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.
Myrrh says:
April 3, 2011 at 7:54 am
Phil – the atmosphere is the liquid gas Air, it has volume and it has weight, it is pressing down on us with a tremendous amount of pressure, around 14 lbs/square inch, around a ton pressing down on our shoulders. The molecules of Air, mainly nitrogen and oxygen, do not move very fast at all through all this.
Most probable velocity of N2 at room temperature is 422 m/sec or 1520 km/hr,
RMS speed is 517 m/sec or 1861 km/hr. He and H2 travel so fast that they can exceed the Earth’s escape velocity!
Anthony do yourself and everyone on here a big favor and get rid of this anti-science nonsense!
DavidH and RichardE
Thanks for the replies.
DavidH
Isn’t this from Ira
“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”
Very different to
“The chicken on the counter heats up until it reached a new equilibrium temperature where it is radiating as much heat as it is absorbing from the beam”.
Is this new equilibrum similar to a CO2 changing the adiabatic heating capacity of earth. Rather than further additional heat or energy being added by backradiation.
Joel Shore says:
April 2, 2011 at 3:27 pm
“Hans,
I have no idea why you introduced the concept of radiation density; it only complicates things to the point where you can make nonsense assertions that are far enough removed from simple understanding that perhaps some people won’t realize what nonsense it is.”
No, it gives a good insight on how to look at radiation coming from a surface. It’s not complicated but much easier as all these posts about the GHG effect. I did not make nonsense assertions, it’s all explained on the Hyperphysics page.
“If you have that object radiating into the vacuum of space, it is going to come into equilibrium by finding the temperature where the radiation it emits by the Stefan-Boltzmann Law matches the incoming power”
Following the gradient and cause and effect, the sun determines the surface temperature, then the surface will radiate.
“Now instead, try surrounding that object by a box or shell that absorbs some of the radiation. That surrounding shell will heat up and by virtue of having a nonzero temperature, it will emit radiation. Now the object will be receiving not only the power it was originally receiving but additional power from the radiating shell; there is no way this can be avoided. Hence, in order to reach a state where energy in = energy out, it will have to emit more power. It does this by raising its temperature. It’s really that simple”
The net power equals sigma *A (T1^4 – T2^4) so the power reduces when the shell gets warmer. You mix energy and power here. Energy in-out will always be the same, but not the power. Besides the in-out balance is First Law that just is what it is but tells you nothing about what will happen, it’s the Second Law that tells you what will happen.
Without a gradient (delta T) nothing happens and in = out = zero.
And think of cause and effect, it’s not power creating temperature but temperature creating power. Without temperature difference there will be no power, without power there will still be a temperature.
“No…It is impossible for it to be in equilibrium. It was in equilibrium at 400 K when it didn’t have the walls at whatever elevated temperature that they are at emitting radiation at it. How can it possibly remain in equilibrium now that it is receiving more energy? That’s mathematically impossible. If it receives more energy, it has to emit more energy…and it does this by increasing its temperature.”
You forgot this part: ‘In reality the walls won’t even become 400 K but 399,9…K at the most as long as the box radiates on the outside (and the RED inside spreads out), so the box will never get as warm as the energy source!’
But if you insist, the limit would be where the surroundings outside this box would also become 400 K and the power becomes zero thus equilibrium.
My point was that this unrealistic case would be the ultimate end, while no heating of the source by the box ever took place because for that to happen the box needs to exceed 400 K.
” So many seem to forget that you need a gradient(delta) in temperature, pressure, concentration or density to get a flow. And if the gradient becomes zero the flow stops.
Exactly…which is why your solution can’t be the steady-state. If there is no net energy flow away from the soldering iron, which is still plugged into the wall and converting electrical energy into thermal energy, what is it going to do? It is going to heat up until the heat leaving it does balance the thermal energy that it is generating! How could it do otherwise?”
