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My understanding is that dark surfaces are better than bright surfaces at radiating and absorbing heat.

But if you have a bright shiny surface like a metal house radiator and you then paint it black the colour of the shiny surface still exists underneath the black paint. The new black surface may be better at radiating the heat but the heat still has to be transferred from the shiny surface to the black paint before it can be radiated from the black paint. Surely adding any layer to the outside of a radiator will reduce its ability to radiate heat.

If you get two empty tins (bean tins) with silvered surfaces inside and out. Then paint the outside of one of them matt black. Then fill both with boiling water and pop a lid on with a thermometer in each and observe the rates of temperature decline. You will notice that the black tin cools at a faster rate. I don't understand why. The heat is now transferring faster through the silvered metal of the blackened tin before it even reaches the black paint.

Why is this?

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    $\begingroup$ Note that a very thin layer of insulation may actually increase heat dissipation because of the increase in surface area. $\endgroup$
    – Andrew Morton
    Commented Sep 2, 2016 at 8:10

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Because radiated heat is heat transferred by light (that is, radiation). The paint is in physical contact with the radiator, so the paint gets heating by conduction, and the paint then cools by radiating heat (as well as a little conduction with the air, and a lot of convection).

Unless the radiator is getting hot enough to glow in optical wavelengths the optical color doesn't matter. If it is hot enough to glow, then black will perform better than white because the ability to emit radiation is the same, from a physics perspective, as the ability to absorb it. If the radiator isn't that hot, then what matters is the "color" in wavelengths you can't see, and shiny metal is good at reflecting light in many wavelengths, so is bad at absorbing/emitting light.

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    $\begingroup$ I'm a dumbo, and that's ok, but where does the black aspect come in , in your answer, can you tell me?. Are you taking it for granted, or could you elaborate a bit. The question is all about black, I think the answer should mention it too. Thanks $\endgroup$
    – user108787
    Commented Aug 25, 2016 at 17:24
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    $\begingroup$ Added second paragraph. $\endgroup$
    – Sean E. Lake
    Commented Aug 25, 2016 at 17:27
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    $\begingroup$ Ah , so while the heat is radiating slowly from the silvered tin and conducting slowly into the air, heat is being transferred out of the painted tin faster by conduction because the paint is a better conductor than air. The black paint then rapidly radiates this newly acquired heat energy away. $\endgroup$
    – Kantura
    Commented Aug 25, 2016 at 17:29
  • $\begingroup$ Black body absorbs all photons of any wavelength end emitts photons of wavelengts depending only on its temperature, see Planck's law. We can see black radiator black because it is too cold to radiate in visible spectra. Grey body absorbs some photons (according to its emissivity) of all wavelengths. Colour body absorbs/emits photons of specified wavelengths. $\endgroup$
    – Crowley
    Commented Sep 1, 2016 at 17:09
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Completing the other answer, radiator color only has a direct impact on the emitted wavelengths of the radiator. A radiator at 373K (100 C / 212F) radiates most energy in the Mid-wavelength infrared (3–8 µm) range, outside the visible spectrum.

The only significant feature of the radiator coat at it's radiating wavelength is transparency. The less it absorbs or reflects light, the larger the effective radiating layer.

As an approximation, only the molecules inside the radiative penetration depth region of the material have photons which will reach the exterior of the material, instead of being absorbed inside. So for a given temperature the amount of heat radiated will be approximately proportional to penetration depth at the radiated frequency.

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