Within an insulated system, \(418.6 \mathrm{kJ}\) of heat is conducted through a copper rod from a hot reservoir at \(+200.0^{\circ} \mathrm{C}\) to a cold reservoir at \(+100.0^{\circ} \mathrm{C} .\) (The reservoirs are so big that this heat exchange does not change their temperatures appreciably.) What is the net change in entropy of the system, in \(\mathrm{kJ} / \mathrm{K} ?\)

Short Answer

Expert verified
Answer: The net change in entropy of the system is \(0.237\,\mathrm{kJ/K}\).

Step by step solution

01

Convert temperatures to Kelvin

First, we need to convert the given temperatures to Kelvin. To do this, add 273.15 to each Celsius temperature: \(T_h = 200.0^\circ\mathrm{C} + 273.15 = 473.15\mathrm{K}\) \(T_c = 100.0^\circ\mathrm{C} + 273.15 = 373.15\mathrm{K}\)
02

Calculate the change in entropy for the hot reservoir

Use the formula \(ΔS_h = \dfrac{Q}{T_h}\) to find the change in entropy for the hot reservoir, where the heat transfer, Q, is negative because it loses heat: \(ΔS_h = \dfrac{-418.6\mathrm{kJ}}{473.15\mathrm{K}} = -0.885\,\mathrm{kJ/K}\)
03

Calculate the change in entropy for the cold reservoir

Use the same formula but with a positive heat transfer, as the cold reservoir gains heat: \(ΔS_c = \dfrac{418.6\mathrm{kJ}}{373.15\mathrm{K}} = 1.122\,\mathrm{kJ/K}\)
04

Calculate the net change in entropy for the system

Add the change in entropy for both reservoirs to find the net change in entropy of the system: \(ΔS_{net} = ΔS_h + ΔS_c = -0.885\,\mathrm{kJ/K} + 1.122\,\mathrm{kJ/K} = 0.237\,\mathrm{kJ/K}\) The net change in entropy of the system is \(0.237\,\mathrm{kJ/K}\).

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Most popular questions from this chapter

A container holding \(1.20 \mathrm{kg}\) of water at \(20.0^{\circ} \mathrm{C}\) is placed in a freezer that is kept at \(-20.0^{\circ} \mathrm{C} .\) The water freezes and comes to thermal equilibrium with the interior of the freezer. What is the minimum amount of electrical energy required by the freezer to do this if it operates between reservoirs at temperatures of $20.0^{\circ} \mathrm{C}\( and \)-20.0^{\circ} \mathrm{C} ?$
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Consider a heat engine that is not reversible. The engine uses $1.000 \mathrm{mol}\( of a diatomic ideal gas. In the first step \)(\mathrm{A})$ there is a constant temperature expansion while in contact with a warm reservoir at \(373 \mathrm{K}\) from \(P_{1}=1.55 \times 10^{5} \mathrm{Pa}\) and $V_{1}=2.00 \times 10^{-2} \mathrm{m}^{3}$ to \(P_{2}=1.24 \times 10^{5} \mathrm{Pa}\) and $V_{2}=2.50 \times 10^{-2} \mathrm{m}^{3} .$ Then (B) a heat reservoir at the cooler temperature of \(273 \mathrm{K}\) is used to cool the gas at constant volume to \(273 \mathrm{K}\) from \(P_{2}\) to $P_{3}=0.91 \times 10^{5} \mathrm{Pa} .$ This is followed by (C) a constant temperature compression while still in contact with the cold reservoir at \(273 \mathrm{K}\) from \(P_{3}, V_{2}\) to \(P_{4}=1.01 \times 10^{5} \mathrm{Pa}, V_{1} .\) The final step (D) is heating the gas at constant volume from \(273 \mathrm{K}\) to \(373 \mathrm{K}\) by being in contact with the warm reservoir again, to return from \(P_{4}, V_{1}\) to \(P_{1}, V_{1} .\) Find the change in entropy of the cold reservoir in step \(\mathrm{B}\). Remember that the gas is always in contact with the cold reservoir. (b) What is the change in entropy of the hot reservoir in step D? (c) Using this information, find the change in entropy of the total system of gas plus reservoirs during the whole cycle.
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