Guest Post by Ira Glickstein
This series began with a mechanical analogy for the Atmospheric “Greenhouse Effect” and progressed a bit more deeply into Atmospheric Windows and Emission Spectra. In this posting, we consider the interaction between air molecules, including Nitrogen (N2), Oxygen (O2), Water Vapor (H2O) and Carbon Dioxide (CO2), with Photons of various wavelengths. This may help us visualize how energy, in the form of Photons radiated by the Sun and the Surface of the Earth, is absorbed and re-emited by Atmospheric molecules. 
DESCRIPTION OF THE GRAPHIC
The animated graphic has eight frames, as indicated by the counter in the lower right corner. Molecules are symbolized by letter pairs or triplets and Photons by ovals and arrows. The view is of a small portion of the cloud-free Atmosphere. (Thanks to WUWT commenter davidmhoffer for some of the ideas incorporated in this graphic.)
- During the daytime, Solar energy enters the Atmosphere in the form of Photons at wavelengths from about 0.1μ (micron – millionth of a meter) to 4μ, which is called “shortwave” radiation and is represented as ~1/2μ and symbolized as orange ovals. Most of this energy gets a free pass through the cloud-free Atmosphere. It continues down to the Surface of the Earth where some is reflected back by light areas (not shown in the animation) and where most is absorbed and warms the Surface.
- Since Earth’s temperature is well above absolute zero, both day and night, the Surface radiates Photons in all directions with the energy distributed approximately according to a “blackbody” at a given temperature. This energy is in the form of Photons at wavelengths from about 4μ to 50μ, which is called “longwave” radiation and is represented as ~7μ, ~10μ, and ~15μ and symbolized as violet, light blue, and purple ovals, respectively. The primary “greenhouse” gases (GHG) are Water Vapor (H2O) and Carbon Dioxide (CO2). The ~7μ Photon is absorbed by an H2O molecule because Water Vapor has an absorption peak in that region, the ~10μ Photon gets a free pass because neither H2O nor CO2 absorb strongly in that region, and one of the 15μ Photons gets absorbed by an H2O molecule while the other gets absorbed by a CO2 molecule because these gases have absorption peaks in that region.
- The absorbed Photons raise the energy level of their respective molecules (symbolized by red outlines).
- The energized molecules re-emit the Photons in random directions, some upwards, some downwards, and some sideways. Some of the re-emitted Photons make their way out to Space and their energy is lost there, others back down to the Surface where their energy is absorbed, further heating the Earth, and others travel through the Atmosphere for a random distance until they encounter another GHG molecule.
- This frame and the next two illustrate another way Photons are emitted, namely due to collisions between energized GHG molecules and other air molecules. As in frame (2) the Surface radiates Photons in all directions and various wavelengths.
- The Photons cause the GHG molecules to become energized and they speed up and collide with other gas molecules, energizing them. NOTE: In a gas, the molecules are in constant motion, moving in random directions at different speeds, colliding and bouncing off one another, etc. Indeed the “temperature” of a gas is something like the average speed of the molecules. In this animation, the gas molecules are fixed in position because it would be too confusing if they were all shown moving and because the speed of the Photons is so much greater than the speed of the molecules that they hardly move in the time indicated.
- The energized air molecules emit radiation at various wavelengths and in random directions, some upwards, some downwards, and some sideways. Some of the re-emitted Photons make their way out to Space and their energy is lost there, others back down to the Surface where their energy is absorbed, further heating the Earth, and others travel through the Atmosphere for a random distance until they encounter another GHG molecule.
- Having emitted the energy, the molecules cool down.
DISCUSSION
As in the other postings in this series, only radiation effects are considered because they are the key to understanding the Atmospheric “Greenhouse Effect”. I recognize that other effects are as important, and perhaps more so, in the overall heat balance of the Earth. These include clouds which reflect much of the Sun’s radiation back out to Space, and which, due to negative feedback, counteract Global Warming. Other effects include convection (wind, thunderstorms, …), precipitation (rain, snow) and conduction that are responsible for transferring energy from the Surface to the Atmosphere. It is also important to note that the Atmospheric “Greenhouse Effect” and a physical greenhouse are similar in that they both limit the rate of thermal energy flowing out of the system, but the mechanisms by which heat is retained are different. A greenhouse works primarily by preventing absorbed heat from leaving the structure through convection, i.e. sensible heat transport. The greenhouse effect heats the earth because greenhouse gases absorb outgoing radiative energy and re-emit some of it back towards earth.
