This question has come to me from my friend in fact: he noted that the heating in the pub is painted black. I replied that it's better for heat emission.

I don't know where did I know that from. And he disagreed, asking me: "Why would black emit more heat than white?" I didn't know. We could, however, agree on fact, that black absorbs more heat than white.

I quickly created a thought experiment to proof by contradiction that the black must emit more light than white:

Assume that white and black both emit the same amount of light. Put a black and white object in an area. Assume that any light (or heat) emitted by the objects can only be absorbed by them. Imagine that the white object emits light and heat. The black one will absorb considerable amount of it. When the black object emits the light the light one will reflect a great portion of it - which can be then absorbed back by the black one.

Written like this, it seems that second law of thermodynamics is being broken by this concept. However, my friend has also thought something to oppose me:

If black objects emits more heat than the white one, why the black objects are hotter when put in the sun? Shouldn't they emit the extra light they absorbed?

I understand that it's not all that simple. There's an article that says I'm right but it doesn't really satisfy me.

For start, infrared is no color at all - how could it have to do anything with black, white or any other color? Shouldn't an "infrared" painted stuff reflect most heat?

I ask for an answer that sufficiently explains why the black and white things absorb/emit heat as they do. And, if questions post by both mine and my friend's arguments are asked, I will be very happy.

  • $\begingroup$ An important wrinkle here is the appearance of the emissivities in both the expression for radiated energy and that for absorbed energy. I think there could be a good brain teaser for my modern physics class in this... $\endgroup$ Commented Jun 12, 2014 at 23:57
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    $\begingroup$ A fantastic question. It's a bit half baked and imprecise: but that's absolutely fine. A good question is half baked by definition - if there's no baking to do, there's no question to be asked. It shows someone thinking really carefully and groping for the right concepts. I think you will go far, Tomáš Zato! $\endgroup$ Commented Jun 13, 2014 at 0:30

2 Answers 2


You can't argue with a black object and a white object alone, as I think you partially understand in trying to build your thought experiment. You need a little bit more to define things properly. See whether the following helps.

Imagine a black object at a temperature $T_0$ and a white object also at $T_0$ inside a perfectly isolating box full of blackbody radiation at some higher temperature $T_1>T_0$ (i.e. without the black and white objects, this radiation is in thermodynamic equilibrium).

To understand exactly what would happen, you would have to describe the "colour" of your objects with emissivity curves that show emissivity as a detailed function of frequency. So your "black" and "white" would need to be defined in much more detail. You would also have to define the surface areas of the two objects and what they are made of (i.e. define their heat capacities). But all of this only effects the dynamics of how the system reaches its final state, i.e. these details only influence how the system evolves. What it evolves to is the same no matter what the details: the box would end up with everything at the same temperature such that the total system energy is, naturally, what it was at the beginning of the thought experiment. "Blacker" as opposed to "Whiter in this context roughly means "able to interact, per unit surface area, with radiation more swiftly": the blacker object's temperature will converge to that of the radiation more swiftly than does that of the whiter object, but asymptotically the white object "catches up". Blacker objects absorb more of their incident radiation its true, but they also emit more powerfully than a whiter object at the same temperature. The one concept emissivity describes the transfer in both directions. Think of emissivity as being a fractional factor applied to the Stefan-Boltzmann constant for the surface as well as being the fraction of incident light absorbed by the surface relative to a perfect blackbody radiator.

This description is altogether analogous to that of the situation where $T_0<T_1$. Begin with $T_0=T_1$, and you've got thermodynamic equilibrium from the beginning. Nothing happens, of course.

Maybe the following will help thinking about what is a really quite a complex question: it would be a fantastic last question for an undergrad thermodynamics exam BTW: You can abstract detail away by saying lets define object $A$ to be blacker than object $B$ if, when both objects are made of the same material, are the same size and shape, the temperature of $A$ converges to the final thermodynamic equilibrium temperature more swiftly than that of $B$ when they are both compared in the box-radiation-object thought experiment above.

