Heat leaves a heat sink via conduction, convection and radiation. I was taught that black surfaces are best at radiating heat and correspondingly a lot of heat sinks are black. But they also have fins for convection. And large heat sinks have lots of large fins. Convection therefore seems important.

So what happens if for aesthetic reasons I have to paint a heat sink brilliant white? White of course is the best colour for reflecting thermal radiation. Would that then reflect the heat back inside like a foil wrapped turkey? Can some one guesstimate a loss of efficiency or a necessary compensatory increase in rating?

P.S. Domestic radiators are white.

P.P.S. What does "material finish" imply in heat-sink terminology? doesn't quite answer it.

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    \$\begingroup\$ This might be better asked in Mechanical engineering, since it is really a fluids/thermal dynamics questions. But radiation is much less efficient that convection. So then fin design and shape and airflow will affect the heat flow much more than paint color. \$\endgroup\$ Oct 11, 2017 at 1:09
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    \$\begingroup\$ There is a lot to say about this subject, because everything has influence (a little). A thick layer of limescale does indeed cause a barrier around a heating element. Do you know the electrical wall panels for infrared heating? They are just a big sheet of metal that is heated. The manufacturers say that it can be painted in every color without having influence on the efficiency. The radiators in a central heating system are also painted. That has no influence. So go ahead, use paint that is ment for high temperatures, but don't put a thick layer on it. \$\endgroup\$
    – Jot
    Oct 11, 2017 at 3:18
  • \$\begingroup\$ If you have a lot of heatsinks that need to be white, you could contact someone like Phanteks or Alpenföhn who make CPU coolers with white heatsinks and negotiate some deal for the use of their processes. \$\endgroup\$ Oct 12, 2017 at 11:26
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    \$\begingroup\$ At room temperatures (~300 K) the peak emission is at about 10 um, according to Stefan-Boltzmann's Law. At this frequency any paint is "black", even a white one, maybe except some shiny metallic paints. \$\endgroup\$ Nov 3, 2017 at 7:17

12 Answers 12


Yes black has the highest emissivity (it also absorbs radiation best- reciprocity - or Kirchoff's law of thermal radiation). Note that it has to be "black" at the relevant wavelengths, which do not necessarily correspond to being black in the visible spectrum.

That means that radiative heat transfer will be maximized if the emissivity approaches 1 (black body). If your heat sink "sees" mostly cooler things it will have better cooling, and if it sees hotter things it will not cool as well.

However, radiative heat transfer is typically not all that significant compared to conducted heat transfer and (usually most important for semiconductor heat sinks in normal environments) convection heat transfer. So usually the color is not all that important compared to the fluid dynamics design of how air flows over the fins and how the heat is conducted to the fins. The fins mostly "see" other fins, so radiation has even less of an effect.

Exceptions exist for those of us designing electronics that has to survive in a vacuum and/or in space or very high altitudes, and if the item being heat sinked (or what it sees) is very hot, radiation can become more important (4th power of temperature).

An example situation where a shiny (low emissivity) heat sink could be superior would be a voltage regulator heatsink in direct view of a heater, incandescent lamp, or vacuum tube.

Any color of dye you like can be applied when anodizing, or none at all, which is called "clear" anodize. Normally the oxidized aluminum (it's not a coating) is a quite thin insulating layer, but in some cases it can be made more than a few mils thick.

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Edit: Let's do a back-of-the envelope calculation to see how significant radiation is. I'll assume a model 530002B02500G heatsink from Aavid Thermalloy. It has a natural convection rating of 2.6 degrees C per watt, which I believe is rated at a rise of 70 degrees C over ambient.

So if your ambient temperature is 25 degrees C and the heat sink is at 95 degrees C the total power dissipated will be 27W.

How much of that is due to radiation? We can treat the heat sink (for radiative coupling purposes only*) as a block of dimensions 64mm x 25mm x 42mm (ignoring the notch) which represents a surface area of 0.011 square meters.

