Absorption of radiation from a 'cooler' source

In summary: Radiation from the coffee will cause the particles of the two cups of coffee to vibrate and transfer energy to each other. The two cups of coffee will then become warmer, but the coffee will have lost energy in the form of heat.
  • #36
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.
 
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  • #37
DaleSpam said:
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
Arfur Bryant said:
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
Arfur Bryant said:
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
 
<h2>What is meant by "absorption of radiation from a 'cooler' source"?</h2><p>"Absorption of radiation from a 'cooler' source" refers to the process in which an object absorbs electromagnetic radiation from a source that has a lower temperature than itself. This can result in a transfer of energy from the cooler source to the object, causing it to increase in temperature.</p><h2>Why does absorption of radiation from a 'cooler' source occur?</h2><p>This occurs due to the laws of thermodynamics, specifically the second law which states that heat will naturally flow from a hotter object to a cooler one. When a cooler source emits radiation, it is absorbed by objects with higher temperatures, causing them to warm up.</p><h2>What factors affect the absorption of radiation from a 'cooler' source?</h2><p>The absorption of radiation from a 'cooler' source can be affected by various factors such as the temperature difference between the source and the object, the type of material the object is made of, and the wavelength of the radiation being emitted.</p><h2>Can objects absorb radiation from a 'cooler' source even if they are not in direct contact?</h2><p>Yes, objects can still absorb radiation from a 'cooler' source even if they are not in direct contact. This is because electromagnetic radiation can travel through space and be absorbed by objects that are within its range.</p><h2>What are some real-world examples of absorption of radiation from a 'cooler' source?</h2><p>Some examples include the Earth absorbing radiation from the Sun, causing it to warm up, or a person standing near a fire absorbing heat from the fire, causing their body temperature to increase. Another example is a greenhouse, where the glass walls absorb radiation from the cooler atmosphere outside, trapping heat inside and causing the temperature to rise.</p>

What is meant by "absorption of radiation from a 'cooler' source"?

"Absorption of radiation from a 'cooler' source" refers to the process in which an object absorbs electromagnetic radiation from a source that has a lower temperature than itself. This can result in a transfer of energy from the cooler source to the object, causing it to increase in temperature.

Why does absorption of radiation from a 'cooler' source occur?

This occurs due to the laws of thermodynamics, specifically the second law which states that heat will naturally flow from a hotter object to a cooler one. When a cooler source emits radiation, it is absorbed by objects with higher temperatures, causing them to warm up.

What factors affect the absorption of radiation from a 'cooler' source?

The absorption of radiation from a 'cooler' source can be affected by various factors such as the temperature difference between the source and the object, the type of material the object is made of, and the wavelength of the radiation being emitted.

Can objects absorb radiation from a 'cooler' source even if they are not in direct contact?

Yes, objects can still absorb radiation from a 'cooler' source even if they are not in direct contact. This is because electromagnetic radiation can travel through space and be absorbed by objects that are within its range.

What are some real-world examples of absorption of radiation from a 'cooler' source?

Some examples include the Earth absorbing radiation from the Sun, causing it to warm up, or a person standing near a fire absorbing heat from the fire, causing their body temperature to increase. Another example is a greenhouse, where the glass walls absorb radiation from the cooler atmosphere outside, trapping heat inside and causing the temperature to rise.

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