The steady state is not my solution. It’s yours, by the idea of heating a power source by a shell that needs to create a steady state and then go beyond that and create a power flow in reverse.
I explained that this is unrealistic, so don’t try to reverse things and make it my solution.
Ira Glickstein, PhD says:
March 30, 2011 at 2:02 pm
Your quote of a quote:
“”But in fact it does not warm up; it’s temperature remains exactly the same. The reason why is very simple to understand but extremely important to physics: the blackbody is already in radiative thermal equilibrium with a hotter source of energy, the higher radiative energy spectrum light from the light-bulb. You cannot make something warmer by introducing to it something colder, or even the same temperature! You can only make something warmer, with something that is warmer! This reality is called the 2nd Law of Thermodynamics, and is so central and fundamental to modern physics it cannot be expressed strongly enough.””
Your reply to the quote of the quote:
“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.”
———————————————————————————————————;
False.
The temperature of an ideal black body determines it’s cut off frequency.
An ideal black body absorbs all incoming coherent electromagnetic radiation, re-emits the coherent electromagnetic radiation below the cut off frequency, and converts the coherent electromagnetic radiation above the the cut off frequency into incoherent high frequency white noise – or heat.
Only electromagnetic radiation above the cut off frequency warms the black body.
Also note, the black body model excludes the possibility of back radiation.
I suggest you re-read the article.
Two wrongs don’t make a right (but 3 lefts do.)
RJ says:
No…It has nothing to do with changing the thermal energy of a gas by adiabatic compression, if that was what you are asking.
Hans says:
Power is just energy per unit time, so your last sentence is basically non-sensical. And, yes, the power does get smaller as the temperature of the shell increases…That’s the whole point. Now take that to its logical conclusion and you will understand why the temperature of the object must increase.
Okay, so to interpret what you are saying charitably: You are saying if the temperatures are the same, there will be no heat flow…i.e., the power will be zero. So what, pray tell, is going to happen to the energy that the object is receiving (e.g., from the sun in the case of the earth or from the conversion of electrical energy into thermal energy in the case of the soldering iron)? You have just told me there is no energy flow away from the object. So, how is it not going to increase its thermal energy and hence its temperature?
I agree … Your solution is not a steady-state, which leads me to wonder why you have presented it to us as what will happen. What you are telling me is something that might occur at some instant of time for some contrived circumstances. However, that is not the subject of our discussion…We are talking about what state the system will find where its temperature is steady with time. The point being that such a temperature will be higher in one case than the other.
Hans, you are really talking nonsense here and you are only embarrassing yourself by keeping it up. With Myrrh, I can understand that he just doesn’t have enough knowledge to educate himself, but I think in your case that you do. So, why don’t you just man up and admit that you are wrong? You’ll actually earn a little bit of respect that way.
Richard E Smith;
The temperature of the chicken cannot be higher than that determined by the quantity of thermal radiation it is receiving whether it inside an oven or not.>>>
RJ;
“Some of the re-emitted Photons make their way… back down to the Surface where their energy is absorbed, further heating the Earth”
Very different to
“The chicken on the counter heats up until it reached a new equilibrium temperature where it is radiating as much heat as it is absorbing from the beam”.
Exactly the same, and for exactly the same reason. For simplicity we’ll use a pair of black body chickens at room temperature of 20 C. By SB Law they are radiating 418 w/m2. Put one chicken in the oven, which we’ll assume is also at room temperature. Chickens are radiating 418 w/m2 and the room and everything in it is radiating at 418 w/m2. Equilibrium.
Turn on magical 500 watt beams, one aimed at Counter Chicken and one through the special window at Oven Chicken. Being black body chickens, they absorb 100%. Counter Chicken increases in temperature until it reaches 918 w/m2 or about 84 degC. Where does the extra 500 watts it is radiating go? Everywhere. It get’s absorbed by the environment the Counter Chicken is in, but the area in our fantasy kitchen of walls, ceiling, floor, is so large by comparison the effect is negligible.