That said, how does this visualization help us understand the issue of “CO2 sensitivity” which is the additional warming of the Earth Surface due to an increase in atmospheric CO2? Well, given a greater density of CO2 (and H2O) molecules in the air, there is a greater chance that a given photon will get absorbed. Stated differently, a given photon will travel a shorter distance, on average, before being absorbed by a GHG molecule and be re-emitted in a random direction, including downwards towards the Surface. That will result in more energy being recycled back to the Surface, increasing average temperatures a bit.
Steve says:
March 30, 2011 at 1:07 pm
Is “back radiation” really required to explain the greenhouse effect? It seems to me that it just confuses the issue.
With a pure nitrogen atmosphere, almost all infrared radiation would pass through. The molecules of this atmosphere are primarily heated by direct contact with the surface of the sphere, and convection of this thin layer with the upper layers.
Now replace a mass of that nitrogen with a like mass of CO2. Now you have layers of the atmosphere (top to bottom) heated by infrared radiation that was previously not heating any portion of the atmosphere. The atmosphere is warmer – period. No need to talk about whether or not back radiation will heat up the surface or not, because it is obvious that the atmospheric blanket itself is warmer than it was before. (If you really want to make it warmer, exchange H2O for either of the other two gases.)
That’s a great point you raise.
But Steve, sorry, you are wrong. If ignoring the tiny radiation bands in nitrogen, the pure nitrogen atmosphere would over time be very, very hot. During the daytime with no clouds and oceans the atmosphere would heat greatly at the surface and disperse upward by conduction and convection as you said. As the Earth turns and night comes, all convection would ceases and the atmosphere cannot lose any heat by radiation as you said. Day by day the atmosphere would get hotter and hotter until the entire atmosphere would reach almost the same temperature as the daytime soil. The only cooling occurring would be by strictly conduction near the surface at night in contact with a frosty soil for it would be radiating unimpeded.
CO2 and all GHGs are a great fast conductors of heat. It is the over-half portion of radiation upward to space that keeps us from being literally cooked.
Joel Shores simple world consists of an earth without oceans and a land surface that does not retain heat.
Clouds, rain and snow (which give a much more continuous absorption/emission possibilities than gas molecules) apparently do not exist.
The moon comes closer to Joel’s model but even there his simplistic assumptions do not give a realistic account of actual temperatures.
According to Joel the so called 33K “greenhouse effect” is entirely due to CO2 and H2O gaseous molecules.
Perhaps we should leave Joel with his simple fables.
It seems pointless to explain how absurd his conclusions are as he is religiously attached to them.
Probably already requested, but how about a page that has each graphic and each explanation below it? That way you can focus on the information in each graphic and not be distracted by the constant changing. Thank you.
where is the low level boundary layer? Pretty much all IR radiation coming from the surface that can be absorbed by CO2 is absorbed by CO2 at ground level. The actual photons that CO2 can absorb beyond this must be generated from the heat in the atmosphere itself.
Ira Glickstein, PhD says:
March 30, 2011 at 8:13 am
from Wikipedia
If an ideal thermally conductive blackbody was the same distance from the Sun as the Earth is, it would have a temperature of about 5.3 °C. However, since the Earth reflects about 30% (or 28%) of the incoming sunlight, the planet’s effective temperature (the temperature of a blackbody that would emit the same amount of radiation) is about −18 or −19 °C, about 33°C below the actual surface temperature of about 14 °C or 15 °C. The mechanism that produces this difference between the actual surface temperature and the effective temperature is due to the atmosphere and is known as the greenhouse effect.
While I certainly do not trust Wikipedia completely, since anyone may post almost anything to obscure topics, I know from personal experience that important scientific topics are monitored by domain experts who protect them from spurious information. Reed Coray, please cite a respected source for your “7ºC” estimate, and same for mkelly regarding your “15 C max”.
