Why does a dense, hot, opaque body produce blackbody radiation?

Question

I read from a textbook that a dense, hot and opaque body produces black body radiation, for example the heated filament of an incandescent light bulb.

Why is it so?

The explanation given is

if a body is opaque, then the protons, neutrons, electrons, and photons which it contains frequently interact, and attain thermal equilibrium.

But a blackbody is supposed to absorb all radiation, and a body being opaque just means that it does not transmit radiation through it. It can still reflect radiation.

It also seems to suggest that photons inside the body is trapped and thus interact to reach thermal equilibrium. If they are trapped, how do they escape to produce radiation?

Answer 1

But a blackbody is supposed to absorb all radiation, and a body being opaque just means that it does not transmit radiation through it. It can still reflect radiation.

Yes, an ideal blackbody must absorb all radiation incident upon it and be in thermal equilibrium. Not only does it absorb all radiation incident upon it, but is self-absorbs any radiation is produces in its interior.

A blackbody cannot be reflective since the reflected spectrum need not be a blackbody spectrum.

It also seems to suggest that photons inside the body is trapped and thus interact to reach thermal equilibrium. If they are trapped, how do they escape to produce radiation?

Clearly, this is an idealised situation. Near the surface of a blackbody, radiation produced inside the body will escape. However, if the interior of the blackbody has attained thermal equilibrium, then the spectrum of the escaping radiation will closely mimic an ideal blackbody.