Heat loss due to radiation (assuming emissivity of 1) is

\$ q = \sigma A (T_H^4 - T_C^4)\$, where \$\sigma\$ is the Stefan-Boltzmann constant ~5.7E-8 (and the temperatures are in Kelvin)

Substituting in the values we get a heat flow of 6.4W due to radiation at 95 degrees C heatsink temperature and 25 degrees C ambient, so less than 25% is due to radiation under optimal conditions for maximizing radiation loss. More likely we've got some forced convection going on and the radiation heat loss is less again. A heat sink that is closer to a cube would have less heat loss due to radiation as well. Not quite low enough to be ignored, but not dominant.

  • For radiation, the convolutions of the heat sink "see" other heat sink surfaces mostly so a block of the outer dimensions is correct for radiation (to a first approximation). They actually have an effect in making the effective emissivity closer to 1.0 than the surface itself since some of the light that is not absorbed will bounce into other surfaces and get another chance to be absorbed (and the same in reverse, of course for radiation of heat- but it's easier to imagine the absorbing of light because we can see visible light and cannot see the IR wavelengths that the heat sink is emitting at reasonable temperatures - if your heat sink is glowing red, yellow or blue-white you probably have other problems).
  • \$\begingroup\$ not all that significant Are you able to ball park this (for non interstellar heat sinks)? Less than 50%? \$\endgroup\$
    – Paul Uszak
    Oct 11, 2017 at 10:01
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    \$\begingroup\$ +1 for '"black" at the relevant wavelengths, [does not] necessarily correspond to being black in the visible spectrum.' \$\endgroup\$
    – owjburnham
    Oct 11, 2017 at 14:50
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    \$\begingroup\$ Nice, using your numbers I calculated you'd need a heat sink temperature of ~470 K to radiate ~25 watts. One other place you wouldn't want a black heat sink (beside your incandescent light) is if the heat sink is exposed to sunlight. \$\endgroup\$ Oct 11, 2017 at 17:00
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    \$\begingroup\$ @Matt I may be wrong, but I think the point he is making is that just because something looks black to the human eye (absorbs visible radiation) doesn't mean it equally absorbs (e.g.) infrared radiation equally well. So it's not necessarily acting as a "black body" without further proof than the naked eye. Hence not "black" in the "black body" sense except in the visible spectrum. \$\endgroup\$
    – John D
    Oct 11, 2017 at 20:43
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    \$\begingroup\$ @CJDennis Okay, added some words to attempt to clarify. \$\endgroup\$ Oct 12, 2017 at 0:53

Do not paint heat sinks. The layer of paint will act as an insulator between the metal and the air, reducing its ability to dissipate heat.

Anodizing a heat sink is less of an issue. The anodization layer is much thinner than paint (a few micrometers in aluminum, for instance), so it presents a much lower thermal resistance than paint would.

The color of the paint will not have a significant effect on cooling unless different colors of paint are thinner or thicker than others, or if the heat sink is exposed to direct sunlight.

  • \$\begingroup\$ What about domestic radiators then? They come painted from the factory, and I think that it's something like thick enamel. It's certainly nothing clever like anodising. \$\endgroup\$
    – Paul Uszak
    Oct 11, 2017 at 10:05
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    \$\begingroup\$ Domestic radiators, despite the name, heat the room via convection: en.wikipedia.org/wiki/Radiator_(heating) \$\endgroup\$ Oct 11, 2017 at 10:24
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    \$\begingroup\$ @PaulUszak, without the paint, the radiator would be too efficient, so people touching it would likely sustain burns. So the insulation effect here is intentional. \$\endgroup\$ Oct 11, 2017 at 12:22
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    \$\begingroup\$ also: aesthetics \$\endgroup\$
    – A C
    Oct 11, 2017 at 14:10
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    \$\begingroup\$ @SteveMelnikoff, Heat sinks in electronic devices transfer heat to the surrounding air by exactly the same mechanism that transfers heat from the cast iron radiator to the air in my bedroom. The air touches the heat sink/radiator, and it picks up heat by conduction. Then, the hot air moves out of the way so that more cool air can take its place. The mechanism that moves the air can be convection (if it's a passive heat sink, or if I leave the ceiling fan switched off in my bedroom), or it can be forced by a fan. \$\endgroup\$ Oct 11, 2017 at 16:09

I used to sell high performance coatings including coatings for thermal management, and this is by no means a simple question. For example one vendor made two coatings that were both a matte black with a recommended dry film thickness of less than two thou and one was designed to be an insulator and the other was designed to be a thermal dissipative coating. Recommended use cases were exhaust headers and brake calipers respectively.

