simple one in a cell phone will do fine. CCD’s work up to around 1 micrometer.
One question I am looking for a good answer.
At sea level, when an ir photon is emitted from earth in a upward direction, what is the distribution of its path length before it excites(is absorbed) by a CO2 molocule and is thermalized into the atmosphere. Believe I know the answer, but looking for indepenant confirmation.
davaidmhoffer says:
CO2 is reasonably well mixed throughout the troposphere,
————-
This is not true and is just flat out wrong. In real air, there is no unifrom distribution of mass of the atmosphere in space and time as shown by weather maps. High pressure cells have more regional mass and dry air than do low pressure cells with moist air. And these are constantly moving sometimes quite rapidly. Humidity lowers the density of dry air by a much as 5%.
Tropical air at ca 30 deg C and with 100% humidity has 80% of the mass per unit volume than does dry cold air at STP ( 0 deg C and 1 atm pressure).
Comprised of nitrogen, oxygen, the inert gases, which are the fixed gases, and CO2, purified dry air (PDA) at STP has presently 390 ml, 17.4 millimoles, 766 mg or 0.000766 kg of CO2 per cubic meter and has a density of 1.2929 kg per cu per meter. PDA does not occur in the earth’s atmosphere. The composition of PDA (i.e, rel amounts of the fixed gases and CO2) is fairly uniform thru out the earth’s atmosphere and is independent of site, temperature, pressure, humidity which includes water vapor and clouds except for minor local variation in particular with prespect to CO2.
If PDA is cooled to -53 deg, the amount of CO2 is 21.6 mmole cubic meter and concentration is 390 ppmv. If PDA is heated to 45 deg C, the is concentration is still 390 ppmv but there is only 12.5 mmoles of CO2 per cubic meter.
GCM calculations generally use the concentration of CO2 in ppmv which is the incorrect metric and thus are fatally flawed. The correct metric is mass (or millimoles) of CO2 per unit volume.
The water droplets of clouds contain CO2 which can be released if they dissipate or transport CO2 to the surface if they turn into raindrops. How much CO2 is sequestered in the clouds? Probably a lot.
enough says:
February 14, 2011 at 5:28 pm
One question I am looking for a good answer.
no real good answer I’m afraid. Depending on the wavelength of the photon, it might travel from surface to space without a hint of a capture OR it might not make it 2 centimeters without a sure thing capture. Most of the critical wavelength bands fall somewhere in between. at lower altitudes (higher pressures) the individual lines are spread out essentially forming bands. At higher altitudes and lower pressures, the lines become much sharper taller and narrower. Molecules here are less likely to absorb or emit a photon whose wavelength is further from the peak.
mikael,
that’s a very restricted definition of heat energy. energy can add velocity to a molecule or it can increase the internal energy state. IR is merely an electromagnetic form of energy. The same goes for radio waves, light, IR, uv, xrays and gamma rays.
J. Bob says:
February 14, 2011 at 8:47 am
eadler says
“If the mean error is know, the readings can be adjusted. If the mean error remains constant, than the temperature anomaly, which is what is being sought will not suffer in accuracy.”
And those “if’s” can be significant. Much of the mean error “adjustment” ability depends on the sensor, electronics, environment, calibration protocols, and how well these protocols are ACCUALLY followed.
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 normallyl done by using computer programs. Calibration of the equipment is not really necessary. The new data base, which is the subject of this thread will be corrected for equipment discontinuities in the same way as the previous thermometer data bases have done it.
This is getting off topic of this thread. There are other threads on this web site that deal with corrections of the temperature record, including the new data base being developed by Muller.
One question I am looking for a good answer.
no real good answer I’m afraid. Depending on the wavelength of the photon, it
Sorry, question was not to the point, If an IR photon in the co2 band is emitted at sea level, what is its path length before being absorbed by a co2 molecule.