I believe the Wikipedia argument is correct. However, it compares apples to oranges. In particular, I’ll assume Wikipedia’s 30% number is correct. However, I then ask: “What part of the Earth reflects 30% of the sun’s energy?” From the papers I’ve read, it’s primarily the clouds and/or particulate matter in the atmosphere, not the surface of the Earth. When quantifying atmospheric effects on Earth surface temperature, shouldn’t you compare the measured Earth surface temperature (i.e., the temperature with an atmosphere) …to… an Earth model that is devoid of an atmosphere? Wikipedia’s model for the Earth surface temperature includes an atmosphere in that it’s the atmosphere that is reflecting 30% of the solar energy. So Wikipedia is comparing measured Earth surface temperatures to a model of Earth surface temperature that includes an atmosphere; and using the temperature difference to claim that the atmosphere raises the Earth surface temperature 33 °C . I call foul. It’s kind of like John Kerry saying he voted against it before he voted for it. (Sorry, I couldn’t resist).
An Earth without an atmosphere will have an albedo much nearer 0 than 0.3. However, assume for a moment that the albedo of an atmosphereless Earth is “x”. Then as I discussed previously, if you’re going to use Planck’s graybody radiation law to determine the rate energy is emitted from that surface, the surface emissivity must be the same as one minus the surface albedo. If true, for an atmosphereless Earth albedo of “x”, the emissivity of that atmosphereless Earth will be “(1-x).” The energy absorbed by the Earth from the sun is equal to the product of (a) the solar energy density (Watts per square meter) at the Earth’s surface, (b) the surface area of the Earth perpendicular to the direction of radiation (pi times the radius of the Earth squared), and (c) one minus the albedo–i.e., (1-x). The rate energy is radiated from the surface of the atmosphereless Earth will be the product of (a) the surface area of the Earth (4 times pi times the radius of the Earth squared), (b) the Stefan-Boltzmann constant, (c) the Earth surface temperature to the fourth power, and (d) the emissivity of the Earth–i.e., (1-x). For steady-state conditions, these two rates are equal. Since the factor “(1-x)” appears on both sides of the Equation, the temperature of an “atmosphereless” Earth is independent of the Earth’s surface albedo (one minus emissivity). Thus, the temperature of an atmosphereless Earth is the same as the temperature of a blackbody Earth–using Wikipedia’s value: 5.3 °C, not -18 ºC or -19 ºC.
“Then as I discussed previously, if you’re going to use Planck’s graybody radiation law to determine the rate energy is emitted from that surface, the surface emissivity must be the same as one minus the surface albedo. ”
Not quite! The surface emissivity FOR VISIBLE LIGHT must be 1-albedo.
The surface albedo FOR IR must be 1.
There is no a priori reason the reflection for visible and IR must be the same.
wayne says:
March 30, 2011 at 2:44 pm
“But Steve, sorry, you are wrong. If ignoring the tiny radiation bands in nitrogen, the pure nitrogen atmosphere would over time be very, very hot. During the daytime with no clouds and oceans the atmosphere would heat greatly at the surface and disperse upward by conduction and convection as you said. As the Earth turns and night comes, all convection would ceases and the atmosphere cannot lose any heat by radiation as you said.”
I didn’t say that a nitrogen atmosphere cannot lose heat by radiation – is that what you are saying? I said that a nitrogen atmosphere would warm less than a CO2 atmosphere via the radiation emitted from a cooling earth surface. A molecule will radiate at wavelengths according to it’s emission spectrum, which is roughly equivalent to it’s absorption spectrum. Warming via conduction does not prevent cooling via radiation – the nitrogen in our atmosphere (most of the gas in it) cools every night. Are you proposing that the nitrogen gas in our atmosphere cannot cool off unless it collides with another type of atmospheric gas?
Bryan says:
No…Retention of heat is not relevant to the considerations that I mentioned. The way that it comes into it is that the ocean helps to keep the temperature more uniform, which means that the fourth-root of the average of T^4 is closer to the average of T, so the greenhouse effect is close to 33 K. If oceans were not there, then the temperatures would be more variable (between day and night, seasons, etc.) and the average surface temperature in the absence of the greenhouse effect would tend to be even lower.
I never said that. Clouds are part of the reason why the atmosphere is not transparent to IR radiation and hence contribute to the greenhouse effect.
Reed Coray says:
Sort of true…although really you are forgetting a couple of things. One is that I believe the albedo of the earth in the absence of clouds is still about 8% or so. The second is that such a colder earth would presumably have more ice on it and would therefore have even a higher albedo than this.
