Question

The outer surface of a spacecraft in space has an emissivity of 0.6 and an absorptivity of 0.2 for solar radiation. If solar radiation is incident on the spacecraft at a rate of \(1000 \mathrm{W} / \mathrm{m}^{2},\) determine the surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed.

Short answer

Question: Determine the surface temperature of a spacecraft when the radiation emitted equals the solar energy absorbed, given that the emissivity is 0.6, the absorptivity is 0.2, and the incident solar radiation is \(1000 \mathrm{W} / \mathrm{m}^{2}.\) Answer: Approximately 289 K.

Step-by-step solution

Calculate the absorbed solar energy

To find the absorbed solar energy, we need to multiply the incident solar radiation by the absorptivity. In this case, the incident solar radiation is \(1000 \mathrm{W} / \mathrm{m}^{2},\) and the absorptivity of the outer surface is 0.2. Therefore, $$ E_{abs}=(1000 \mathrm{W} / \mathrm{m}^{2})(0.2) = 200 \mathrm{W} / \mathrm{m}^{2}. $$

Use the Stefan-Boltzmann law

We can use the Stefan-Boltzmann law to calculate the emitted energy. The formula for the Stefan-Boltzmann law is: $$ E_{emit} = \epsilon \sigma \ T^4, $$ where \(E_{emit}\) is the emitted radiation energy, \(\epsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant \((5.67 \times 10^{-8} \mathrm{W} / \mathrm{m}^{2} \mathrm{K}^{4})\), and T is the surface temperature in Kelvin.

Set the absorbed energy equal to the emitted energy and solve for the surface temperature

Since we are looking for the surface temperature where the emitted radiation equals the absorbed energy, we can set \(E_{abs} = E_{emit}\) and solve for T: $$ 200 \mathrm{W} / \mathrm{m}^{2} = (0.6)(5.67 \times 10^{-8} \mathrm{W} / \mathrm{m}^{2} \mathrm{K}^{4}) \ T^4. $$ First, we'll isolate T^4: $$ T^4 = \frac{200 \mathrm{W} / \mathrm{m}^{2}}{(0.6)(5.67 \times 10^{-8} \mathrm{W} / \mathrm{m}^{2} \mathrm{K}^{4})} \approx 5882352.94 $$ Now we'll take the fourth root to find the temperature: $$ T = \sqrt[4]{5882352.94} \approx 289 \mathrm{K} $$ Therefore, the surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed is approximately 289 K.

Key concepts

Absorptivity

Absorptivity is a measure of how well a material absorbs energy when exposed to radiation. It is a dimensionless quantity that ranges from 0 to 1, where 0 means no absorption and 1 signifies complete absorption of the incident radiation. In the context of our spacecraft, the absorptivity is given as 0.2, which means that 20% of the incident solar radiation is absorbed by the surface.
This property is crucial because the energy absorbed contributes to the temperature rise of the object. When we calculate the absorbed solar energy (\(E_{abs}\)), we multiply the incident solar radiation (\(1000 \text{W/m}^2\)) by the absorptivity, resulting in an absorbed power of \(200 \text{W/m}^2\). This value is a starting point to determine the surface temperature of the spacecraft.

Emissivity

Emissivity, similar to absorptivity, is a measure of a material's ability to emit thermal radiation. It is also a dimensionless quantity, ranging from 0, for a perfect reflector, to 1, for an ideal black body that emits maximum thermal radiation possible for a given temperature. The emissivity for the spacecraft's outer surface is 0.6, indicating that it can emit 60% of the thermal radiation that a perfect emitter would at the same temperature.
Emissivity plays a key role in calculating the emitted energy using the Stefan-Boltzmann law. In our problem, we use the spacecraft's emissivity to determine how much energy it emits (\(E_{emit}\)) as a function of its surface temperature.

Solar Radiation

Solar radiation is the energy we receive from the Sun, which can be thought of as a stream of photons carrying energy that can be absorbed, transmitted, or reflected by objects in space. In our exercise, the solar radiation incident on the spacecraft is specified as \(1000 \text{W/m}^2\). This energy source is the primary factor that affects the spacecraft's temperature when in space and exposed to sunlight.
For the spacecraft, the absorbed portion of this solar radiation will heat it up. This is why understanding both the incident solar radiation and the material's absorptivity is vital for accurate thermal management in space applications.

Thermal Radiation Equilibrium

Thermal radiation equilibrium occurs when the energy absorbed by an object is equal to the energy it emits. At equilibrium, the object's temperature remains stable because the input and output of energy are balanced. For the spacecraft, we reach this equilibrium when the energy emitted (\(E_{emit}\)) equals the absorbed solar energy (\(E_{abs}\)).
Mathematically, we express this condition as \(E_{abs} = E_{emit}\). The point of thermal equilibrium is crucial because it enables us to calculate the temperature at which the spacecraft will stabilize in its given environment, given that other forms of heat transfer are negligible.

Surface Temperature Calculation

The surface temperature calculation involves finding the temperature at which an object radiates as much energy as it absorbs. Using the equation derived from thermal equilibrium and the Stefan-Boltzmann law, we calculated that the spacecraft's surface temperature in space (where emitted and absorbed radiation are equal) is approximately 289 K (16°C).
The calculation takes into account the measured solar radiation, the material properties of the spacecraft's surface (absorptivity and emissivity), and the Stefan-Boltzmann constant that relates the emitted energy to the fourth power of the temperature. These elements help us understand and predict the thermal behavior of objects like spacecraft in the harsh environment of space.