Absorption of radiation from a 'cooler' source

[Dale]

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.

 

[Arfur Bryant]

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

 

[Nugatory]

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.

 

[Jano L.]

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.

 

[Arfur Bryant]

Guys,

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

Kind regards,

AB