This got me thinking about the nature of blackness. Two not-particularly-related questions:
What is really meant by “absorb” here? If no visible light is emitted by a black object (let’s not talk about black holes but conventional objects that are black in color), is all the visible light emitted as EMR in other frequencies, like IR?
What is going on with a glossy black surface, like a freshly waxed black Alfa Romeo? It certainly looks black but it also reflects like a mirror–reflects no light but reflects all light. What is the resolution of this apparent paradox?
All objects with a temperature emit light. When a visible light photon is absorbed by an object (whether black or otherwise) it’s converted to heat which is subsequently re-radiated as longer-wave EM, primarily IR. The answer to your second question is that no black object reflects NO light; there is always some portion of the incident light which is reflected–the surface merely appears black by comparison. What light that is reflected is more or less evenly distributed across the spectrum, otherwise the surface would have a color cast to it. If an object is smooth and shiny, like a mirror, it will reflect an image. In the case of a shiny black object, that image is considerably dimmer than a mirror’s, but it’s just as clear.
Roughly, there are two components to light scattering from a surface: specular and diffuse. The specular is just the normal reflection, like from a mirror, or shiny metal. Diffuse scatter is like what you get from flat paint. The light scatters equally in all directions*. For the glossy black, the diffuse scatter is very low, close to zero. The specular reflection is small also, but larger than the diffuse scatter.
*It doesn’t have to be equal in all directions. If you look at the reflection of a bright light in an LCD panel, you might notice that the diffuse scatter is brighter when it is near the specular point.