What of Oven Chicken? Oven Chicken is in an Oven made of…Oven Stuff, which is also a black body. Its not very much larger than the chicken. It starts out at 20 C, just like the chicken. So Black Oven is radiating 418 w/m2 and so is Oven Chicken, equilibrium. Until we hit Oven Chicken with that 500 watt beam. Oven Chicken rises in temp up to 918 w/m2 and 84 degrees. But that’s much hotter than Black Oven, which is at 20 degrees and only radiating 418 w/m2. So Black Oven is getting 418 w/m2 on the outside from the environment, but from the inside it is getting more because the Oven Chicken is at 84 degC. Now the area of the inside walls of the Black Oven is much larger than that of the Oven Chicken. Let’s say 10 times. So the inside walls of the Black Oven are getting an extra 500 w/m2, but that’s spread out over ten times as much area, so 50 w/m2. Since outside of Black Oven is receiving 418 and radiating 418, but inside Black Oven is receiving 468 but only radiating 418…Black Oven’s temperature must rise. How much?
Black Oven has TWO surfaces that can radiate heat, one inside, and one outside. So being made of some magical Black Body Stuff so thin that the outside surface area is pretty much the same as the inside surface area, Black Oven must increase in temperature until it is radiating an extra 25 w/m2. That would be 418 w/m2 (room temp) plus 25 w/m2 (from Oven Chicken) a temperature of 24.3 degC. Now Oven Chicken has an equilibrium problem. Oven chicken is radiating at 918 w/m2, but stupid Black Oven is radiating at 25w/m2 more than room temperature. PLUS, it has an area 10 times that of Oven Chicken, so Oven Chicken is now getting ANOTHER 250 w/m2.
Oven Chicken’s temp now goes up to 126 degC and it is radiating 1443 w/m2. But that’s 250 w/m2 going back toward Black Oven, with a 10 times surface area….so Black Oven is now getting ANOTHER 25 w/m2 more than before. But it has two surfaces to radiate from, so another 12.5 w/m2. So Black Oven has to get to 455.5 w/m2 or 26.4 degC.
Now Black Chicken has a problem…. extra 125 w/m2 = 134.8 DegC.
Now Black Oven is out of whack, extra 6.25 w/m2 = 27.4 DegC
Now Black Chicken is getting another 62.5 w/m2….
Then Black Oven get’s an extra 3.125 w/m2
Black Chicken 31.25 w.m2
Black Oven 1.66 w.m2
And so on until the numbers get so small they don’t matter. Exactly the same process as Ira described for GHG’s warming the earth. Exactly the numbers from SB Law. Exactly in accordance with 2nd Law of Thermodynamics. And most importantly…
Verified by experimentation. Build an apparatus like this, measure the effective black body characteristics of both Actual Chicken and Actual Oven, and you’ll be able to apply an Actual Heat Source and predict in advance the Actual Final Temperature with in fractions of a degree. Been done, many times, repeatedly, since 1879 when Stefan first formulated the equation. Theory and practical verfication match precisely.
Sorry if that bursts your bubbles, argue cold things can’t heat warm things until you are blue in the face, the theory has been verified by experimentation thousands of times for more than a century. I consider the claims of the AGW Alarmongerers to be a total farce, but that has to do with a whole lot of other factors. On this issue they have it right. Stop getting your panties in a twist about this and hit them on the dozens and dozens of things they have wrong instead.
At each step the amount of temp
Richard E Smith says:
The flow of radiation is not “blocked” by the box. The box absorbs the radiation. Then, because it has a nonzero temperature, it emits radiation. You don’t have to call it “back radiation” if that term upsets you…In fact, there are people (e.g., this retired meteorologist http://www.ems.psu.edu/~fraser/Bad/BadGreenhouse.html ) who have recommended avoidance of that term because it tends to sow confusion. I always thought he was a bit too militant on this point, but I am beginning to see his reasoning now that I see how the term “back radiation” leads some people to say completely nonsensical things like “reflection or backradiation merely reflects exactly the same amount of energy so there can be no additional heating”.