I did have heatsinks coated and did some experimenting and found at low temperature differentials the bare heatsink worked best followed by the insulated one and the radiant coating was worst. At high temperature differentials the emissive coating showed it’s worth outperforming bare metal and the insulative coating also became more effective at slowing heat transfer.

I also played around with other coatings such as conventional paints which were all insulators until they caught on fire, and a thick insulating coating which was the best insulator until it melted.

Conclusion: It depends on your temperature, needed heat transfer and other factors Which coating is the best for your needs, but using a standard paint will never help a heatsink.


Shiny metal surfaces are extremely good at reflecting, extremely poor at emitting radiation. Anything (almost anything) you paint them with will improve heat radiation.

Black-painted surfaces will absorb sunlight better, because most of the energy in sunlight is or around the visible spectrum. But unless your heat sink is so hot that it glows, it won't be emitting much radiation in the visible spectrum, so its color to the eye is completely irrelevant.

Passively ventilated heatsinks that are not enclosed in a reflective container emit a significant amount of radiation. I'm sorry, I don't have the numbers at hand, but it is less than 50%. For parallel fins that are the same colour, radiation is balanced by absorption, so internal fin colour doesn't matter very much, but for exposed surfaces it does matter.

For this reason, you will see that the forced-air heatsinks inside your computer often do not have black anodized pins.

Exposed aluminium heatsinks have to be painted/anodized anyway, to prevent corrosion, otherwise you'll have to include instructions like "not to be used with 1000 yards of the seashore"


The property you are probably looking for is emissivity.

There are 3 main processes through which heat is transferred:

  • Conduction (movement of heat inside the material)
  • Convection (movement of heat through movement of the heated material)
  • Radiation

The effectivity of each of these processes depends on the materials in question and the environment temperature. The total heat flow is the sum of the heat flow from each of the processes at the air/heatsink boundary.

The radiative heat flow component is proportional to the emissivity of the surface, the temperature to the fourth, and the effective surface area.

Have a look at this handy list of emissivities at room temperature.

The highest emissivities are found in Ice, Water, Glass, Limestone, and Paint (including white); note that all of those are not black. In fact, it seems like apart from polished metals, all materials have a near-ideal emissivity.

Also, as noted in most other answers, a layer of paint will act as a great insulator (for conduction inside the heatsink).


Some papers, like "heat transfer coefficient of painted films" (1956) find "overall coefficients of heat transfer from pipes coated with paints consisting of linseed oil as the vehicle and titanium dioxide, lamp black, or aluminum, were 4 to 15 per cent higher than the over-all coefficient for bare pipe". TiO2 is apparently best there.

They suggest that the paint film conducts better than the oxide layer that would appear if not painted, and that the paint also replaces some gasses trapped in the oxide film.

So if you have an iron heatsink (I understand the paper tests on iron pipe), paint might even COOL the heatsink, if applied in a good way such that you get reduced oxide layers and a thin paint layer. Extrapolating, if your heatsink would have or develop an isolating oxidation layer over time, painting it is good.

For aluminum, there is oxide, but the oxide layer is only 4nm and also Al2O3 is not a serious thermal isolator. So in that paint is not needed for keeping the oxides away. Additional paint will then only decrease the cooling efficiency

If you really need to paint it, make sure the paint gives minimal thermal isolation and apply a thin layer.


Here are some rules of thumb, with ballpark figures.

Painting an extruded aluminium heatsink will increase its coefficient of emissivity from .09 to about .94. At the temperatures we are talking about, colour doesn't matter at all. What counts is the dullness of the surface. The thickness of the paint layer affects both radiation and conduction.

Paint is less conductive than metal, so do not paint your heatsink unless radiation dissipates a significant proportion of the heat generated by the source. This is likely only under conditions of natural convection. If there is a fan on the heatsink, don't paint it.

