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The only way that the station can exhaust thermal energy is by radiation, which it does using thin, 1.5 m -by-3.1 m panels that have a working temperature of about 6 ℃

Electronics and inhabitants of the International Space Station generate a significant amount of thermal energy that the station must get rid of. The only way that the station can exhaust thermal energy is by radiation, which it does using thin, 1.5 m -by-3.1 m panels that have a working temperature of about 6 ℃. You may want to review (Pages 384 -388)


How much power is radiated from each panel? Assume that the panels are in the shade so that the absorbed radiation will be negligible Assume that the emissivity of the panels is 1.0. 

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Answer #1

From stefan's law, the heat radiated is given by

Q = Ae@T^4

A = area = 1.5x3.1 = 4.65 m^2 ; e = emissivity = 1; @ = 5.67x10^-8 ; T = 273 + 6 = 279K

Q= 4.65 x 1 x 5.67x10^-8 x 279^4

= 1597.54 W

= 1.6 kW

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