The outer surface of a spacecraft in space has an emissivity of 0.8 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.

158.32 K = -114.83 °C

Explanation:

Since P = P' where P = power absorbed and P' = power radiated

P = αAQ where α = absorptivity = 0.3, A = area of spacecraft and Q = rate of incident solar radiation = 950 W/m²

Also, P' = εσAT⁴ where ε = emissivity of spacecraft = 0.8, σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m²K⁴, A = area of spacecraft and T = surface temperature of spacecraft.

So, P = P'

αAQ = εσAT⁴

T⁴ = αQ/εσ

T = ⁴√(αQ/εσ)

Substituting the values of the variables into the equation, we have

T = ⁴√(0.3 × 950 W/m²/(0.8 × 5.67 × 10⁻⁸ W/m²K⁴))

T = ⁴√(285 W/m²/(45.36 × 10⁻⁸ W/m²K⁴))

T = ⁴√(6.2831 × 10⁸ K⁴))

T = 1.5832 × 10² K

T = 158.32 K

In Celsius T(C) = 158.32 - 273.15  = -114.83 °C

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