Absorption of radiation from a 'cooler' source

  • #26
Look up radiosity.
Did that. Not sure that it addresses the difference in wavelength though...

AB
 
  • #27
You may have to look to quantum theory here. No matter what energy state an electron / atom is in, a quantum of energy may knock it into a higher energy state. The probabilities may be small.
Factchecker,

Can I have your opinion on this statement? (From hyper-physics)

["In the interaction of radiation with matter, if there is no pair of energy states such that the photon energy can elevate the system from the lower to the upper state, then the matter will be transparent to that radiation."]

AB
 
  • #28
Dale
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DaleSpam,

A blackbody is an idealised concept. I am talking about the real world. Yes, a blackbody can absorb all radiation but this has no relevance to my question.
There are lots of things that are reasonable approximations to blackbodies. In any case, going to a greybody complicates the analysis, but doesn't change the conclusion. All you have to do is do the analysis at each frequency, factor in the emissivities of the two bodies at that frequency, and integrate over the frequencies. The end result is the same.

But lets pretend you were <not> talking about blackbodies. Now...

In your situation 2), you state that the warm body would cool slower due to the absorption of radiation from the cooler body. This is exactly what my original question asked. If the warmer body absorbs that radiation for energy gain, how can this happen when the frequency of the receiving radiation cannot elevate the electron orbit?
Not all thermal absorption is due to elevating atomic orbits.

For example, in a star the most common interaction will probably be inelastic scattering off a free charge carrier. In other cases an absorbed photon will increase molecular energy by bending or rotating or some other degree of freedom. Or in a solid a photon can be absorbed to produce several phonons in the lattice. Or a photon can produce standing waves inside a cavity. Etc.

It is a mistake to think of a photon's interaction with matter so narrowly. The point is that if a greybody is emitting at a given wavelength then it can also absorb at that same wavelength.

Further, if what you say is true, what would happen if more cooler bodies were added, surrounding the warmer body? Would the warmer body cool even slower until - eventually - the number of cooler surrounding bodies is such that the warmer body is absorbing so much energy it starts to warm up!
The rate at which it cools down would indeed decrease, but it would never start to warm up. Once the warm body is completely enclosed by cooler bodies (they cover 4π steradians) then the addition of more cooler bodies makes no further difference.

At this point you have engineered a situation whereby adding more cooler objects to a system actually increases the temperature of the warmer body!

This cannot be correct.
I agree.
 
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  • #29
AlephZero
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Does the fire (filament) heat up because it absorbs this 'cooler' radiation from the chair?
You need to take everything in your scenario into account here. If the fire is in a closed room (whatever the size of the room) it will be absorbing the heat radiated back from the walls of the room. Adding a chair into the scenario is just a detail.

Suppose the electrical power into the fire is a constant 1000W.

If the fire was in outer space (and ignoring the existence of the rest of the universe) it would reach a constant temperature where it radiated 1000W, the same as the electrical power.

If it was in a room where its surroundings radiatie say 10W of heat back to the fire, it will be at a slightly higher constant temperature where it radiates 1010W. The net amount of heat entering the room is still 1000W.

Of course there must be some way for the net 1000W of heat to get out of the "closed" room (e.g. conduction through the walls and convection into the air outside), otherwise the fire and the room will never reach a constant temperature. They will both increase in temperature "for ever" (at least, until something melted or caught fire), because of the 1000W of electrical heat input.
 
  • #30
Dalespam,

Thank you for that reply.

AB
 
  • #31
You need to take everything in your scenario into account here. If the fire is in a closed room (whatever the size of the room) it will be absorbing the heat radiated back from the walls of the room. Adding a chair into the scenario is just a detail.

Suppose the electrical power into the fire is a constant 1000W.

If the fire was in outer space (and ignoring the existence of the rest of the universe) it would reach a constant temperature where it radiated 1000W, the same as the electrical power.

If it was in a room where its surroundings radiatie say 10W of heat back to the fire, it will be at a slightly higher constant temperature where it radiates 1010W. The net amount of heat entering the room is still 1000W.

Of course there must be some way for the net 1000W of heat to get out of the "closed" room (e.g. conduction through the walls and convection into the air outside), otherwise the fire and the room will never reach a constant temperature. They will both increase in temperature "for ever" (at least, until something melted or caught fire), because of the 1000W of electrical heat input.
AlephZero,

I understand what you are saying. However, are you really sure that the filament of the fire gets hotter? That was my question. Because, if it does, then it appears to me that the 2nd Law of Thermodynamics has been broken. Remember that the only thermal energy in the room comes from the filament. If it truly absorbs the 10W from the walls, then the filament has heated itself. How can something heat itself?

The chair is not to be dismissed. Adding the chair provides an example of a cooler body in addition to the walls which had been at equilibrium with the fire (as I stated). My question addresses whether the 'backradiation' from the chair can heat the filament. Do you agree with this?

However, thank you for a comprehensive reply.

