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?
Asked
Active
Viewed 6,819 times
1 Answers
9
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).
stafusa
- 12,435
-
@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: https://en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation#Theory – 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." (https://goo.gl/DouvRU). (3) Aren't there QM reasons to believe it's true? (https://goo.gl/eBk3DW) (4) Or do you refer to more sophisticated corrections such as http://www.pnas.org/content/114/17/4336.full.pdf ? – stafusa Aug 28 '17 at 02:41
-
(1) The derivation at physics.SE drops the $\lambda$ dependence in both $\alpha$s and both $\epsilon$s, so the algebraic result they get is false. I'm sure you can think of many nonzero functions whose integral over some interval is zero. For the integral they get, one could arrange, for each $\lambda$ that either $\alpha_1(\lambda) = 0$ or $\alpha_2(\lambda) = 0$ and see that the quotient in $\alpha$s is zero regardless of the $\epsilon$s. ... – Eric Towers Aug 28 '17 at 03:52
-
(2) This is false. Any random cubic meter in the interior of the Sun is about as close as one can get to this isotropic thermal bath. The photon spectrum in this volume is nearly Planckian, but with characteristic voids corresponding to hydrogen and helium. This is how helium was discovered. (Be careful. There are also voids caused by "cold" gasses near the surface and in the corona where the closeness to an isotropic thermal bath breaks down. Not all those lines are missing in the interior.) – Eric Towers Aug 28 '17 at 03:57
-
(3) Stefan-Boltzmann assumes the emitter is a blackbody. If the body follows some other emission profile, then the equilibrium condition is that the total energy leaving equals the total energy arriving. Consider the solar spectrum above -- less energy is leaving the Sun system at those dark lines. If we throw the Sun in one of these magic isotropic thermalized boxes, the photon spectrum in the box will always be deficient in those lines because the Sun system readily absorbs but under-emits them. – Eric Towers Aug 28 '17 at 04:02
-
(4) Nope. Nothing that exotic. – Eric Towers Aug 28 '17 at 04:13
-
It's worth pointing to Kirchhoff's Law, where we find "any arbitrary material emitting and absorbing thermal electromagnetic radiation at every wavelength in thermodynamic equilibrium" and " equality of emissivity and absorptivity often does not hold when the material of the body is not in thermodynamic equilibrium." Both the OP and your scenario of an object emitting more than it is absorbing (or vice versa) are not describing equilibrium conditions. – Eric Towers Aug 28 '17 at 04:16
-
@EricTowers, I think it's a great discussion, but at the wrong venue. :) I'm considering posting it as a question later one (maybe today still), and warning you here about it (don't delete the comments just yet). I'd happy to have you putting your arguments into an answer. Hopefully we'd get some feedback from the community as well. – stafusa Aug 28 '17 at 10:12