So when you say, “The temperature of the chicken cannot be higher than that determined by the quantity of thermal radiation it is receiving whether it inside an oven or not”, then you are correct. However, it is receiving more thermal radiation when it is in the oven because the oven walls are at a temperature higher than the surroundings (but lower than the chicken) and they radiate, just as any object does.
Right on Phil – all these molecules travelling at escape velocity… we obviously live in a vacuum in which no sound is heard except the sound of one hand clapping.
Well done. You’ve explained it nicely for me.
Myrrh says:
April 3, 2011 at 11:35 am
Right on Phil – all these molecules travelling at escape velocity… we obviously live in a vacuum in which no sound is heard except the sound of one hand clapping.
Well done. You’ve explained it nicely for me.>>>
Myrrh, you clearly know nothing about physics. It takes two hands to clap. Traditional Physics. You can CATCH the clap, but that’s a biological clap not a physical clap. A physical clap requires two hands, and any assertion that a clap made with just one hand has any sound at all, not matter how quiet, is pure, unadulterated, claptrap. You’re making a fool of yourself by claiming otherwise. Look it up on any physics site, or any practical reference guide on clapping.
I finally figured out where you have gone wrong here. You think that temperature, per se, is the same as energy.
You think that I assume the metal box in which I hung that 400K soldering iron will get up to 399K or so. I assume nothing of the sort. The box is larger than the soldering iron and much larger than the tip. Indeed, if the box was the size of a tool box, you might be able to hold it in your hands and find it quite warm, but not burning. The tip, of course, would burn your fingers.
When the temperatures stabilize, the radiation from the box to the free air in the room, we both agree, must exactly equal the electrical power in to the soldering iron. Since the box is so much larger surface area than the tip and shank and other hot parts of of the soldering iron, it can, acccording to SB, radiate that amount of energy at a much lower temperature than the 400K tip.
There are a couple commenters in this and others of my Topic threads who I have judged to be incapable of reasoned cross-discussion. Either because they are so fixed in their prejudices or clueless in their science backgrounds. So, I just scroll through their comments without reply, for the most part, unless my eye catches something humorous or otherwise worth reading.
I don’t get paid in cash for this blogging activity, but I do feel rewarded when my Topics score high in Page Views and Number of Responses, and when I rank high in the list of Guest Contributors. So (you know who you are) keep it up!
Of course, the best reward is when a comment teaches me something important that I did not know before, which has occured several times in these Topic threads. THANKS! And thanks also to those WUWT readers who take the time to try to set those who honestly misunderstand on the correct track. And, of course, to those who take the time time to thank me.
Every person adds joy to the world – some by arriving and some by leaving.
Ira said that the warmth we feel from the Sun comes from the Solar energies which are the Visible and two shortwave either side, UV and Near IR, the downwelling energies of the AGW Energy Budget. None of these is Thermal. They cannot be felt as heat. What is felt as Heat is Thermal IR. This is what Ira is refusing to address in my questions. I can only conclude that it is a CON, and the refusal to engage with me on this is because it is knowingly a CON.
Tim, I don’t believe this 1% for Thermal IR downwelling to Earth. The only thing so far I’ve been able to find on actually measuring the temperature of the light waves themselves was an experiment which said to first cover the thermometer bulb with black tape before doing the prism experiment. But, at every colour of light shown on the white paper Thermal IR is also being measured. The experiment however stopped there, and claimed that what was being measured was discrete colours visible and invisible. Nonsense, right? Now, mercury is practically a perfect reflector of visible light, so to get a true reading of each visible colour one would have to do the same experiment without the bulb being black taped, and then this result subtracted from the first. So many scientists here, do you all agree that would give me a true reading of each colour discretely?
