coolforumguy said:
If an object absorbs light, that means its electrons are not being caused to accelerate, and thus should not heat up.
The answer to the question, "Do black objects get hotter in the sun?" is most certainly "yes". It's easy enough to test this for yourself, by leaving a black stone and a white stone side by side in bright sunlight for a few hours... and solar water heaters are painted black because that way they absorb more heat... and if you'll won't find many dark-colored roofs being built in sunny climates... and many more examples.
So the question isn't whether black objects get hotter in the sun, the question is why they do.
Light-colored object: Incoming photon interacts with an electron, increasing its energy by one photon's worth of energy and exciting it to a higher energy level. The structure of the material is such that the when the electron dumps this energy by dropping back to a lower energy level, it releases a photon just like the incoming one. This photon flies away from the material, taking one photon's worth of energy with it so there is no net increase in the energy of the object - one photon's worth of energy came in, one photon's worth of energy left.
This process happens so quickly that we say that there's just one photon that came in and bounced off; and as Vanadium50 points out above, something like a mirror is really good at making this happen.
Dark-colored object: Incoming photon interacts with the electron, increasing its energy by one photon's worth of energy and exciting it to a higher level. The structure of the material is such that when the electron dumps this energy by dropping back to a lower level, it doesn't release a photon that flies away. Instead, the electron directly or indirectly interacts with its nucleus and its neighboring atoms to set them vibrating - and that's heat. One photon's worth of energy came in, one photon's worth of energy didn't go out; and the total amount of energy in the object has to go up.
In practice, these interactions are complicated and very difficult to model correctly. For example, a smooth surface will usually be more reflective than a rough surface even if the the material is the same; it is possible to calculate this effect exactly, and it's been done, but it's a LOT of work. Thus, we generally use the approach that DaveC426913 described above: Add up the energy of the incoming photons, subtract the energy of the outgoing photons, and the difference has to be amount of energy that's left behind to heat up the object.