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

Answer: The equilibrium temperature of the space vehicle surface is approximately 410 K.

Step-by-step solution

Understand the solar properties

We are given the solar flux (\(F_{sun}\)) as \(1400 \mathrm{~W} / \mathrm{m}^{2}\), solar absorptivity (\(\alpha\)) as \(0.4\), and thermal emissivity (\(\epsilon\)) as \(0.6\). \(F_{sun}\) represents the intensity of sunlight on the surface of the space vehicle.

Calculate the absorbed power

When sunlight strikes the surface of the space vehicle, a part of the sunlight is absorbed by the surface, which is determined by the solar absorptivity. The absorbed power (\(P_{absorbed}\)) can be calculated using the formula $$P_{absorbed} = \alpha \times F_{sun}.$$Plug in the values provided:$$P_{absorbed} = 0.4 \times 1400 \mathrm{~W} / \mathrm{m}^{2} = 560 \mathrm{~W} / \mathrm{m}^{2}.$$

Recall the Stefan-Boltzmann Law

The Stefan-Boltzmann Law states that the power emitted by an object (due to radiation) is $$P_{emitted} = \epsilon \cdot \sigma \cdot A \cdot T_{eq}^4,$$where \(\epsilon\) is the emissivity, \(\sigma = 5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^2 \mathrm{~K}^4\) is the Stefan-Boltzmann constant, \(A\) is the area, and \(T_{eq}\) is the equilibrium temperature in Kelvin.

Equate the absorbed and emitted powers

At equilibrium, the absorbed power should be equal to the emitted power. Therefore, we have $$560 \mathrm{~W} / \mathrm{m}^{2} = 0.6 \cdot 5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^2 \mathrm{~K}^4 \cdot A \cdot T_{eq}^4.$$Since we are dealing with the surface, the area \(A\) will cancel out on both sides. Thus, we have:$$T_{eq}^4 = \frac{560 \mathrm{~W} / \mathrm{m}^{2}}{(0.6)(5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^2 \mathrm{~K}^4)}.$$

Solve for equilibrium temperature

Finally, solve for the equilibrium temperature (\(T_{eq}\)) by taking the fourth root on both sides of the equation:$$T_{eq} = \sqrt[4]{\frac{560 \mathrm{~W} / \mathrm{m}^{2}}{(0.6)(5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^2 \mathrm{~K}^4)}} \approx 410 \mathrm{~K}.$$The correct answer is (c) \(410 \mathrm{~K}\).

Key concepts

Solar Absorptivity

Solar absorptivity refers to a material's ability to absorb solar radiation. In simpler terms, it measures how much of the incoming sunlight is absorbed by the surface. Materials with high absorptivity will soak up a significant amount of sunlight, converting it into heat. On the contrary, materials with low absorptivity will let most of the solar radiation pass through or reflect it away.

• An absorptivity of 1 means the material absorbs all incoming solar radiation.
• If absorptivity is 0, the material doesn't absorb any solar energy.
• In our example, the surface has an absorptivity of = 0.4, meaning it absorbs 40% of the incoming solar energy.

Understanding solar absorptivity is crucial for designing spaceships or any structure exposed to the sun. Efficiently controlling how much sunlight a surface absorbs can influence temperature regulation and energy management.

Thermal Emissivity

Thermal emissivity is the measure of a surface's effectiveness in emitting energy as thermal radiation. It is a ratio comparing a material's thermal emission to that of a perfect black body, an ideal surface that absorbs all incident radiation and emits the maximum amount possible.

• A black body has an emissivity of 1, meaning it radiates 100% of the absorbed energy back out.
• If emissivity is 0, it means the material doesn’t emit any absorbed thermal radiation.
• In our context, the surface has a thermal emissivity of = 0.6, indicating it releases 60% of the absorbed energy.

Controlling emissivity helps in managing a vehicle's heat balance in space. By understanding emissivity, scientists and engineers can ensure that the excess heat is effectively radiated away, preventing overheating and ensuring the thermal safety of both passengers and equipment.

Equilibrium Temperature

Equilibrium temperature is achieved when the amount of absorbed energy equals the amount of emitted energy. At this point, the temperature of a surface remains constant because energy input and output are balanced. This is particularly significant in environments like space, where there is no air to conduct heat away, making radiation the sole means of energy exchange for surfaces.

In solving for equilibrium temperature using the Stefan-Boltzmann Law, one sets the absorbed power to be equal to the emitted power. The formula involves the absorbed power, thermal emissivity, and the Stefan-Boltzmann constant to calculate the temperature:

• The formula is given by \(T_{eq}^4 = \frac{P_{absorbed}}{\epsilon \cdot \sigma}\).
• Simplifying this expression allows us to find the temperature, \(T_{eq}\), of the space vehicle at which it neither gains nor loses heat.

Combining these concepts provides a comprehensive understanding of how a vehicle regulates its temperature in space. Achieving an optimal equilibrium temperature is essential for the safe operation of spacecraft by preventing both freezing and overheating under the intense conditions of space.