At any rate, the full statement is that the atmosphere (including clouds) produces two effects: One is about 33 K of warming due to the greenhouse effect and the other is some cooling due to the albedo of the clouds. So, yes, the net effect of the atmosphere is not a full 33 K of greenhouse warming. However, again, the sort of calculations you discuss are under the assumption of no change in surface albedo [actually, you wrongly assumed 0 albedo]…In reality, if you removed CO2 from the atmosphere, not only would you remove a lot of the water vapor too, you would also increase the surface albedo of the earth.
Reed –
On re-reading what you wrote, I realize that you didn’t assume a zero albedo for the earth as I said in my last post but you did assume, as Tim Folkerts points out, that the emissivity is the same in the visible as it is in the far IR…which is indeed not a very good assumption. Emissivities of the earth’s surface in the visible vary widely with the type of surface, whereas emissivities in the far IR tend to be very close to 1 for nearly all surfaces. (I believe I’ve read that the only place that there can be any significant deviations from 1 is in some desert regions.)
“”””
Steve says:
March 30, 2011 at 4:47 pm
I didn’t say that a nitrogen atmosphere cannot lose heat by radiation – is that what you are saying? “”””
Hi Steve. Yes, I was portraying that on what Phil. said in the prior post. I would agree with that. Nitrogen, except for some small lines near 3 µm, cannot radiate at all. I tend to agree with him though. There are many claiming that all matter radiates when above 0 K as you just said. If all matter radiates, and an atmosphere of pure N2 radiates, then this is no different than what a GHG is portrayed to radiate in all directions and all atmospheres and all molecules always cause a “greenhouse effect”.
Tim Folkerts says:
March 30, 2011 at 4:09 pm
“Then as I discussed previously, if you’re going to use Planck’s graybody radiation law to determine the rate energy is emitted from that surface, the surface emissivity must be the same as one minus the surface albedo. ”
Not quite! The surface emissivity FOR VISIBLE LIGHT must be 1-albedo.
The surface albedo FOR IR must be 1.
There is no a priori reason the reflection for visible and IR must be the same
I agree. Real-world materials can exhibit emissivities (albedos) that are a function of frequency. For blackbodies, Planck’s radiation law contains a frequency dependent term of the form (f^3)*(df)/[e^(hf/kT) – 1]. When integrated over all frequencies (0 to infinity), the result is radiated power that is proportional to the fourth power of the temperature T. If to account for a different “visible light” emissivity and “IR” emissivity an additional frequency dependent factor g(f) is included in the law, then the integral over all frequency will only be proportional to the temperature to the fourth power if g(f) is constant. The T^4 rule may be a good approximation, but it won’t be theoretically correct. As such, if you’re going to compute the radiated energy using a temperature dependence of T^4, aren’t you implicitly assuming an emissivity (albedo) that is the same at all frequencies? If you’re going to use an emissivity that is a function of frequency, then shouldn’t you specify that dependence?
I’m curious. What do you believe is the primary source of solar reflection: (a) the surface of the Earth, or (b) the Earth’s atmosphere and matter in the atmosphere, or (c) something else? Do you agree or disagree that if you’re going to determine the effect of the atmosphere on Earth surface temperature by (a) treating the measured Earth surface temperature as the “Earth with atmosphere temperature”, and (b) using a model to compute the atmosphereless Earth surface temperature, the model should be entirely devoid of atmospheric effects?
If someone wanted to (a) model the position/frequency dependence of Earth surface absorption (no atmosphere) and the position/frequency dependence of Earth surface emission (no atmosphere), (b) compute the “average” Earth surface temperature using that model, and (c) argue that the effect of the Earth’s atmosphere is the difference between the average measured Earth surface temperature and the model average Earth surface temperature, I would have no objection. But that is not how the 33 degree number is computed. The 33 degree number is based on a reception albedo that, I believe, represents an “atmospheric albedo”, not an Earth surface albedo.
Ira Glickstein, PhD says:
March 30, 2011 at 2:02 pm
“Having made that error, the author continues:
One might think that, because the blackbody is now absorbing more light, even if it is its own infrared light, then it should warm up. …
Yep, it should warm up and (if it is a true blackbody in the longwave region) it will absorb that reflected longwave energy and, all else being equal, will warm up.”
