The outer surface of a spacecraft in space has an emissivity of 0.6 and an absorptivity of 0.2 for solar radiation. If solar radiation is incident on the spacecraft at a rate of 1200 W/m2, determine the surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed. The surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed is K.

Respuesta :

Answer:

T surface = 3.9°C

Explanation:

given data

emissivity  0.6

absorptivity = 0.2

solar radiation is incident rate =  1200 W/m²

solution

we get here surface temperature by equality of emitted and absorbed heat rate that is

Q (absorbed) = Q (heat )  .................1

α Qinc = [tex]\epsilon * \sigma *A*T^4(surface)[/tex]  

T surface = [tex]\sqrt[4]{\frac{\alpha Qinc}{\epsilon *\sigma * A} }[/tex]       ..........................2

put here value and we get

T surface = [tex]\sqrt[4]{\frac{0.2*1000}{0.6*5.67**10^{-8}} }[/tex]  

T surface = 276.9 K

T surface = 3.9°C

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