Question

If the Earth is treated as a spherical black body of radius \(6371 \mathrm{~km}\), absorbing heat from the Sun at a rate given by the solar constant (1370. W/m \(^{2}\) ) and immersed in space with an approximate temperature of \(T_{s p}=50.0 \mathrm{~K},\) it radiates heat back into space at an equilibrium temperature of \(278.9 \mathrm{~K}\). (This is a slight refinement of the model in Chapter 18.) Estimate the rate at which the Earth gains entropy in this model.

Short answer

Question: Estimate the rate at which Earth gains entropy given the Earth's radius \(R = 6371\,\text{km}\), the solar constant \(S = 1370\,\text{W/m}^2\), the temperature of space \(T_{space} = 50.0\,\text{K}\), and Earth's equilibrium temperature \(T_{Earth} = 278.9\,\text{K}\).

Step-by-step solution

Calculate the cross-sectional area and surface area of Earth

To determine the heat absorbed by Earth, we need its cross-sectional area, and to find the heat emitted back to space, we need its surface area. Let's find both of these with the given Earth's radius \(R = 6371\,\text{km}\). Cross-sectional area: \(A_c = \pi R^2\) Surface area: \(A_s = 4\pi R^2\)

Calculate the rate of heat absorption and heat emission

Given the solar constant \(S = 1370\,\text{W/m}^2\), the rate of heat absorbed by Earth can be computed as follows: Heat absorption rate: \(Q_{abs} = A_c \times S\) The equilibrium temperature of Earth is \(T_{Earth} = 278.9\,\text{K}\). Given the temperature of space \(T_{space} = 50.0\,\text{K}\), we can use the Stefan-Boltzmann law to find the rate at which Earth radiates heat back to space. The Stefan-Boltzmann constant \(\sigma = 5.67\times10^{-8} \,\mathrm{W} \cdot \mathrm{m}^{-2} \cdot \mathrm{K}^{-4}\). Heat emission rate: \(Q_{em} = A_s \times \sigma \times (T_{Earth}^4 - T_{space}^4)\)

Calculate the rate at which Earth gains entropy

Now, we will compute the rate at which Earth gains entropy. To do this, we need to find the entropy gain due to heat absorption and the entropy loss due to heat emission. Entropy change can be computed using the formula: \(\Delta S = \frac{Q}{T}\) So, the entropy gain due to heat absorption: \(\Delta S_{abs} = \frac{Q_{abs}}{T_{space}}\) And the entropy loss due to heat emission: \(\Delta S_{em} = \frac{Q_{em}}{T_{Earth}}\) The net rate at which Earth gains entropy is the difference between the entropy gain due to heat absorption and the entropy loss due to heat emission. Entropy gain rate: \(\Delta S_{net} = \Delta S_{abs} - \Delta S_{em}\) With all the steps outlined above, you can now substitute the given values and compute the rate at which Earth gains entropy in this model.

Key concepts

Stefan-Boltzmann law

The Stefan-Boltzmann law is a key principle in understanding how objects emit thermal radiation. It tells us how power (or energy per time) is radiated from a body that has a certain surface area and temperature. The law is expressed with the formula:
\[ Q = A \cdot \sigma \cdot T^4 \]
Where:

• \(Q\) is the power emitted,
• \(A\) is the surface area of the emitting body,
• \(\sigma\) is the Stefan-Boltzmann constant,
• \(T\) is the absolute temperature in Kelvin.

This equation shows us that the energy emitted depends significantly on the temperature of the body, specifically to the fourth power. This means that even small increases in temperature can lead to large increases in emitted energy. For Earth, this law helps us calculate how much heat it emits back into space when taking into account its temperature and surface area.

heat absorption

Heat absorption in the context of Earth involves taking in energy from the Sun. This process is fundamental in maintaining the climate system and supporting life. The Earth absorbs solar energy across its sunlit hemisphere. To calculate this absorbed energy, we use the solar constant, which represents the average solar energy received per unit area.

Absorbed heat is typically estimated by:

• using the cross-sectional area of Earth as seen from the Sun,
• multiplying it by the solar constant.

This gives the rate at which energy is absorbed. The absorbed heat is vital for sustaining temperature balances and driving atmospheric processes on Earth.

heat emission

Once the Earth absorbs solar energy, it doesn't just keep this energy indefinitely. Instead, it re-radiates energy back into space to maintain thermal balance. This re-radiation is governed by the Stefan-Boltzmann law, taking into account the Earth's surface area and temperature.

The process involves:

• Using the entire surface area of Earth in the calculations (since it radiates in all directions),
• Employing Earth's temperature to the fourth power to compute effective emission.

The emission computation considers the difference between Earth's surface temperature and the temperature of space, making this calculation crucial in understanding Earth's energy dynamics.

black body radiation

Black body radiation refers to the way an idealized object, or a "black body," emits and absorbs thermal radiation. Ideal black bodies absorb all incoming radiation without reflecting any, and they emit radiation uniformly in all directions when at equilibrium.

Key properties of black bodies include:

• They are perfect emitters and absorbers of radiation,
• Their emission depends only on their temperature, not material or surface characteristics.

In the study of planetary science, Earth's treatment as a black body helps simplify calculations of temperature equilibrium without atmospheric effects. It sets a foundational model for how Earth deals with energy from the Sun and exchanges heat with space, critical for understanding global temperature regulation and climate.