Ira, so when will this cycle of selfheating stop?
Lets slice a blackbody cavity which is at a equilibrium temperature of 288 Kelvin in half and replace the other half with a mirror. How much warmer will this cavity become?
Nitrogen can’t radiate heat?
Since when? Everything above absolute zero radiates energy.
Atmospheric nitrogen has kinetic energy and therefore must radiate heat.
The atmosphere is heated by molecular-kinetic effects i.e. molecular collisions. Atmospheric temperature is a function of pressure and density not radiative effects.
bananabender says:
March 30, 2011 at 7:08 pm
Nitrogen can’t radiate heat?
Since when? Everything above absolute zero radiates energy.
Atmospheric nitrogen has kinetic energy and therefore must radiate heat.
No that isn’t true for gases, N2 doesn’t have a dipole and therefore doesn’t emit in the IR.
davidmhoffer says:
March 30, 2011 at 9:39 am
bananabender;
So….the fresh cool air streaming into the centre of the fire would cool it….convection through the “venturi” is exhausting hot air OUT of the fire…this process sounds like net heating to you? The volatiles evaporate DESPITE the net cooling of the convection process because the combined heat of the various surfaces radiating at each other is high enough to overcome the cooling via convection. Don’t believe me? Build yourself a decent sized fire and turn a fan or leaf blower on it to put 10 or 20 times the air through and around the hole thing than convection could. Watch what happens and report back.
For every phenomenon there is a simple and elegant explanation that is totally and utterly wrong.
Your anecdotes are amusing folklore but they show absolutely no understanding of basic scientific principles.
Have you ever used a bunsen burner? You vary the flame temperature of this burner by altering the amount of air mixing with the gas. The more air you add the more completely the gas burns and the hotter the flames. Fanning a fire is exactly the same principle.
Air is an extremely poor heat conductor. The cooling effect of increased air flow into a fire is far less than the increased heat produced by more efficient combustion.
You can hold your finger 1cm from the side of a bunsen flame and not get burnt. Gases have almost no mass so they radiate very little energy. A bunsen transfers energy directly from the flame by conduction not radiation.
Really? What are these science degrees you possess? Clearly not physics or chemistry since you only “studied” them, and you no more understand how radiation works than you do Scouting 101. Oh yeah, you didn’t have that credential either.
I was employed as chemist in the food and beverage industries after I graduated from university. I regularly used various analytical spectroscopy methods including AAS, GC-MS, spectrophotometry, IR and near IR spectroscopy . Spectroscopy is based entirely on radiative physics.
During the daytime with no clouds and oceans the atmosphere would heat greatly…
No need to talk about whether or not back radiation will heat up the surface or not…
and therefore amplifies the heat at the ground from -18 C up to +15 C….
Apparently, heat = “temperature”
but the mechanisms by which heat is retained are different….
that the temperature of a gas containing a given amount of heat…
The atmosphere and the ocean may retain heat, but the atmospheric heat retention is infinitesimal compared to the heat retention of the oceans….
Apparently, heat = “total thermal energy”
Some would say the word “net” is redundant since heat is usually a macroscopic concept describing the net energy flow….
As most of the heating of the atmosphere is by conduction and convection …
ice reflects a lot more sunlight than ocean or unfrozen land surface thus reducing the amount of solar surface heating…
Apparently, heat = “a flow of energy”
As long as the discussion contains such different ideas about what is perhaps the single most important word in this discussion, I can’t see and hope of resolving anything. We will all be talking past each other as we each try to make out own points using our own understanding of the words.
P.S. In thermodynamics, the 3rd definition (a flow of energy) is the accepted definition. If you see “Q” in an equation, it means how much energy was TRANSFERRED, not how much energy is CONTAINTED. If you see “Q” on a diagram, it is always associated with an arrow, not an object. Heat (and similarly work) are not “state variables” and do not describe the properties of any system.
wayne says:
March 30, 2011 at 6:09 pm
“Nitrogen, except for some small lines near 3 µm, cannot radiate at all.”
That determines the wavelengths at which it cools, it doesn’t limit the ability to cool off. It will still radiate according to it’s temperature. Nitrogen will radiate a lot of energy in a narrow band, instead of a moderate amount of energy across multiple bands.