Myrrh says:
April 4, 2011 at 7:54 am
Ira said that the warmth we feel from the Sun comes from the Solar energies which are the Visible and two shortwave either side, UV and Near IR, the downwelling energies of the AGW Energy Budget. None of these is Thermal. They cannot be felt as heat. What is felt as Heat is Thermal IR. This is what Ira is refusing to address in my questions. I can only conclude that it is a CON, and the refusal to engage with me on this is because it is knowingly a CON.
They are all able to warm an object when absorbed. The refusal to engage you is because you won’t listen and keep on trotting out the same garbage.
Phil – what is the problem here? It is absolutely bog standard science fact that the warmth we FEEL is Thermal IR. That’s how we, scientists every one of us, can tell. The heat we feel from a fire is THERMAL IR. We each of are able to test for ourselves whether something is giving off heat. If it’s giving off heat it’s Thermal IR. It’s that bloody simple.
Something doesn’t have to be hot enough to glow red to be able to feel the Thermal IR it’s giving off, a hot ring on a stove set to simmer won’t glow red, cooling embers of a fire.
We cannot feel heat from UV. It can burn us if intense enough, but we do not feel it as heat, it is not Thermal IR. It doesn’t even penetrate the skin! It barely gets through to the edge of the first layer of skin (of three layers), the epidermis. UV may well be heating a small millimetre of the surface of the Earth, but that’s about it. If that’s “converting to heat energy, Thermal IR from the Earth upwelling”, then just how much is it?
Visible light doesn’t penetrate that much further, we can’t feel it as heat. Near IR is not hot either.
So, I’m still demanding proof that it’s these Solar energies of AGW that are actually heating the Earth and producing all that Thermal IR upwelling.
Unless you can provide this, your AGW Energy Budget is pie in sky, fiction.
And, I think it absolutely deplorable that you treat my questions about this with such arrogant disdain, vaunting yourselves as scientists and belittling me. This is AGW’s basic premise on which all other energy calculations are made. You have a scientific obligation to answer this. If you don’t, then you are either con artists or, not scientists.
Myrrh says:
No, the energy budget is based on measured quantities. Whether or not you can “feel” visible light, which involves a lot of neurological and other complex issues in addition to physics is irrelevant, although there is no reason to believe you are right on that anyway.
We don’t have any responsibility whatsoever. Out of some vague sense of wanting to help educate someone who is seriously confused, we have tried to engage you and we have been met with belligerent assertions of nonsense at every turn. You have made the choice to wallow in your own ignorance despite our best efforts to try to set you on a course to understanding. You are clearly too ideologically committed to what you believe to be disabused of that ignorance. It is ridiculous to claim that we have some sort of obligation to do something that you are making impossible for us to do.
davidmhoffer says that a blackbody chicken in an oven emitting all the radiation it is absorbing will be heated up further by its own emissions returning to it from the inside of the oven it is sitting in. (David’s oven is assumed to have a thin blackbody skin.) So a cooler object (the oven walls) is heating a warmer object (the chicken).
In his example the blackbody chicken receives 500 watts from a heat source, but the oven walls radiate half back to the chicken warming it up. The chicken is now emitting 750 watts and in David’s example, if I understand it correctly, half of this extra 250 watts is re-emitted back to the chicken and then half of this is re-emitted (250+125+62.5 etc) i.e. an extra 500 watts at no extra expense and the chicken is emitting twice as much energy as is being provided by the only heat source. (David’s decimal points and zeros are a bit wayward, but that is what I think he is saying. It certainly resembles standard greenhouse physics.)
I have some observations about this.
When the chicken is receiving 750 watts why does it not emit the whole of the 750 back to the walls? When half of this is returned we would then have an additional 375 watts. Add this to the 750 and we now have 1125. Half of 1125 is then re-emitted and so on. Such heating could go indefinitely.
