If black is the best absorber and radiator, why does it get hot? I can guess that it’s emission peaks at a higher temperature than white. So when a light is turned onto a black and white piece of paper, the initial condition is not at equilibrium. As black absorbs the light energy, it’s temp rises and then it re-emits photons at a lower frequency. Is there a quantitative description of this process somewhere that will tell me what the equilibrium temp will be?
I rationally expect this should be independent of the heat capacity of the material (black is black no matter what makes it black) but my intuition says no - a space shuttle thermal tile painted black should feel cooler than iron painted black if both objects are at thermal equilibrium under the same light source.
I read these posts
If a black body is a perfect absorber, why does it emit anything?
Why is black the best emitter?
And although they are clear, I cannot tease out the answer to my question.
 A: 
Is there a quantitative description of this process somewhere
that will tell me what the equilibrium temp will be?

The Stefan-Boltzmann law says how much radiation
an ideal black body will emit:
$$j_{\text{emitted}}=\sigma T^4 \tag{1}$$
where

*

*$j_{\text{emitted}}$ is the total emitted power per area,

*$\sigma = 5.67\cdot 10^{-8} \frac{\text{W}}{\text{m}^2\text{K}^4}$
is the Stefan-Boltzmann constant,

*and $T$ is the absolute temperature.

When the black body is in equilibrium with the incoming light,
then the absorbed light power is equal to the emitted radiation power.
$$j_{\text{absorbed}}=j_{\text{emitted}}  \tag{2}$$
From (1) and (2) we get
$$T=\left(\frac{j_{\text{absorbed}}}{\sigma}\right)^{1/4}$$
So it is as you rationally expected: The equilibrium temperature is
independent of the heat capacity of the material.
Example:
Let's assume a light intensity of $1000\text{ W/m}^2$
(typical for cloudless sunny day).
Then the black body will be heated to temperature $T=364$ K,
which is $91$ °C.
A: For surfaces surrounded by air and exposed to sunlight or other e-m radiation, the main reason for black ones getting hotter is that they absorb radiation but lose heat mainly by convection, which process is much less dependent on surface colour. Thus a body with a matte black surface has to get hotter than one with a shiny surface in order to lose heat at the greater rate that it is absorbing, to reach equilibrium.