“If all matter radiates, and an atmosphere of pure N2 radiates, then this is no different than what a GHG is portrayed to radiate in all directions and all atmospheres and all molecules always cause a “greenhouse effect”.”
The greenhouse effect is not a matter of the gases radiating, but absorbing. The surface of the earth radiates at a much larger variety of wavelengths than some gases can absorb, so some radiation goes directly to space without warming the atmosphere. If surface radiation is absorbed by an atmospheric molecule, it’s going to radiate that energy back out (within it’s emission spectrum), but not until it’s gained some vibrational/stretching/rotational energy in it’s bonds – it has to heat up to cool down.
bananabender says:
March 30, 2011 at 7:08 pm
“The atmosphere is heated by molecular-kinetic effects i.e. molecular collisions. Atmospheric temperature is a function of pressure and density not radiative effects.”
For convection to occur there must be a collision of masses. The vacuum of space doesn’t have enough mass for the upper atmosphere to collide with. The sun’s energy is radiated to the earth, and the earth radiates this energy back out into space as it cools.
Atmospheric pressure is a function of temperature. Of course our atmosphere (as do most) behaves as a near ideal gas. There is no fixed volume – it can only expand against the force of gravity. The outward force of temperature balanced by the downward force of gravity determines the pressure and volume. There is no volume constraint, so changes in temperature generate changes in pressure. Do you think that the night side of earth is exposed to a lower pressure from outer space, so the temperature goes down?
Steve says:
March 30, 2011 at 4:47 pm
Warming via conduction does not prevent cooling via radiation – the nitrogen in our atmosphere (most of the gas in it) cools every night. Are you proposing that the nitrogen gas in our atmosphere cannot cool off unless it collides with another type of atmospheric gas?
I must admit I have been on both sides of this subject; does our atmosphere radiate as a whole, all molecules included, or only IR active molecules, and both according to temperature. My interests of decades has been in astrophysics but I have never seen a definitive spectrum of purely nitrogen or argon or for pure hydrogen to indicate that those symmetric gases in our galaxy are opaque to IR. Neither have I searched for them. I am like many and have only seen the lab spectrums of nitrogen that show no interaction in those frequencies but the one bump in then near IR.
That is why I was careful to answer back and explain my assumption on that earlier comment.
I have been to countless university sites on the subject and have found totally conflicting reports, usually because the word ‘gases’ is not separated out. So, you could flip me back so easily if you could just show me a spectrum with lines or continuum ranges that are identified to be nitrogen or argon for that matter. I just can’t find any so I stay in line with the apparent evidence, that whatever our atmosphere is radiating, it is only via IR active molecules.
And surely you are not assuming Kirchhoff’s law does not apply in this fine point on a frequency by frequency basis. For if it radiates it must absorb equally. We all know for certain that solids and liquids radiate close to a gray body manner but I’ll ask you, why do think nitrogen gas is always radiating energy away in all directions?
The whole “burning logs” thread seems to have too many variables. Let’s make it more controlled.
Three separate identical vertical logs. The logs are each surrounded by identically-sized tubes a few inches bigger in diameter open at the top & bottom to allow convection.
* Tube 1 is glass, which will not melt, but is mostly transparent to Visible and near IR
* Tube 2 is oxidized metal, with an emissivity near 1.
* Tube 3 is shiny metal, with an emissivity near 0 (ie almost a perfect mirror).
I predict that Log 3 will burn fastest, due to all the reflected photons (which will look like a black body at the temperature of the glowing log) . These reflected photons will help raise the temperature of the log.
I predict that Log 2 will burn next fastest, due to all the emitted photons at the temperature of the tube (which will be cooler than the log, but warmer than the room).
I predict that Log 1 will burn slowest, since there will be very few photons striking it, so it will not get any boost in temperature.
With longwave IR it would not be a mirror, but an absorbing/re-emiting thin, black-painted metal shell surface that would return about half the energy it absorbed (the other half radiating from the other side of the shell and away from the source). So, if the system was 100% efficient (which it could not be) it would be the sum of 1 + 1/2 + 1/4 + 1/8 and so on, which you must know has a limit at 2. Thus, if the longwave IR source was completely surrounded by the shell, the temperature could potentially increase to correspond to double energy, but, in the real world, it would reach equilibrium well before that.
