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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.

Qmechanic
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aquagremlin
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2 Answers2

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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.

Thomas Fritsch
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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.

Philip Wood
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