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An ideal blackbody absorbs all incident radiation. Josef Stefan found that the intensity $R$ (power per unit area) radiated by an ideal blackbody is given by

$$ R = \sigma T^4 $$

Q1) Since an ideal blackbody absorbs all incident radiation, and if it is in thermal equilibrium, doesn't that mean that $R$ is just whatever the incident intensity of radiation coming in is? So like whatever radiation hits it is what is given off? In other words if I shine a flashlight on an ideal blackbody, and I know that the flashlight gives off 100 W per m$^2$, and my ideal blackbody is a sheet of 3 m$^2$, then the emitted power from the black body is 100*3 = 300 W, or an intensity of 300 W / 3 m$^2$ = 100 W/m$^2$ which is the same as the flash light?

Q2) Does a blackbody mean that the object is black? I understand the keyhole thing where you shine light in a keyhole and its trapped there, making it essentially a blackbody. But what if an actual black object absorbed all incident radiation. That means that 1) it will keep absorbing radiation and getting hotter and hotter forever. or 2) the black object will absorb all incident radiation (we said it was an ideal black body) but then it emits radiation given by the equation above. Does that mean that the color black actually has a frequency of light?

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  • $\begingroup$ I suggest you read the answer to this question. Black body is presumed to take all the radiation in and "thermalize" it and emit again. I.e., you cannot trace the emitted light to the shone upon in any way, that's the whole idealization of a "black body". Nevertheless, it can shine quite a lot and it does so with different colours. It can be red-hot, yellow-hot, white-hot... See incandescence. $\endgroup$ – Void Aug 30 '14 at 20:30
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Question 1:

The requirement of thermal equilibrium means that as much radiation power is absorbed as is emitted. Note this does not mean the intensity of the emitted light must match that of the absorbed light, unless the blackbody is uniformly illuminated over its entire emitting surface.

An example in astrophysics might be an object which radiates as a blackbody, but is only illuminated from one side. In the example you give; doesn't a sheet have two sides? In these cases, the emitted blackbody radiation will have about half the intensity, but be emitted over twice the area that absorbs the incoming light.

Question 2:

Blackbodies don't have to be black, they just have to absorb all radiation that is incident upon them. The Sun is a (good approximation to) a blackbody, but I think we can agree it isn't black. In fact, based on the scenario given in Q1, you know that the blackbody can't be black, because it must radiate at all frequencies (with a Planck spectrum), such that it gives off as much power as it absorbs.

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Since an ideal blackbody absorbs all incident radiation, and if it is in thermal equilibrium, doesn't that mean that R is just whatever the incident intensity of radiation coming in is?

No, since the incident radiation may be confined to several wavelengths (the light may be filtered). The emitted radiation from a black body should follow the Planck spectrum. The amount of energy given off (even if it takes longer) will be equal to the energy absorbed.

But what if an actual black object absorbed all incident radiation. That means that 1) it will keep absorbing radiation and getting hotter and hotter forever. or 2) the black object will absorb all incident radiation (we said it was an ideal black body) but then it emits radiation given by the equation above. Does that mean that the color black actually has a frequency of light?

If an object absorbs all incident light, it will appear black. When a black body absorbs light, the frequency distribution of the emitted radiation is rearranged to Planck spectrum. Usually, this means the energy is given off in the infra-red region which you cannot see (the black body will feel warm).

Black does not have any frequency associated with it.

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