AB
 
  • #32
AlephZero
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Remember that the only thermal energy in the room comes from the filament
No. You are adding 400 joules of energy per second to the filament of the heater, from whatever is generating the electricity supply (plus whatever is radiated from the rest of the room and its contents)

The equilibrium temperature of the filament is whatever it needs to be, so that it loses the same amount of heat as it absorbs.
 
  • #33
Dale
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The back radiation doesn't heat the filament, it just makes it cool more slowly. The 1000 W of external power is what heats the filament. Because the filament cools more slowly, the filament must burn hotter to radiate the 1000 W.

The second law is not violated, no net energy is spontaneously going "uphill".
 
  • #34
No. You are adding 400 joules of energy per second to the filament of the heater, from whatever is generating the electricity supply (plus whatever is radiated from the rest of the room and its contents)

The equilibrium temperature of the filament is whatever it needs to be, so that it loses the same amount of heat as it absorbs.
AlephZero,

First of all, thank you. I am getting closer to understanding this subject.

To be pedantic, I still say the only 'thermal' energy is from the filament. The electrical source is not thermal energy.

As for the equilibrium temperature, this goes back to my first post. Is the backradiation from the wall absorbed by the filament for net gain? If yes, you are correct. If no, the filament does not burn hotter. Would that be fair?

So, to make sure I understand, if you have two light bulbs with identical filament and identical electrical supply, but one bulb surface is frosted, according to you the frosted bulb filament would burn hotter, yes?
 
  • #35
The back radiation doesn't heat the filament, it just makes it cool more slowly. The 1000 W of external power is what heats the filament. Because the filament cools more slowly, the filament must burn hotter to radiate the 1000 W.

The second law is not violated, no net energy is spontaneously going "uphill".
DaleSpam,

Thank you for helping me to understand.

However, if the filament burns hotter because it is cooling more slowly (your last sentence), then surely the backradiation has - in fact - heated the filament. Yes?
 
  • #36
Dale
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Not in the sense of the 2nd law. The net energy transfer is still from hot to cold.

The second law does not require that any interaction between a hot and a cold body be identical to an interaction between a hot and a 0 K thermal bath. It only requires that the net energy transfer through heat always be from hot to cold. That is the case in this situation.

Cooling more slowly is not heating. No insulation manufacturer claims that their insulation heats the home.
 
  • #37
Not in the sense of the 2nd law. The net energy transfer is still from hot to cold.

The second law does not require that any interaction between a hot and a cold body be identical to an interaction between a hot and a 0 K thermal bath. It only requires that the net energy transfer through heat always be from hot to cold. That is the case in this situation.

Cooling more slowly is not heating. No insulation manufacturer claims that their insulation heats the home.
Again, thank you for this. It was your term '...would burn hotter' that made me think you were saying the backradiation caused the filament to heat up.

Interesting point though, now I think about it, but if AlephZero's numbers means the filament increases to 1010W (by absorbing the 10W backradiated from the wall), then the wall would now be receiving more energy than it was originally. Therefore it is now emitting more backradiation (I know that is probably not an official term) to the fire. This cycle would continue ad infinitum. Doesn't this mean a perpetual (although incrementally small) increase in energy being emitted at the filament for no extra demand from the electrical supply? This would seem to be unlikely, leading me to question whether the backradiation is actually absorbed by the filament.

Does that make sense?

AB
 
  • #38
Nugatory
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Interesting point though, now I think about it, but if AlephZero's numbers means the filament increases to 1010W (by absorbing the 10W backradiated from the wall), then the wall would now be receiving more energy than it was originally. Therefore it is now emitting more backradiation (I know that is probably not an official term) to the fire. This cycle would continue ad infinitum. Doesn't this mean a perpetual (although incrementally small) increase in energy being emitted at the filament for no extra demand from the electrical supply?
No. The electrical supply is delivering 100 Joules every second (that's what 100 Watts means). If you look at the surface of any volume (any shape, any size, includes or doesn't include part or all the walls) that completely encloses the filament and wait long enough for the system to reach equilibrium, you'll find that the net flow of energy through that surface is 100 Joules per second.
 
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  • #39
Jano L.
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Interesting point though, now I think about it, but if AlephZero's numbers means the filament increases to 1010W (by absorbing the 10W backradiated from the wall), then the wall would now be receiving more energy than it was originally. Therefore it is now emitting more backradiation (I know that is probably not an official term) to the fire. This cycle would continue ad infinitum. Doesn't this mean a perpetual (although incrementally small) increase in energy being emitted at the filament for no extra demand from the electrical supply? This would seem to be unlikely, leading me to question whether the backradiation is actually absorbed by the filament.

Does that make sense?

AB
Yes, but each next scattering will produce weaker and weaker radiation. The net result in stationary radiation transfer will be such that the net power leaving the electrically or otherwise supplied source of heat radiation will be finite and constant.
 
  • #40
Guys,

OK, thanks very much for your time. It has helped me a lot.

Kind regards,

AB
 

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