Why does a good emitter of heat radiation appear black at room temperature?

Answer 1

It's because the reflectivity of a surface at any given wavelength must equal its absorptivity. If it were otherwise, an objected in thermal equilibrium within a container of equal temperature would absorb either more or less radiation at that wavelength than it emits, contradicting the equilibrium assumption.

Answer 2

Heat is given off as infrared waves. Since infrared wavelengths are longer then the longest visible light wavelengths we can see (which are red), then we don't see a thing (hence black). Our eyeballs just aren't capable of seeing them. Now if you heat something up enough, it will get hot enough to give off wavelengths in the visible range. This is why metal will glow red if heated enough.