Are heat radiation and absorption symmetric?

A friend of mine heard a popular science show on the radio. A caller asked what is better to wear on a hot day, white clothes or black clothes. The answer given was that it did not matter because although black absorbs more readily it also radiates it more readily. My friend said of course that is true, because these things should be symmetric.

My thoughts were that although I do not know about rates of heat radiation, it does not seem to me that processes must be symmetric. For instance, in thermodynamics, entropy always increases, which does not seem symmetric to me.

My question has two parts
  • In this specific case, are the two processes symmetric?
  • In general, are physical processes symmetric?
 

Borg

Science Advisor
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You're not going to get any answers by posting this in the New Member Introductions thread.
 
You're not going to get any answers by posting this in the New Member Introductions thread.
Thanks
 

Khashishi

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They are symmetric in a certain sense; however, not in the way that the radio program suggests. They are just wrong. A black shirt will generally make you hotter, if you are out in the sunlight.

So, in what sense are they symmetric? The emissivity and absorptivity coefficients are the same. The coefficients are functions of wavelength (or frequency) with value between 0 and 1. The thermal emission and absorption are both reduced by this same factor. So, a very shiny material will have a coefficient of close to 1 for visible wavelengths, so it will absorb only a small fraction of the incoming light (and reflect the rest). It will also emit a small fraction of what a black shirt does. A black shirt will absorb most of the incoming light, but will also emit close to a black body spectrum. (Look that up). But the Sun puts out way more thermal radiation than your shirt, even when far away, so decreasing the radiation transfer of the Sun and your shirt by the same fraction will decrease the overall temperature of the shirt.
 

Bandersnatch

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In this sense of symmety - i.e., heat absorbed = heat radiated, the situations are indeed symmetric, under certain assumptions (thermodynamic equilibrium).
Not all processes need to be symmetric, though. For example, before reaching equilibrium, and object will emit/absorb more radiation than it absorbs/emits.

However, symmetry does not mean the situations in this case are equivalent. While a black stone and a white stone left on the sun will end up emitting the same amount of energy as each of them absorbs, these won't be the same amounts for both stones. The black stone will reach higher equilibrium temperature (the amount of emitted radiation depends on temperature) than the white one.

This would suggest that wearing black is disadvantageous in hot climates. But there are more factors at play. For example, in loose-fitting clothes (think Bedouin robes), where the material does not touch the skin directly, the absorbed heat can be reradiated without much conduction.
Additionally, white material allows deeper penetration of radiation, which offsets its higher reflectivity.

A black shirt will generally make you hotter, if you are out in the sunlight.
However, see here:
Why do Bedouins wear black robes in hot deserts?
http://www.nature.com/nature/journal/v283/n5745/pdf/283373a0.pdf
(paywalled)
 
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