If a black body is a perfect absorber, why does it emit anything?

Say what? Which part of "absorbs" does this go with?

The key to understanding this is to carefully note the phrase "in thermal equilibrium".

This means that the rates of absorption and emission are the same.

If a body were at a lower temperature than the environment, the rate of absorption would be higher and the body would then heat up.

If a body were at a higher temperature then the environment, the rate of emission would be higher and the body would then cool down.

But, in thermal equilibrium, the temperature is constant and, thus, the rates of absorption and emission must be equal.

Now, put this all together:

• A black body is an ideal absorber, i.e., a black body does not reflect or transmit any incident electromagnetic radiation.
• An object in thermal equilibrium with the environment emits energy at the same rate that it absorbs energy.

Then, it follows that, a black body in thermal equilibrium emits more energy than any other object (non-black body) in the same thermal equilibrium since it absorbs more energy.

Imagine several various objects, including one black body, in an oven and in thermal equilibrium. The black body will 'glow' brighter than the other bodies.

"OK, that's nice. It's an object that absorbs (takes in itself and stores/annihilates forever) any electromagnetic radiation that happens to hit it."

Right! But that frontslash is important, and you want the first option. It can absorb the energy as heat and it never reflects anything. What black-body theory then goes on to detail is what the black-body emits. This isn't ambient light--a black body would emit radiation even if it never absorbed any, as long as its temperature were nonzero.

As you note, the name is kind of a misnomer. Black bodies (interesting ones) hardly ever appear black. To look at, the sun is basically a black body with a temperature of 5500K. As you may have noticed, the sun is not black$^\mathrm{[citation \ needed]}$.