The outer surface of a spacecraft in space has an emissivity of 0.44 and a solar absorptivity of 0.3. If solar radiation is incident on the spacecraft at a rate of 950 W/m2, determine the surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed.

Respuesta :

Answer:

[tex]T=326.928K[/tex]

Explanation:

From the question we are told that:

Emissivity [tex]e=0.44[/tex]

Absorptivity [tex]\alpha =0.3[/tex]

Rate of solar Radiation [tex]R=0.3[/tex]

Generally the equation for Surface absorbed energy is mathematically given by

 [tex]E=\alpha R[/tex]

 [tex]E=0.3*950[/tex]

 [tex]E=285W/m^2[/tex]

Generally the equation for Emitted Radiation is mathematically given by

 [tex]\mu=e(\sigmaT^4)[/tex]

Where

T=Temperature

 [tex]\sigma=5.67*10^8Wm^{-2}K_{-4}[/tex]

Therefore

 [tex]\alpha*E=e \sigma T^4[/tex]

 [tex]0.3*(950)=0.44(5.67*10^-8)T^4[/tex]

 [tex]T=326.928K[/tex]

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