An electronic package with a surface area of \(1 \mathrm{~m}^{2}\) placed in an orbiting space station is exposed to space. The electronics in this package dissipate all \(1 \mathrm{~kW}\) of its power to the space through its exposed surface. The exposed surface has an emissivity of \(1.0\) and an absorptivity of \(0.25\). Determine the steady state exposed surface temperature of the electronic package \((a)\) if the surface is exposed to a solar flux of \(750 \mathrm{~W} /\) \(\mathrm{m}^{2}\), and \((b)\) if the surface is not exposed to the sun.

We need to find the energy balance equation for power conservation between the energy going in and the energy going out through the package's surface. The energy going in will be the power \(\mathrm{1 \ kW}\) from the electronics and the absorbed fraction of solar flux \((750 \mathrm{~W} / \mathrm{m}^{2} \times 0.25)\) when exposed to the Sun. In both cases, the surface will be radiating energy out according to the Stefan-Boltzmann law \((e \sigma T^4 \times A)\), where \(\sigma = 5.67 \times 10^{-8} \mathrm{W} / (\mathrm{m}^{2} \ \mathrm{K}^{4})\) is the Stephan-Boltzmann constant, \(e\) is the emissivity, and \(A\) is the surface area. For case (a), we have the following energy balance equation: $$P_{in} + P_{solar} = P_{out}$$ For case (b), the energy balance equation is: $$P_{in} = P_{out}$$

Since we have the energy balance equation for the case when the package is exposed to solar flux, we can now substitute the given values and solve for the steady-state temperature \(T_a\). $$P_{in} + P_{solar} = e \sigma T_a^4 A$$ $$1000 + (750 \times 0.25) = 1.0 \times 5.67 \times 10^{-8} \times T_a^4 \times 1$$ $$1000 + 187.5 = 5.67 \times 10^{-8} \times T_a^4$$ $$T_a^4 = \frac{1187.5}{5.67 \times 10^{-8}}$$ $$T_a = \sqrt[4]{\frac{1187.5}{5.67 \times 10^{-8}}}$$ $$T_a = 335.97 \ \mathrm{K}$$ The steady-state temperature when the surface is exposed to solar flux is \(335.97 \ \mathrm{K}\).

Similarly, using the energy balance equation for the case when the package is not exposed to the Sun, we can solve for the steady-state temperature \(T_b\). $$P_{in} = e \sigma T_b^4 A$$ $$1000 = 1.0 \times 5.67 \times 10^{-8} \times T_b^4 \times 1$$ $$T_b^4 = \frac{1000}{5.67 \times 10^{-8}}$$ $$T_b = \sqrt[4]{\frac{1000}{5.67 \times 10^{-8}}}$$ $$T_b = 314.63 \ \mathrm{K}$$ The steady-state temperature when the surface is not exposed to the Sun is \(314.63 \ \mathrm{K}\). So we have the steady-state temperatures for (a) and (b) as: (a) \(T_a = 335.97 \ \mathrm{K}\) and (b) \(T_b = 314.63 \ \mathrm{K}\).