I am not familiar with the details of a blackbody cavity, but I can imagine something like an electric soldering iron. With a given power input, the tip, in free air, would reach a temperature of X. If you put the whole soldering iron in a metal box, and suspended the metal box in free air, the tip would reach a higher temperature, all else being equal. I include the last caveat because I believe that the resistance element in the soldering iron would increase its resistance as it heated up, which, at constant voltage input, would reduce the current somewhat, and therefore the input power. But, if you had a Variac (remember those – I used to own one) and you increased the voltage to get the exact same power input as before, the tip would increase in temperature compared to where it was in free air. You could calculate the new temperature using the Stefan Boltzmann (fourth power) Law. The new tip temperature would correspond to nearly twice the power.
Note however that no extra energy is being created. At equilibrium, the metal box will radiate (and convect) away the exact amount of power as is being input to the soldering iron, energy in = energy out. The only difference will be that the tip of the soldering iron will be warmer when enclosed in the metal box than when it was in free air.
Got it?
Gravity is a continual force and gas molecules are in motion, In any process where you pressurize a sufficient amount of gas, its temperature will increase. In processes where the gas is pressurized from one side, the temperature will be highest at the surface where where pressure is highest. This can be demonstrated with a centrifuge, or the leading edges of an aircraft. There’s nothing novel, or in violation of the laws of Physics here.
This would be true if you replace “scientist” with “greenhouse hypothesis enthusiast”.
This violates the second law of Thermodynamics.
What you should realize, is that you have no physical explanation of what you believe to be “basic facts”. and that it is highly probable that the misguided one is yourself.
From this thread I have discovered that:
– I can heat my house for free by filling the roof space with CO2.
– Pure N2 has a temperature of absolute zero because it can’t radiate energy.
– Chimneys make fires cooler.
– Igloos heat the occupants by radiating “cold” energy.
– Two 50W light bulbs placed together are brighter than one 100W bulb.
– CO2 always absorbs energy at ground level and radiates energy at the edge of space.
– Chemists are all morons who think intramolecular bond stretching and orbital electron excitation are different things.
Sarcasm off.
wayne says:
March 30, 2011 at 9:07 pm
“My interests of decades has been in astrophysics but I have never seen a definitive spectrum of purely nitrogen or argon or for pure hydrogen to indicate that those symmetric gases in our galaxy are opaque to IR.”
I think you misspoke there. Nitrogen gas is primarily transparent to IR, not opaque, just like it is primarily transparent to visible light. If nitrogen gas were primarily opaque to IR, imagine how difficult it would be to take a thermographic image in an atmosphere that is almost 80% nitrogen gas. It would be like trying to take a photograph in the fog.
“So, you could flip me back so easily if you could just show me a spectrum with lines or continuum ranges that are identified to be nitrogen or argon for that matter.”
In the infrared ranges? I can’t find those either, which is understandable. Typical infrared spectrometers use nitrogen gas as a purge for a very good reason – it doesn’t interfere with the readings. I have found a few articles, unfortunately behind paywalls, that reference it’s absorption/emission spectrum in the far infrared:
http://link.aip.org/link/doi/10.1063/1.431648
http://iopscience.iop.org/0022-3700/10/3/018
http://www.informaworld.com/smpp/content~db=all?content=10.1080/00268978400102221
“Why do think nitrogen gas is always radiating energy away in all directions?”
Because outer space is much colder than the temperature of the atmosphere, and nitrogen gas makes up almost 80% of the atmosphere. I feel it get cold at night, so the heat is going somewhere. Since it doesn’t get continuously hotter every day, the heat must be leaving the atmosphere, right?
