| ▲ | moi2388 4 hours ago | |||||||
How does this happen? Why would something which reflects light on the outside get hotter than one which absorbs light? This makes no intuitive sense to me. | ||||||||
| ▲ | mapt 4 hours ago | parent [-] | |||||||
The radiative equilibrium temperature is a function of (watts in - watts out), with a fourth power law shoved in there somewhere. A blackbody at radiative equilibrium absorbs whatever visible light you throw at it, and then spits it back out according to a distribution law that mostly places it in the thermal infrared bands (at temperatures we've familiar with, anyway). Remove convection/conduction as heat transfer methods, and you end up with two numbers dictating radiative balance: Percent reflectivity in the bands it's exposed to Percent emissivity in the bands it's emitting The balance between these dictates temperature, and they're generally inversely correlated. Mirrors are good reflectors, but very poor emitters. | ||||||||
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