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If a black object and a shiny object were left in vacuum and bombarded with thermal radiation would the shiny object eventually reach the same temperature as the black object?

The reason I ask is I am having a hard time wrapping my head around Kirchhoff's law and Boltzmann's law and thermal radiation. I know black objects absorb more heat and get hotter than shiny objects, but if absorbance is equal to emittance than the color should have no effect on the temperature over an infinite amount of time. Is that correct?

Qmechanic
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Demni
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  • Yes, that is correct. The easiest way to see this is to have the black body and the shiny body in a perfectly reflecting box. This is a closed system and there is nothing happening that would prevent the objects inside to reach thermal equilibrium. If you want, you can use the Clausius formulation of the second law to determine that the only possible physical process that is consistent with thermodynamics leads to an equalization of the two temperatures. – CuriousOne Jun 03 '15 at 18:18
  • Take a look at this other question: Why is black the best emitter?. – DanielSank Jan 23 '19 at 01:14

1 Answers1

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If the two bodies are put in equilibrium radiation, they will both adapt their temperature to common temperature. The black object will do so faster.

Ján Lalinský
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