You may be interested to read how satellite measurements pertaining to our temperature records are primarily readings in the microwave band, which correlates to heat radiated by oxygen gases (including ozone):
http://en.wikipedia.org/wiki/MSU_temperature_measurements
bananabender says:
March 30, 2011 at 7:51 pm
Your anecdotes are amusing folklore but they show absolutely no understanding of basic scientific principles. >>>
My anecdotes are an attempt to show via real world example how the laws of physics work. The equation to describe the amount of power in watts per square meter (P) that a black body will radiate at a given temperature (T) in degrees Kelvin is as follows:
P = 5.67 * 10^-8 * T^4
If you will take note Mr Banana, there is no variable in that equation pertaining to the temperature of the environment the black body exists in. In other words, the amount of energy being radiated by the black body is 100% governed by its temperature in degrees K. It matters not one whit if it is warmer or colder than its immediate environment, that is the amount of energy that it will radiate. The energy it radiates is in the form of photons, which travel at the speed of light (for purposes of this discussion they are light) and have a mass approaching a limit of zero. They exhibit the characteristics of both particles and waves, and are actually neither and both at the same time. Light exhibits a number of properties that demonstrate it exists physically outside of certain parameters we normally take for granted. Consider as an example a person on a high speed train reading a newspaper by the dome light above. If they were to calculate the speed with which photons emerge from the bulb and travel to the newspaper, the observer would arrive at a speed of exactly 1 times the speed of light. A stationary observer however would not see the light as going “straight down”. The stationary observer would see the light going down and sideways at the same time. If the stationary observer calculates the speed of the photons they will get… 1 times the speed of light. Before you start screaming that this violates simple pythagorean geometry and the laws of motion, please be advised that the speed of light is in fact immutable, the traveller on the high speed train does not pass through time as fast as the stationary observer, and hence they both see the light travelling at the same speed. If you want to argue with me as to this being correct or not, I shall cite one reference only:
Mr Albert Einstein
His theory of relativity provided for the quantification of precisely how much time a body in motion at a given velocity would lose in comparison to a stationary body, and this has been confirmed through experimentation multiple times. In brief, photons ain’t what you seem to think, they radiate from a surface according to the Stefan-Boltzman equation (above) and they don’t give a tinkers damn how fast you think they are going, they only have one speed, and that’s the speed they go until they collide with something, and they will collide with that something no matter what the temperature of that something is. Would you like me to explain Plank’s constant next?
“Have you ever used a bunsen burner? You vary the flame temperature of this burner by altering the amount of air mixing with the gas. The more air you add the more completely the gas burns and the hotter the flames. Fanning a fire is exactly the same principle. “>>>
Well, having discredited yourself as a physicist, chemist, and Scout, you now want to move on to Bunsen burners? Yes, I’ve used one. Many times. If you open the vent at the bottom all the way to let in as much air as possible, the flame will most likely blow out. At best you will get a crackling sound with the bottom of the flame starting about a quarter of an inch above the lip of the burner. Close down the vent until you get a tall silent flame instead. This results in the oxidizing molecules travelling as straight upward in a narrow column as possible in order to concentrate the heat best on the bottom of your flask or other object you happen to be heating. They’re very handy for heating up and bending glass tubing in this manner, but I recommend you keep the first rule of chemistry labs in mind when you try. Hot glass looks exactly the same as cold glass.
“I was employed as chemist in the food and beverage industries after I graduated from university. I regularly used various analytical spectroscopy methods including AAS, GC-MS, spectrophotometry, IR and near IR spectroscopy . Spectroscopy is based entirely on radiative physics.>>>
Congrats. I see you neatly side stepped the question, which I will repeat. You claim to have three degrees. What are they? Not physics or chemistry obviously or you wouldn’t be trying to build your credibility by quoting all the cool instruments you used at a job after university. Guess what? I can drive a car. That doesn’t make me an automotive engineer. I can use a computer, that doesn’t mean I can design one. Operating lab equipment doesn’t provide you with any understanding of the physics the equipment is based on. Unless you actually understand the physics, you’re just following the instructions in the manual and for all you know its just a big, fancy, thermometer.
If you still think that operating a thermometer makes you a physicist, I may be willing to go another round. But at least come up with a tough one that doesn’t make you look foolish in the end, Mr not-a-Scout/chemist/physicist but I operated a spectrometer at work once and have three degrees.
BigWaveDave says:
March 30, 2011 at 10:28 pm
Atmospheric pressure is a function of temperature. Of course our atmosphere (as do most) behaves as a near ideal gas. There is no fixed volume – it can only expand against the force of gravity. The outward force of temperature balanced by the downward force of gravity determines the pressure and volume. There is no volume constraint, so changes in temperature generate changes in pressure. Do you think that the night side of earth is exposed to a lower pressure from outer space, so the temperature goes down?
I will rephrase my comment. The atmosphere can be sufficiently heated by molecular kinetic effects without needy any radiation-absorbing “GHGs”. Radiative energy transfer only becomes more important than conduction and convection at the outer edge of the atmosphere.