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Given two heatsinks identical to each other except in color, how can the differential efficiency of thermal radiation be determined? For example, if both heatsinks are aluminum, with one anodized as black and the other with just its bare aluminum surface, then what is the percentage difference in thermal radiative efficiency?

BACKGROUND: I am an electrical engineer working mostly in test engineering. About 20 year ago, I was tasked with designing an electrical test load, which itself used fans and large heatsinks to remove heat from the test item ( a 10 KW 3-phase generator from a U.S. miliary jet ). I did ask for advice on heatsinks from co-workers who were mechanical engineers. Although we did use bare aluminum heatsinks, I never thought at the time to ask if a black-anodized radiator would be more efficient. Would the difference be $1%$ or $2%$, or $15%$?

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At the temperature that heatsinks connected to solid state devices operate, the primary heat transfer mechanism is not radiation, it is conduction to the air and then convection. Because of this, the color of the heatsink will not make a measurable difference in its heatsinking capabilities.

Of far greater importance is the cleanliness of the heatsink fins. When they get covered with dust and the spaces between the fins get loaded with lint, their effectiveness drops significantly. Rather than being concerned with the color of the heatsink, you should worry instead about filtering the airflow through the heatsink assembly to prevent it from getting loaded with dirt and lint.

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