David says that his thought experiment has been proven thousands of times and the second law is not broken because radiative equilibrium is reached as 500 watts enter the oven and 500 go out. However, there seems to be the creation of energy inside the oven. The effect is the same as having two heating elements of 500 watts. Please point me to one of these experiments because it seems to me that if it is true, we only need to provide 500 watts to produce 1kw (or more?) and save a lot of energy in the cooking or heating process.
The thought experiment seems to show that radiative insulation is impossible.
Further, if instead of blackbody oven walls we had one that reflected all the radiant emissions from the chicken (like the solar oven that started this debate) the mutual heating between the chicken and the oven would produce so much thermal energy that it would explode making rather a sorry mess of the kitchen (not to mention the chicken). I know David does not think that this can happen (obviously) but it is not apparent to me why not, given that he believes that radiation from the surroundings of the chicken can heat the chicken up thereby adding more radiation from the chicken and so on – a perpetuum mobile.
Myrrh;
And, I think it absolutely deplorable that you treat my questions about this with such arrogant disdain, vaunting yourselves as scientists and belittling me.>>>
Having attempted reasoned, factual, logical and detailed explanation founded upon theory verified by experimentation, and having failed utterly to get a single point through to you, we have concluded that you are protected by some sort of Magical Armour of Density.
Our attacks on this seemingly impervious protection are now taking the form of sarcasm, belittling, and anything else anyone can think of to try an find a chink in the armour through which to slip a tiny sliver of knowledge.
I believe myself that the effort may be futile, as it is well known that what you don’t know can’t hurt you, making some people invulnerable by default. That is how the Magical Armour of Density works. The Magical Cloak of Invisibility works in the opposite fashion, but has the same effect. Instead of giving the wearer such high Density that it can aborb nothing, the Magical Cloak of Invisibility simply allows everything to pass right through the wearer and so the wearer still absorbs nothing.
“I know David does not think that this can happen (obviously) but it is not apparent to me why not, given that he believes that radiation from the surroundings of the chicken can heat the chicken up thereby adding more radiation from the chicken and so on – a perpetuum mobile”.
Agree. I’m still waiting for an explanation on this point.
Davidmhoffer -Avoiding giving me the specific information I have asked for is no reply.
I am demanding a specific scientic proven fact, if you can’t provide it say so. Ad hominem attacks on me because of your own ignorance of the answer is getting tiresome.
“For simplicity we’ll use a pair of black body chickens at room temperature of 20 C. ”
Ya just gotta laugh!
cf. the spherical cow http://en.wikipedia.org/wiki/Spherical_cow
Myrrh, you are being ridiculed because your responses are lacking.
For example, your response to my question was “Visible light is easily reflected…What we see as colour in our world is what is reflected. The green light is reflected by plants, which absorb blue and red light for photosynthesis, so in reflecting back at us we see green.”
Which didn’t answer my question at all. What I asked you was what do you think happens to the energy of the visible light that isn’t reflected? Yes, everything around us reflects visible light. Now imagine every surface covered with a thin mirror, with the sun overhead. The difference in energy between that reflection and what you actually see when you look around is, approximately, the energy of the visible light that you need to account for.
Richard E Smith says:
April 4, 2011 at 10:28 am
I know David does not think that this can happen (obviously) but it is not apparent to me why not, given that he believes that radiation from the surroundings of the chicken can heat the chicken up thereby adding more radiation from the chicken and so on – a perpetuum mobile.
It’s feedback not a perpetuum mobile, the principle is used by engineers in radiation transfer calculation all the time!
A classic example that I have given here before is of a thermocouple measuring the temperature in a gas turbine combustor exhaust. If a bare thc is used it will heat up and reach an equilibrium temperature (in radiative equilibrium with its surroundings). If a tubular silica radiation shield is put around it the measured temperature increases because it is now in Eq with the shield which is in Eq with the cooler surroundings. The radiation from the shield heats up the thc even though it is cooler than the thc.
Joel Shore says:
April 3, 2011 at 11:24 am
“Hans says:
The net power equals sigma *A (T1^4 – T2^4) so the power reduces when the shell gets warmer. You mix energy and power here. Energy in-out will always be the same, but not the power.
