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Imagine we have a tungsten wire that we heat up to 1000 kelvins inside a tube with a special coating that reflects 99.99% of tungsten's emitted light due to thermal radiation back to the wire.

My intuition says that the wire will absorb some of the redirected light and it will cause it to heat up and release more light creating a positive feedback loop till it melts and evaporates(if the special coating doesn't do it first).

Are there any fundamental violations in my reasoning?

Edit:

For clarification i assumed that an electric current was used to heat up the wire to 1000 kelvins and the same current is always applying through the whole process.

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  • $\begingroup$ This cannot work simply because of energy conservation. If you perfectly reflect the light back at it and it reabsorbs it successfully, it will simply not cool off (since it cannot radiate energy), but there is not reason for it to heat up (where would the extra energy come from ?) $\endgroup$
    – Frotaur
    Commented Aug 2, 2023 at 15:58
  • $\begingroup$ Is whatever heated the wire initially still acting on the wire (like electric current)? $\endgroup$
    – BowlOfRed
    Commented Aug 2, 2023 at 16:06
  • $\begingroup$ Yes i assumed electric current $\endgroup$
    – Kos
    Commented Aug 2, 2023 at 16:07
  • $\begingroup$ The energy loss in your system is by radiation - think of your reflective coating as insulation. Now your question is similar to what temperature does my heater get to if I insulate it from its surroundings? $\endgroup$
    – D Duck
    Commented Aug 2, 2023 at 19:41

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Are there any fundamental violations in my reasoning?

Yes, conservation of energy.

The energy gained from reabsorbing radiation can't be higher than the energy lost to radiation.

To address your edit: If you have some power (say 1000 W) coming in, you will reach thermal equilibrium when radiative heat losses equal 1000W.

By putting a reflective coating and recouping 99% of losses, then the wire will heat up to whatever temperature allows it to radiate at $\frac{1000}{0.99} = 100,000$W, so that the supplied 1000W keep it in equilibrium.

Of course the more efficient the reflective coating is, the more you have to worry about the coating itself being able to withstand the head and not radiate heat itself.

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