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I was outside with a friend the other day and we ended up talking about clothing choices in the heat, and he repeated the usual statement that one should wear white instead of black in order to stay cool.

However, I seem to remember the relevant statement about black and white objects (identical except for colour) being that the black object absorbs and emits radiation at a greater rate than the white surface. If this is the case, yes, someone wearing a black shirt will heat up (or cool down) faster than if they wore a white shirt, but once they reach equilibrium with their environment, won't they both be the same temperature?

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marked as duplicate by John Rennie, Martin, Kyle Kanos, Qmechanic Jun 10 '15 at 17:05

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    $\begingroup$ Possible duplicate: physics.stackexchange.com/q/78616 $\endgroup$ – Bosoneando Jun 9 '15 at 23:57
  • $\begingroup$ Black will absorb both body heat and external heat better than white, white will just let you cook yourself by reflecting your body heat back at you. Sorry phone won't let me paste link but seems reasonable to me. $\endgroup$ – user81619 Jun 10 '15 at 0:01
  • $\begingroup$ possible duplicate of Radiation– white vs black house, hot or cool? $\endgroup$ – John Rennie Jun 10 '15 at 9:30
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A blackbody is more efficiently coupled to its environment by thermal radiation than an object with poorer emissivity.

In the sun, a black t-shirt will absorb heat (by radiation) more efficiently than a white t-shirt, or a tshirt made of shiny aluminum foil (ow).

In the shade, a black t-shirt will radiate heat more efficiently than a white t-shirt.

However the temperature difference between the sun and the shirt (about 5500 K) is enormously more than the temperature difference between the shirt and its shady, room-temperature surroundings (perhaps 10 K). So the primary method for cooling off in the shade is probably convection, which is the same no matter what color your shirt is.

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