Question

A solar flux of \(1400 \mathrm{~W} / \mathrm{m}^{2}\) directly strikes a spacevehicle surface which has a solar absortivity of \(0.4\) and thermal emissivity of \(0.6\). The equilibrium temperature of this surface in space at \(0 \mathrm{~K}\) is (a) \(300 \mathrm{~K}\) (b) \(360 \mathrm{~K}\) (c) \(410 \mathrm{~K}\) (d) \(467 \mathrm{~K}\) (e) \(510 \mathrm{~K}\)

Short answer

The equilibrium temperature of the space vehicle's surface in space at 0 K is approximately 410 K.

Step-by-step solution

Understand the equilibrium condition

In equilibrium, the energy absorbed by the surface is equal to the energy radiated by the surface. Therefore, we can express this condition as follows: $$\text{Absorbed Energy} = \text{Radiated Energy}$$

Calculate the absorbed energy

The energy absorbed by the surface can be calculated using the solar flux and the solar absorptivity. The formula for this is: $$\text{Absorbed Energy} = \text{Solar Flux} * \text{Solar Absorptivity}$$ Substitute the given values: $$\text{Absorbed Energy} = (1400 \frac{\mathrm{W}}{\mathrm{m}^{2}}) * 0.4$$ Now, calculate the absorbed energy: $$\text{Absorbed Energy} = 560 \frac{\mathrm{W}}{\mathrm{m}^{2}}$$

Calculate the radiated energy

The energy radiated by the surface due to thermal radiation can be written as: $$\text{Radiated Energy} = \text{Emissivity} * \sigma * T^{4} * A$$ Where: - Emissivity is the thermal emissivity of the surface (0.6 in this case) - \(\sigma\) is the Stefan–Boltzmann constant (\(5.67 × 10^{-8} \frac{\mathrm{W}}{\mathrm{m}^{2}\mathrm{K}^{4}}\)) - T is the surface temperature in Kelvin, which we are trying to find - A is the surface area Since we are calculating the temperature and the question is asking the equilibrium surface temperature, we will consider the radiated and absorbed energy per unit area. Thus, A in both cases cancels out.

Equate absorbed and radiated energy and solve for T

Now, we can equate the absorbed and radiated energies: $$560 \frac{\mathrm{W}}{\mathrm{m}^{2}} = 0.6 * 5.67 × 10^{-8} \frac{\mathrm{W}}{\mathrm{m}^{2}\mathrm{K}^{4}} * T^{4}$$ Divide both sides by `\(0.6 * 5.67 × 10^{-8} \frac{\mathrm{W}}{\mathrm{m}^{2}\mathrm{K}^{4}}\)`: $$T^{4} = \frac{560 \frac{\mathrm{W}}{\mathrm{m}^{2}}}{0.6 * 5.67 × 10^{-8} \frac{\mathrm{W}}{\mathrm{m}^{2}\mathrm{K}^{4}}}$$ Calculate \(T^{4}\): $$T^{4} = 1.64 × 10^{8} \mathrm{K}^{4}$$ Now, take the fourth root on both sides: $$T = \sqrt[4]{1.64 × 10^{8}\mathrm{K}^{4}}$$ Calculate T: $$T = 410 \mathrm{K}$$ Thus, the answer is: **(c) \(410 \mathrm{~K}\)**