Power is just energy per unit time, so your last sentence is basically non-sensical.”
What is non-sensical? You have 100 J radiated away in 1 hour and get 100 J out, you have 100 J radiated in 10 hours and get 100 J out. So the energy in = out always no matter what and is just a given fact, the power is what results from the temperatures.
“And, yes, the power does get smaller as the temperature of the shell increases…That’s the whole point. Now take that to its logical conclusion and you will understand why the temperature of the object must increase.”
Less power is the conclusion. It seems to me that you think there exists a conservation of power principle that will dictate a temperature, and that doesn’t exist. First you have a temperature and the Second Law that tells you if any energy will flow and how fast and in which direction, then the First Law for bookkeeping. It flows from high to low, from source to receiver and the power will have to adjust to things happening at the end, not the other way around.
Any restriction placed in a flow will reduce the flow and there is no physical effect that wants to restore it to the original flow.
I have a pump pushing 2 m^3/s through a pipe, but it’s to much so I have a valve to reduce it to 1 m^3/s. But you want to tell me that this is impossible, because the pressure will go up because I had 2 m^3/s so I will always get 2 m^3/s because there is still electricity turning the pump.
” Besides the in-out balance is First Law that just is what it is but tells you nothing about what will happen, it’s the Second Law that tells you what will happen.
Without a gradient (delta T) nothing happens and in = out = zero.
And think of cause and effect, it’s not power creating temperature but temperature creating power. Without temperature difference there will be no power, without power there will still be a temperature.
Okay, so to interpret what you are saying charitably: You are saying if the temperatures are the same, there will be no heat flow…i.e., the power will be zero.”
You think there will be power?
“So what, pray tell, is going to happen to the energy that the object is receiving (e.g., from the sun in the case of the earth or from the conversion of electrical energy into thermal energy in the case of the soldering iron)? You have just told me there is no energy flow away from the object. So, how is it not going to increase its thermal energy and hence its temperature?”
You keep pulling things out of context don’t you? Am I describing the sun – earth situation with ‘Without a gradient (delta T)’ you think? Surely your smarter than that.
So tell me now is there a power flux in Q = sigma *A (293^4 – 293^4)? Is there a temperature? So what drives what?
Richard E smith says:
First of all, the walls are not reflecting the energy from the chicken. They are absorbing energy and they are (re-)emitting energy. They, like the chicken, will find the temperature at which the amount of energy they are emitting balances the amount that they are absorbing and that will be the steady-state.
What david is describing to you is the iterative process by which one can figure out what is going to happen in this case. It is no more a “perpetuum mobile” than my describing the course of a runner in a 1 mile race as, “He first runs the first half mile, then runs the next quarter mile, then runs the next eight mile, …” It is a converging series that converges to the answer that satisfies the known laws of conservation of energy and of radiative emission from objects.
One could just write down the equations that lead to the steady-state solution to this process without going through the iterative technique that he describes. However, the iterative technique is useful in that it describes how things actually evolve over time…i.e., how the chicken in the oven once it has heated up to the same temperature as the one on the counter is still receiving more energy than it is emitting and thus must heat up more over time. It also makes clear that, contrary to your claim that this somehow violates the conservation of energy, such a higher temperature is in fact ***REQUIRED*** by conservation of energy. When the chicken in the oven is only as hot as the one on the counter got then it is still emitting less than it is absorbing…and hence, unless energy was magically disappearing, it will continue to heat up further.
Again, it is actually required by conservation of energy that the chicken heat up higher. It is your proposed solution, not his, that violates conservation of energy, as it would require the destruction of energy to prevent the chicken in the oven from getting hotter than the one on the counter. Furthermore, there is no more “creation of energy inside the oven” than there is “creation of energy” when I put on a coat to keep myself warmer than I would be if I didn’t put a coat on.