Nelson,
Sorry to be argumentative here, but this does not make sense to me. If a black car radiates heat just as much as it absorbs heat, then why is it hotter than a white car?
It seems to me that the car in the sun is an extreme case, though. One usually doesn't put his or her amp out in the sun. If one did, however, the black heatsink would definitely be hotter than a shiny one, much like the car example. But lets look at another extreme, two heatsinks, one black and one shiny, in complete darkness. What difference does the color make if there is no light to reflect off of the heatsinks. They would both have no color, except for their blackbody radiation (usually in the infrared). The equation for blackbody radiation is only dependent on temperature, and has nothing to do with paint. Hence, both heatsinks would give off the same amount of radiation.
So, what am I getting at?:
In bright light (sunlight) the black heatsink will be hotter.
In no light (darkness) both heatsinks would be the same temperature.
If you are right Nelson, then at some point in the middle (medium light) the black heatsink will have to radiate more heat than it absorbs, as opposed to the shiny one. What is the mechanism for this change? It does not seem clear to me.
I realize that radiative heat loss is not as productive as conductive heat loss and painting a heatsink black will not be too big of a problem. But, unless someone can explain it to me it seems to be worse (however slight) than a shiny heatsink.
I am shooting from my own intuition and a little bit of physics and I may be totally wrong. Can you spot an error or oversight in my logic?
-Dan