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

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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

01

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: \(A_s = 4\pi R^2\)
02

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: \(Q_{em} = A_s \times \sigma \times (T_{Earth}^4 - T_{space}^4)\)
03

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.

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Most popular questions from this chapter

Which of the following processes always results in an increase in the energy of a system? a) The system loses heat and does work on the surroundings. b) The system gains heat and does work on the surroundings. c) The system loses heat and has work done on it by the surroundings. d) The system gains heat and has work done on it by the surroundings. e) None of the above.

Consider a Carnot engine that works between thermal reservoirs with temperatures of \(1000.0 \mathrm{~K}\) and \(300.0 \mathrm{~K}\). The average power of the engine is \(1.00 \mathrm{~kJ}\) per cycle. a) What is the efficiency of this engine? b) How much energy is extracted from the warmer reservoir per cycle? c) How much energy is delivered to the cooler reservoir?

A water-cooled engine produces \(1000 .\) W of power. Water enters the engine block at \(15.0^{\circ} \mathrm{C}\) and exits at \(30.0^{\circ} \mathrm{C}\). The rate of water flow is \(100 . \mathrm{L} / \mathrm{h}\). What is the engine's efficiency?

Suppose an atom of volume \(V_{\mathrm{A}}\) is inside a container of volume \(V\). The atom can occupy any position within this volume. For this simple model, the number of states available to the atom is given by \(V / V_{A}\). Now suppose the same atom is inside a container of volume \(2 V .\) What will be the change in entropy?

If liquid nitrogen is boiled slowly-that is, reversiblyto transform it into nitrogen gas at a pressure \(P=100.0 \mathrm{kPa}\), its entropy increases by \(\Delta S=72.1 \mathrm{~J} /(\mathrm{mol} \mathrm{K}) .\) The latent heat of vaporization of nitrogen at its boiling temperature at this pressure is \(L_{\text {vap }}=5.568 \mathrm{~kJ} / \mathrm{mol}\). Using these data, calculate the boiling temperature of nitrogen at this pressure.

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