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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.

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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.

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  • $\begingroup$ thank for the calculation of the equilibrium temp. of an ideal black body. As my question specifically asked for a quantitative solution, you did answer that. However I was hoping for an equation that gave me insight into how much hotter a black object would get than a white object. My initial sentence speculated that black objects get hotter than white ones. At equilibrium the calculation predicts the same temperature for a white colored object as a black object (color is not included in the calculation above) but experience shows me that black objects get hotter in the sun than white objects $\endgroup$ – aquagremlin Jul 12 at 23:00
  • $\begingroup$ In this question for example physics.stackexchange.com/questions/122911/… there is the qualitative discussion that white is a poor absorber of light compared to black. But it is also a poor emitter. If emission=absorption at equilibrium, then I would rationalize that the temp of white objects should be the same as black objects at equilibrium. Experience tells me otherwise. $\endgroup$ – aquagremlin Jul 12 at 23:09
  • $\begingroup$ David Hammen's comment in that thread 'So we need something else to explain why black objects get hotter than do white ones. The answer lies in the fact that absorptivity and emissivity are frequency and temperature dependent for real objects. ' alludes to an answer but I cannot find a quantitative solution for how many degrees hotter will a piece of paper get if I paint it black. $\endgroup$ – aquagremlin Jul 12 at 23:09
  • $\begingroup$ I do not want you to be upset that i am 'changing the question', I did mark yours as the answer. If you feel I was vague or the question was poorly stated, I will ask it differently. $\endgroup$ – aquagremlin Jul 12 at 23:13
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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.

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  • $\begingroup$ thank you for your answer but please refer to my comment above about the difference between colors, rather than surface texture. $\endgroup$ – aquagremlin Jul 12 at 23:01

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