Question

If the Earth is treated as a spherical blackbody 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 that has the approximate temperature \(T_{\mathrm{sp}}=50.0 \mathrm{~K},\) it radiates heat back into space at an equilibrium temperature of \(278.9 \mathrm{~K}\). (This is a refinement of the model in used in Example \(18.8 .\) ) Estimate the rate at which the Earth gains entropy in this model.

Short answer

Answer: According to the simplified model, the Earth actually loses entropy at a rate of approximately -2.8587 x 10^15 J/Ks. However, it's important to note that this model does not consider all the factors influencing temperature and entropy.

Step-by-step solution

Calculate the absorbed power

We'll first find the rate at which heat is absorbed by Earth. To do this, we'll use the solar constant and the Earth's radius. The formula for the absorbed power, \(P_{abs}\) is: \(P_{abs} = \pi r^{2}S,\) where \(r\) is the Earth's radius (\(6371 \times 10^3 \, m\)), and \(S\) is the solar constant (\(1370 \, W/m^2\)). Let's calculate the absorbed power: \(P_{abs} = \pi (6371 \times 10^3 \, m)^2(1370 \, W/m^2) \approx 1.7416 \times 10^{17} \, W\)

Calculate the radiated power

Next, we'll find the rate at which heat is radiated back into space. To do this, we'll use the Stefan-Boltzmann law for blackbody radiation, given by: \(P_{rad} = \sigma AT^4,\) where \(\sigma\) is the Stefan-Boltzmann constant (\(5.6704 \times 10^{-8} W/(m^2K^4)\)), \(A\) is the surface area of Earth (\(4\pi r^2\)), and \(T\) is the Earth's equilibrium temperature (278.9 K). So, the radiated power is: \(P_{rad} = (5.6704 \times 10^{-8} W/(m^2K^4)) (4\pi (6371 \times 10^3)^2 m^2) (278.9 \, K)^4 \approx 1.7416 \times 10^{17} \, W\)

Calculate the entropy change rates

Now, we'll calculate the rate at which Earth gains entropy from the incoming radiation, by using the entropy formula for an isothermal process, given by: \(\Delta S = \frac{\Delta Q}{T}\) Since we are determining the rates, we will use d instead of \(\Delta\). The rate at which Earth gains entropy from the incoming heat, \(dS_{abs}\), is given by: \(dS_{abs} = \frac{dQ_{abs}}{T}\) where \(dQ_{abs} = P_{abs} dt\) Similarly, the rate at which Earth loses entropy due to the outgoing heat, \(dS_{rad}\), is given by: \(dS_{rad} = \frac{dQ_{rad}}{T}\) where \(dQ_{rad} = P_{rad} dt\) Now let's calculate the values: \(dS_{abs} = \frac{1.7416 \times 10^{17} \, W}{278.9 \, K}\) \(dS_{abs} \approx 6.2458 \times 10^{14} \, J/{Ks}\) \(dS_{rad} = \frac{1.7416 \times 10^{17} \, W}{50.0 \, K}\) \(dS_{rad} \approx 3.4833 \times 10^{15} \, J/{Ks}\)

Calculate the net entropy change rate

Finally, we'll find the net rate at which Earth gains entropy, which is the difference between the incoming and outgoing entropy rates: \(dS_{net} = dS_{abs} - dS_{rad}\) \(dS_{net} \approx 6.2458 \times 10^{14} \, J/{Ks} - 3.4833 \times 10^{15} \, J/{Ks} \approx -2.8587 \times 10^{15} \, J/{Ks}\) As the net entropy change rate is negative, the Earth actually loses entropy in this model. However, this is a simplified model and does not consider all the factors influencing temperature and entropy.