# Does an object's color change its rate of cooling?

The motivation for this question comes directly from this thread. The proposition is that the color of something changes how fast it cools (note: specifically the rate of cooling, not taking into account the change in heat absorption). For example: would adding black food coloring to an otherwise light-colored fluid, make it cool faster? The proposed explanation for such an effect is that the fluid becomes a 'better' black body. Is this true? Why or why not?

My current thinking:
The 'color', per se, has no effect on the rate of cooling. For two bodies at the same temperature, any difference in color is due to it's reflection properties, not its emission properties - which will be identical, because they have the same temperature.

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The reflectivity, $r$, and emissivity, $e$, are linked by $r + e = 1$.

Consider an object at the same temperature as it's surroundings. It must emit radiation at the same rate as it absorbs it otherwise it would heat up or cool down. If it heated up or cooled down it would be at a different temperature from it's surroundings and you could use the temperature difference to drive a heat engine and get a perpetual motion device. If the emissivity, $e$, is the same as the absorptivity, $1-r$, then obviously $e + r = 1$.

The actual case of your coffee has some extra wrinkles because it's emitting at infrared wavelengths, and it's far from obvious how black food colouring would affect the emissivity and reflectivity at infra-red (rather than visible) wavelengths. However if you added something to the coffee that reduced the reflectivity in the IR it must also increase the emissivity and therefore the rate of cooling by radiation.

I highlighted by radiation because in the real world coffee cools mainly by evaporation and convection and changing the reflectivity/emissivity will have little effect.

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First: the object is hotter, that's why it's cooling. What you've said makes a lot of sense, thanks; but I'm still a little confused---Kirchoff's law is based on maintaining equilibrium, i.e. the absorptivity must match the emissivity to maintain equilibrium via radiation: i.e. the amount absorbed is the same as lost. Thus, you're still not cooling faster, you're just making up for absorbing more. –  zhermes Mar 14 '13 at 18:26
If the coffee is a lot hotter than the environment you can ignore the radiation it's absorbing from the environment. In that case the heat flow is proportional to the emissivity, so a higher emissivity/lower reflectivity means faster cooling. –  John Rennie Mar 14 '13 at 18:30
But again, isn't the motivation for the connection between absorption and emission to balance the radiation---i.e. to preserve planck emission. So does that mean that a non-'black' body, e.g. white-coffee, will not be emitting as a black-body in the visible portion of the spectrum (where it appears 'white')? –  zhermes Mar 14 '13 at 18:36
I'm not sure what you're asking. Non-black bodies typically have a wavelength dependant emissivity e.g. the Sun is nearly a black body but the emissivity varies from unity wherever there is a spectral line. So the emissivity of coffee almost certainly differs at IR and visible wavelengths. However a black body at 100C (i.e. your coffee) has negligible radiation at visible wavelengths anyway so I'm not sure that this is relevant. You could argue that visible light could heat the coffee like it heats a greenhouse, but I doubt this would be significant for domestic lighting. –  John Rennie Mar 14 '13 at 18:53
If the rate of radiative cooling is different for different colors, which are at the same temperature (i.e. black vs. white coffee both at 100C) --- then the spectrum of one or the other needs to deviate from a blackbody. That's the part I'm confused about. –  zhermes Mar 14 '13 at 21:00