Can a black coating increase the efficiency of a heat dissipator? I have an aluminium profile that contains too many LEDs and heats up around 70°C, which is about +20°C above my expectation. Would an external black coating increase the efficiency of the heat dissipation by the aluminium?
 A: Yes, a black body will radiate more heat away than a white body. That much is correct.
However, this is true for each wavelength individually. A paint that reflects red but absorbs blue will radiate blue light effectively, but not red light. And the overall effect on heat radiation is the emission of a perfect black body times the absorption coefficient for each wavelength.
Now, what kind of wavelengths does a 70°C black body radiate most? Well, most definitely not visible light. It's somewhere in the infrared spectrum. And that is the problem: Black color says something about visible wavelengths only, not about infrared absorption. You can judge the absorption in the visible band of wavelengths, but you need a high absorption in a totally different band.
Thus, when you buy typical black color, you do not know whether it will increase or decrease the heat radiation of a 70°C object. Of course, you might guess that the absorption coefficient of the relevant infrared band is correlated with that of visible light, but that does not need to be the case. I mean, you can buy blue color, and you can buy red color, and each will absorb the light that the other reflects. Your black color will be white in some other band of wavelengths, but in which?
A: You seem to be wanting to make use of black body radiation. The formula for energy dissipated through black body radiation is $\sigma T^4$, where $\sigma=5.670367(13)\times10^{−8}\ \mathrm{W\ m^{-2}\ K^{-4}}$. Plugging in $T=343\ \mathrm K$, we get $785\ \mathrm{W\ m^{-2}}$. However, it would also be absorbing heat from the surroundings. If we model the surroundings as a black body emitting at temperature $300\ \mathrm K$, we get  $459\ \mathrm{W\ m^{-2}}$, for a net of $326\ \mathrm{W\ m^{-2}}$. This means that for every square centimeter, you'll get getting around $33\ \mathrm{mW}$ of cooling. This is probably going to be only a small fraction of the cooling you need. Most of your cooling is coming from conduction to the air, and then convection within the air, and as others have pointed out, a black coating will likely decrease the heat conductivity of the aluminum. It will likely be more productive to increase the surface area and air flow, and decrease the surrounding temperature.
The modelling of the surrounding as being a black body at room temperature is, of course, questionable, but even without it, you'll be getting only $78\ \mathrm{mW\ cm^{-2}}$. (And if it's in an enclosed space, you may be getting less than $33\ \mathrm{mW}$, as it will be heating up the surroundings, and that heat will just be radiated back.) Furthermore, it's quite possible that the effective black body temperature of the surroundings is greater than the temperature of your profile. Unless it's in a dark room, it will be absorbing heat from whatever lighting there is in the room. Thus, its net black body heat exchange may be positive, in which case making it darker would make things worse even without taking into account conductivity. Since we're not actually dealing with perfect black body radiation, there is a possibility that you can decrease its albedo in $343\ \mathrm K$ range without significantly increasing it in the visible range, but that's rather advanced engineering.
A: Your heat sink gains heat from its contact with the hot LEDs and looses it to the environment (air) around it. Since it uses convection to lose heat, any coating that has less thermal conductivity than aluminium will act as an insulator and will cause it to get hotter, which will work against you.
As one of the comments point out, you can increase the amount of heat lost to the environment by increasing the surface area of your heat sink.
You can also look for a material with higher thermal conductivity so heat is transported faster inside the material.
A: There are thermal dissipative coatings which can be used to improve heat removal performance. These have both good IR emissivity and low thermal resistance, and are very different from regular black paint (they are not necessarily black to begin with).
Typically dissipative coatings are only used at high temperatures (hundreds of degrees). At lower temperatures, even specialized coatings only harm heat dissipation by adding thermal resistance.
Temperatures around 70°C are completely normal for modern LEDs, so unless your LEDs came with a datasheet which specifies a lower temperature, or the profile in question should remain cool for other reasons (like contact with skin), I would advise to leave it as it is. If you have to make it cooler, either find a bigger profile, a profile optimized for convection cooling (bigger surface area) or increase the airflow.
A: Possibly. Implicit in your question is the assumption that radiative heat transfer is playing or could play an important role in your configuration (vs. convection). If so, then applying a "black" coating (and thus increasing the emissivity to essentially 1, with the caveat that we're talking about the maximum wavelengths of emission at 70°C) could benefit you. However, note that the coating itself may hinder heat transfer across various interfaces.
I would plug the relevant numbers into the various formulas for heat transfer ($hA_\mathrm{surface}(T-T_\infty)$ for convection, $kA_\mathrm{cross\,section}\Delta T/\Delta x$ for conduction, $\sigma\epsilon A_\mathrm{surface\,facing\,surroundings}(T^4-T^4_\infty)$ for radiation, as described in any introductory heat transfer textbook and at many locations online) to estimate the relative magnitude of convection vs. radiation before attempting to optimize a heat transfer mechanism that might be unimportant.
As an example, the natural convection coefficient $h$ can be very broadly estimated to around $10\,\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{K}^{-1}$ by order of magnitude (I'm sure exceptions exist in certain geometries). From the numbers you've given, one can estimate that changing the emissivity from 0 to 1 (by anodizing the aluminum, for instance) could potentially boost the outgoing heat flux by a detectable amount, but probably less than 100%. Furthermore, as other posters have noted, you may have other options to increase the outgoing heat flux much more substantially—using fins or a fan, perhaps.