The hotter a surface is compared to its surroundings, the more heat will be radiated rather than lost to convection.

An LED panel operating at 60 Celsius with a bare aluminium top surface open to the air will radiate about 10% of the heat generated by the LEDS. This could increase to 50% if the surface is painted, almost doubling the amount of power you can put through the panel.

I have made a spreadsheet that performs some approximate calculations for flat horizontal surfaces: https://smokedprojects.blogspot.com/2021/06/dont-believe-hype-led-panels-run-hot.html


It depends on thermal conductivity and radiation of paint...

If you got paint that radiates well you can mix some grains in it to increase surface area... Some big metal sawdust from aluminium would do the job...


Do not paint the heat sink as it will become less efficient to radiate off heat since you will have an extra layer covering your heat sink. Black heat sink does better at radiate off heat but it's in a small percentage advantage.

If you uses fan to blow your heat sink, the color doesn't really matter much.


" I was taught that black surfaces are best at radiating heat "

That is not correct. A black object absorbs a broader spectrum of radiation than a colored body, but color in and of itself has no effect on the efficiency of a radiator. Many heatsinks are black because black anodizing is a low cost process that prevents oxidation and corrosion. BTW, it also is an insulator, so if it is important that a transistor be in electrical contact with its heatsink, you can't count on physical contact alone.

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    \$\begingroup\$ Radiation efficiency is the same as absorption efficiency. They are the same thing. \$\endgroup\$
    – david
    Oct 11, 2017 at 4:53
  • \$\begingroup\$ @david But radiative efficiency only matters if an object is hot enough to radiate a significant amount of heat -- as in, glowing hot. \$\endgroup\$
    – user39382
    Oct 11, 2017 at 6:35
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    \$\begingroup\$ @duskwuff, no, it doesn't have to be glowing hot. In space, for example, the vast majority of heat loss/gain will be through radiation and as such, the colour of heatsinks in space or vacuum of importance. \$\endgroup\$
    – RJR
    Oct 11, 2017 at 9:05
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    \$\begingroup\$ @duskwuff Most things glow if you view them at the appropriate wavelengths. See "thermal camera" for a simple example... \$\endgroup\$
    – Paul Uszak
    Oct 11, 2017 at 9:58
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    \$\begingroup\$ This is absolutely true. Emissivity is a property of the material, does not depend on its color. I've conducted experiments verifying that a white painted heatsink radiates much better than a bare aluminum one. Likewise, an anodized heatsink radiates much better than bare aluminum, not because anodized heatsinks are often dyed black, but because aluminum oxide has much higher emissivity than aluminum metal \$\endgroup\$
    – user28910
    Oct 11, 2017 at 13:50

A black surface both absorbs and emits better than an equivalent white surface. The best colour choice depends on the environment surrounding the heatsink. If in a cold dark environment then use a black heatsink. If the device is ever expected to encounter a bright environment or radiant heat, use a white heatsink.

  • \$\begingroup\$ Don't we have to assume that a heatsink's temperature > ambient? Otherwise what's the point as heat only travels in one direction? \$\endgroup\$
    – Paul Uszak
    May 30, 2018 at 14:07
  • \$\begingroup\$ Yes, but what matters is whether there is a source of radiant heat nearby (sunlight, or a hot surface), rather than air temperature. I've edited my answer to clarify this, thank you. \$\endgroup\$
    – TopCat
    May 30, 2018 at 14:42

This heatsink calculator* ought to not be made use of in circumstances where the warm resource is much smaller sized than the base of the heatsink. The concentration of the warm resource over an area a lot smaller than the warm sink base is not taken into consideration in this calculator.

[*I am involved with the company that provides this calculator]

  • \$\begingroup\$ I've moved the comments re spamming to chat here. \$\endgroup\$
    – Russell McMahon
    Jul 24, 2020 at 0:51
  • \$\begingroup\$ A degree of "self promotion" of one's own site is acceptable as long as the answer is mainly about being an answer to the question. As @Andrew T noted, please refer to How to not be a spammer and always disclose your affiliation on the post in the future. \$\endgroup\$
    – Russell McMahon
    Jul 24, 2020 at 0:56

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