Answer to Question #225435 in Molecular Physics | Thermodynamics for Rayweela

Question

A spacecraft of surface area of 60m² with a temperature of 600K if the radiation that it receives from the sun is equivalent to a temperature in space of 120K, find the net rate of loss of heat energy from the spacecraft

Answer 1

The power loss per unit area (radiant emittance) is given by the Stefan–Boltzmann law:

$$j_{loss} = \sigma T_1^4$$

where $\sigma = 5.67\times10^{-8}\dfrac{W}{m^2\cdot K^4}$ is the Stefan–Boltzmann constant, and $T_1 = 600K$ is the temperature of the spacecraft.

The power gain per unit area is given similarly:

$$j_{gain} = \sigma T_2^4$$

where $T_2 = 120K$ is the temperature of the space. Thus, net rate loss from all area ($A = 60m^2$ ) is:

$$P = A(j_{loss} - j_{gain}) = A\sigma (T_1^4 - T_2^4)$$

$$P = 60\cdot 5.67\times10^{-8}\cdot (600^4 - 120^4) = 4.4\times 10^{5}W$$

Answer. $4.4\times 10^5W$.