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A perfectly black body absorbs all incident radiation on it at any temperature and emits all radiations at any temperature,and thus at equilibrium absorption=emission.Say the black body is kept in a colder room, the rate of emission must be higher so it can attain thermal equilibrium but since it absorbs all radiations at any temperature and emits all radiation at any temperature, shouldn't its temperature be the same as what it was before?

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  • $\begingroup$ sam as what it was before $\endgroup$
    – AJknight
    Feb 19, 2022 at 15:49
  • $\begingroup$ Absorption doesn’t = emission $\endgroup$ Feb 19, 2022 at 17:08

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The definition of a blackbody does not include it "emits all radiation at any temperature". I'm not even sure what that could mean.

The definition is that it absorbs all radiation.

If a blackbody is put in a colder room, then it will emit more radiation than it absorbs. Roughly speaking, it might emit blackbody radiation at its higher temperature, while absorbing blackbody radiation at the lower temperature of the cold room. Ultimately, its temperature will reduce to match the temperature of the room.

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Absorption equals emission on thermal equilibrium, right. If the black body temperature is higher than the temperature of the environment, emission is higher than absorption, and vice-versa, right again.

In any case, temperature will tend to equilibrium with time, the rate equation being a first order differential equation with $dT / dt$ and fourth powers $T^4$ (see here).

Your source of confusion comes from "it absorbs all radiations at any temperature and emits all radiation at any temperature". You got the temperature dependence wrong. A black body has a radiation pattern as shown in the other answer, alright, we say that it emits radiation at all wavelenghts, with zero-power at $\lambda$ equals zero and infinty, and it has a peak power at $\lambda_{peak}$.

What changes with temperature is the peak, according to Wien's law,

$$\lambda_{peak} = b/T,$$

where $b$ is a constant (Wien's displacement constant). So the higher the temperature the lower the peak emission wavelength. But, overall speaking, it will emit more than absorb while its temperature is higher, and absorb more than emit while its temperature is lower.

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Here is a the spectrum of emission of a blackbody Blacbody spectrum from Wikipedia English

As you can see, the blackbody emits radiation in all wavelengths, but how much it emits for each wavelength depends on the temperature.

For example, in the case of the sun, with a surface temperature of about 6000 K, the peak of its emission is found to be in the visible spectrum.

The same occurs for the absorption, a theoretical blackbody absorbs radiation of every wavelength, however how much it absorbs for a given wavelength varies and it is temperature dependent.

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