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Chapter 15: Problem 54

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 \(螖S_h = \dfrac{Q}{T_h}\) to find the change in entropy for the hot reservoir, where the heat transfer, Q, is negative because it loses heat: \(螖S_h = \dfrac{-418.6\mathrm{kJ}}{473.15\mathrm{K}} = -0.885\,\mathrm{kJ/K}\)
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: \(螖S_c = \dfrac{418.6\mathrm{kJ}}{373.15\mathrm{K}} = 1.122\,\mathrm{kJ/K}\)
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: \(螖S_{net} = 螖S_h + 螖S_c = -0.885\,\mathrm{kJ/K} + 1.122\,\mathrm{kJ/K} = 0.237\,\mathrm{kJ/K}\) The net change in entropy of the system is \(0.237\,\mathrm{kJ/K}\).

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