Thinking about this now, I am not sure whether the above definition would hold for every beginning temperature of the radiation. Maybe there are pairs of surfaces whose relative blackness is different at different beginning temperatures such that $A$ is blacker than $B$ with some beginning temperature whilst the order swaps at a different beginning temperature. I think it is unlikely, but that is probably a different question altogether.

By the way, which pub do you drink in? I might come along.

Afterword on a Heater's Colour:

You ask by implication what is the best colour to paint a heater. This is not a simple question and involves the dynamics of the heater system. It's really an engineering question. I suspect in general it is better for them to be blacker rather than whiter. Here's a glimpse of the kind of factors bearing on the situation.

If you can say a heater has a constant nett input of $P$ watts, then at steady state that's going to be its output to the room, altogether regardless of its colour. There may be a materials engineering implication here: if you paint the heater whiter, and if its dominant heat transfer to the room is by radiation (rather than by convection or conduction), then it has to raise itself to a higher temperature than it would were it blacker so as to radiate $P$ watts into the room. So its materials might not be as longlasting, and it might be more of a fire hazard than it would be were it blacker.

If the heater is the hot water kind, and again if radiative transfer is significant, then the heating system has to run hotter to output power at a given level if the heater is whiter. At a given flow rate and given temperature of heating water, the heat output of heater is lower if it is whiter. You're trying to design the heater to be an "anti-insulator": you want the heat to leak out of the flow circuit in at the heater, not through the lagging on the hot water pipes outside the building channelling the water from the boiler to the heaters. If the hot water pipes leak heat in the same room, then that's no problem.

Recall the quartic dependence of the Stefan Boltzmann law. At room temperatures with a low temperature heater (the hot water kind) $\sigma\,T^4$ is likely to be pretty small compared with other heat transfer mechanisms, in contrast to my idealised scenarios above. So the heater's colour is likely to be pretty irrelevant.

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    $\begingroup$ Thank you for the explanation! I performed an experiment where I put an opened black (internal) box connected to an opened white (internal) box, with plastic in between. My thought was that the black box should heat up, while the white box cooled, but they stayed the same temperature. I assume it is because the black box, though it absorbs more readily, also emits more readily. It could also be that the black color was not black in infrared, or that the plastic reflected infrared light. "Free energy" ideas are not supposed to work, and now I understand why that one doesn't work. $\endgroup$
    – Jonathan
    Commented Jan 14, 2015 at 19:22

Black absorbs more visible light than white, but that doesn't say what happens at other wavelengths. There is no color infrared. Each object has an absorptivity that is a function of wavelength. We find it easy to determine the absorptivity in the visible, much harder at other wavelengths. In some of the infrared, they are both black. In particular, I suspect the radiator is just as good painted white as black, both because you want radiation in the infrared, and because much of the heat is convected away from it.

If you put a black and white object in a mirror room, both heated to the same temperature where the emission is (mostly) in the visible spectrum, the black will absorb a higher fraction of the radiation falling on it. It will also emit more radiation, by the same amount-that is what you are missing. The temperatures will stay the same as they are in thermal equilibrium.

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    $\begingroup$ Then why in sun, the black does not emit the extra amount of light back? And why does it absorb more infrared when it has, in the infrared spectrum, the same color as the white? $\endgroup$ Commented Jun 13, 2014 at 0:04
  • $\begingroup$ Also, how come there are only visible colors? Other animals can see in different ranges than humans do (bees for example). So even "invisible" colors must be able to be painted on something I guess... $\endgroup$ Commented Jun 13, 2014 at 0:05
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    $\begingroup$ For the first, because the object is much cooler than the sun. Wien's displacement law shows the peak wavelength a perfectly emitting (the physics meaning of black) object radiates at. A black piece of paper is still say 320K, so radiates in the far infrared. When I say both radiators are black in the infrared, it means they absorb infrared equally and well. We only name colors in the visible spectrum because that is what we perceive. $\endgroup$ Commented Jun 13, 2014 at 0:10

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