Question

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 \(950 \mathrm{~W} / \mathrm{m}^{2}\), determine the surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed.

Short answer

Answer: The surface temperature of the spacecraft when the absorbed radiation equals emitted radiation is 293.2 K.

Step-by-step solution

Define Given Variables

The given information tells us about the emissivity (\(\epsilon\)), solar absorptivity (\(\alpha\)), and solar radiation \(G_{S}\): \(\epsilon = 0.8\) \(\alpha = 0.3\) \(G_{S} = 950 \mathrm{~W} / \mathrm{m}^{2}\)

Calculate the Solar Energy Absorbed

To find the solar energy absorbed by the spacecraft, we multiply the solar radiation by the solar absorptivity: \(Q_{absorbed} = \alpha G_{S} \Rightarrow Q_{absorbed} = 0.3 \cdot 950 \mathrm{~W} / \mathrm{m}^{2}\) \(Q_{absorbed} = 285 \mathrm{~W} / \mathrm{m}^{2}\)

Use the Stefan-Boltzmann Law

According to the Stefan-Boltzmann Law, the power emitted by the spacecraft's surface due to radiation is given by: \(Q_{emitted} = \epsilon \sigma A T^{4}\) where \(\epsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{W} / \mathrm{m}^{2} \mathrm{K}^{4}\)), \(A\) is the surface area, and \(T\) is the surface temperature of the spacecraft in Kelvin. To find the point where \(Q_{emitted} = Q_{absorbed}\), we can set these two equal: \(\epsilon \sigma A T^{4} = Q_{absorbed}\) Since we're interested in the temperature \(T\), we can isolate that variable: \(T^{4} = \frac{Q_{absorbed}}{\epsilon \sigma A}\)

Calculate the Surface Temperature

Since the surface area, \(A\), cancels out on both sides of the equation, we are left with: \(T^{4} = \frac{285 \mathrm{~W} / \mathrm{m}^{2}}{0.8 (5.67 \times 10^{-8} \mathrm{W} / \mathrm{m}^{2} \mathrm{K}^{4})}\) \(T^{4} = 79336.2\) Now, take the fourth root of this result: \(T = \sqrt[4]{79336.2} = 293.2 \mathrm{K}\) The surface temperature of the spacecraft when the absorbed radiation equals emitted radiation is \(293.2 \mathrm{K}\).