# Why is black a better emitter of heat than white? [duplicate]

I know that black absorbs light and converts it into heat which makes it a good emitter of radiant heat while white reflects it. Let's say if I place 2 cups, 1 black and 1 white, same material, in a dark room, which cools faster? Why is black a better emitter? Is it because it converts light to heat?

The existence of equilibrium demands that emissivity is equal to absorptivity.

As long as a body is, say, emitting more energy than absorbing, its temperature will be decreasing: therefore, as thermal equilibrium imply all parts of the system share the same temperature, for equilibrium to be reachable, the body's emissivity must equal its absorptivity. See the Kirchhoff's law of thermal radiation.

Another point to consider is that being black or white are object characteristics with respect to visible frequencies of light, while at room temperature most of the emission/absorption happens at lower frequencies (infrared). For example, the infrared pictures of the aluminum box below make it evident that the emissivities of its white and black surfaces are very similar (as explained in its manual).

Source: Wikipedia

Answer: Now, if the cups are "black" and "white" at infrared, than the black cup will cool faster, since it's emitting energy through radiation at a higher rate than the white one (and absorbing less then emitting, since the environment is cooler).

• @EricTowers, thanks for spotting that one! It's corrected now. – stafusa Aug 27 '17 at 22:15
• When you say "emissivity must equal its absorptivity", you are saying that two efficiency coefficients must be equal. Are you sure you don't mean "emitted energy must equal its absorbed energy" (or something similar)? – Eric Towers Aug 27 '17 at 22:22
• @EricTowers, yes, I'm rather sure: the coefficients are equal. The energies will be the same only when in equilibrium. For example, the cup is guaranteed to cool down precisely because his coefficients are the same, while the environment, being cooler, has less radiation for the cup to absorb. The Wikipedia article explains it nicely: en.wikipedia.org/wiki/… – stafusa Aug 27 '17 at 22:37
• You have misintepreted Wikipedia. It correctly observes that the emitted energy integral equals the absorbed energy integral. Under the assumption that the photon spectrum has a blackbody distribution, this forces the efficiency coefficients to be equal at every wavelength. However, no body is a blackbody, so this equality of coefficients essentially never happens. Balance of energy flow is the correct equilibrium condition for real bodies. – Eric Towers Aug 27 '17 at 22:49
• @EricTowers, I think you're mistaken, but maybe I am, do you have a reference on that? To me it appears equilibrium is all that's needed. (1) This proof (physics.stackexchange.com/a/329052/75633) appears to be independent of black bodies. (2) Also "Kirchhoff's law applies exactly, though no perfectly black body in Kirchhoff's sense is present." (goo.gl/DouvRU). (3) Aren't there QM reasons to believe it's true? (goo.gl/eBk3DW) (4) Or do you refer to more sophisticated corrections such as pnas.org/content/114/17/4336.full.pdf ? – stafusa Aug 28 '17 at 2:41