Use the Stefan-Bolzmann law and assume an emissivity of 1.0
Radiated power = 2*A*sigma*T^4, where
T = 279 K.
'sigma' is the Stefan-Boltzmann constant, which you should look up.
A = 2.2*4.1 = 9.02 m^2
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, 2.2 m-by-4.1 m panels that have a working temperature of about 6 degrees Celsius.
How much power is radiated from each panel? Assume that the panels are in the shade so that the absorbed radiation will be negligible.
Hint: Don't forget that the panels have two sides!
3 answers
I ended up with 3098.89 W but it's wrong...
(number of sides on the panel)(e)(area)(sigma)(T^4)=P
(2)(1)(9.02)(5.67E-8)(279^4)=6197.78
You probably just put it into your calculator wrong, or forgot to multiply by 2.
(2)(1)(9.02)(5.67E-8)(279^4)=6197.78
You probably just put it into your calculator wrong, or forgot to multiply by 2.
1 answer