Edits to “Statement”, “Solution”

Valter edited Aug 19, 2026, 3:01 PM

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			### Statement

		−	$5.9.14.$ [Insert the problem statement]
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		−	### Solution
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		−	### Statement
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			$5.9.14.$ Show that the efficiency of a heat engine in a cyclic process is maximal when the entropy of the system does not change.

			### Solution

			In the problem statement, the "system" implies a complete closed thermodynamic system, which includes the working fluid of the heat engine as well as the heat reservoirs (the heater at temperature $T_1$ and the cooler at temperature $T_2$).

			Over one complete cycle, the working fluid returns to its initial state. Since entropy $S$ is a state function, the change in entropy of the working fluid itself over a cycle is always zero:
			$$\Delta S_{\text{wf}} = 0$$

			Therefore, the total change in entropy of the closed system over a cycle consists solely of the change in entropy of the heat reservoirs:
			$$\Delta S_{\text{system}} = \Delta S_{\text{heater}} + \Delta S_{\text{cooler}}$$

			Suppose that during a cycle, the engine receives heat $Q_1$ from the heater and rejects heat $Q_2$ to the cooler (where $Q_1, Q_2 > 0$).
			Then the heater loses entropy, and the cooler gains it:
			$$\Delta S_{\text{system}} = -\frac{Q_1}{T_1} + \frac{Q_2}{T_2}$$

			According to the second law of thermodynamics, the entropy of a closed system cannot decrease ($\Delta S_{\text{system}} \ge 0$). Therefore:
			$$\frac{Q_2}{T_2} - \frac{Q_1}{T_1} \ge 0 \implies \frac{Q_2}{Q_1} \ge \frac{T_2}{T_1}$$

			Now let us express the efficiency of the heat engine through the work $W = Q_1 - Q_2$:
			$$\eta = \frac{W}{Q_1} = \frac{Q_1 - Q_2}{Q_1} = 1 - \frac{Q_2}{Q_1}$$

			Let us substitute the obtained thermodynamic inequality $\frac{Q_2}{Q_1} \ge \frac{T_2}{T_1}$ into this expression:
			$$\eta \le 1 - \frac{T_2}{T_1}$$

			<b>Conclusion:</b>
		@@ -35,7 +29,3 @@Solution
			The efficiency reaches its maximum possible value $\eta_{\text{max}} = 1 - \frac{T_2}{T_1}$ (Carnot cycle efficiency) if and only if the inequality turns into an equality:
			$$\frac{Q_2}{T_2} - \frac{Q_1}{T_1} = 0 \implies \Delta S_{\text{system}} = 0$$
			Thus, the efficiency of a heat engine is maximal if the process is completely reversible and the total entropy of the system does not change.
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		−	#### Answer
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		−	[Insert a concise answer or boxed result]

	@@ -1,11 +1,5 @@
	### Statement		### Statement

	$5.9.14.$ [Insert the problem statement]

	### Solution

	### Statement

	$5.9.14.$ Show that the efficiency of a heat engine in a cyclic process is maximal when the entropy of the system does not change.		$5.9.14.$ Show that the efficiency of a heat engine in a cyclic process is maximal when the entropy of the system does not change.

	### Solution		### Solution

	In the problem statement, the "system" implies a complete closed thermodynamic system, which includes the working fluid of the heat engine as well as the heat reservoirs (the heater at temperature $T_1$ and the cooler at temperature $T_2$).		In the problem statement, the "system" implies a complete closed thermodynamic system, which includes the working fluid of the heat engine as well as the heat reservoirs (the heater at temperature $T_1$ and the cooler at temperature $T_2$).

	Over one complete cycle, the working fluid returns to its initial state. Since entropy $S$ is a state function, the change in entropy of the working fluid itself over a cycle is always zero:		Over one complete cycle, the working fluid returns to its initial state. Since entropy $S$ is a state function, the change in entropy of the working fluid itself over a cycle is always zero:
	$$\Delta S_{\text{wf}} = 0$$		$$\Delta S_{\text{wf}} = 0$$

	Therefore, the total change in entropy of the closed system over a cycle consists solely of the change in entropy of the heat reservoirs:		Therefore, the total change in entropy of the closed system over a cycle consists solely of the change in entropy of the heat reservoirs:
	$$\Delta S_{\text{system}} = \Delta S_{\text{heater}} + \Delta S_{\text{cooler}}$$		$$\Delta S_{\text{system}} = \Delta S_{\text{heater}} + \Delta S_{\text{cooler}}$$

	Suppose that during a cycle, the engine receives heat $Q_1$ from the heater and rejects heat $Q_2$ to the cooler (where $Q_1, Q_2 > 0$).		Suppose that during a cycle, the engine receives heat $Q_1$ from the heater and rejects heat $Q_2$ to the cooler (where $Q_1, Q_2 > 0$).
	Then the heater loses entropy, and the cooler gains it:		Then the heater loses entropy, and the cooler gains it:
	$$\Delta S_{\text{system}} = -\frac{Q_1}{T_1} + \frac{Q_2}{T_2}$$		$$\Delta S_{\text{system}} = -\frac{Q_1}{T_1} + \frac{Q_2}{T_2}$$

	According to the second law of thermodynamics, the entropy of a closed system cannot decrease ($\Delta S_{\text{system}} \ge 0$). Therefore:		According to the second law of thermodynamics, the entropy of a closed system cannot decrease ($\Delta S_{\text{system}} \ge 0$). Therefore:
	$$\frac{Q_2}{T_2} - \frac{Q_1}{T_1} \ge 0 \implies \frac{Q_2}{Q_1} \ge \frac{T_2}{T_1}$$		$$\frac{Q_2}{T_2} - \frac{Q_1}{T_1} \ge 0 \implies \frac{Q_2}{Q_1} \ge \frac{T_2}{T_1}$$

	Now let us express the efficiency of the heat engine through the work $W = Q_1 - Q_2$:		Now let us express the efficiency of the heat engine through the work $W = Q_1 - Q_2$:
	$$\eta = \frac{W}{Q_1} = \frac{Q_1 - Q_2}{Q_1} = 1 - \frac{Q_2}{Q_1}$$		$$\eta = \frac{W}{Q_1} = \frac{Q_1 - Q_2}{Q_1} = 1 - \frac{Q_2}{Q_1}$$

	Let us substitute the obtained thermodynamic inequality $\frac{Q_2}{Q_1} \ge \frac{T_2}{T_1}$ into this expression:		Let us substitute the obtained thermodynamic inequality $\frac{Q_2}{Q_1} \ge \frac{T_2}{T_1}$ into this expression:
	$$\eta \le 1 - \frac{T_2}{T_1}$$		$$\eta \le 1 - \frac{T_2}{T_1}$$

	<b>Conclusion:</b>		<b>Conclusion:</b>
	@@ -35,7 +29,3 @@Solution
	The efficiency reaches its maximum possible value $\eta_{\text{max}} = 1 - \frac{T_2}{T_1}$ (Carnot cycle efficiency) if and only if the inequality turns into an equality:		The efficiency reaches its maximum possible value $\eta_{\text{max}} = 1 - \frac{T_2}{T_1}$ (Carnot cycle efficiency) if and only if the inequality turns into an equality:
	$$\frac{Q_2}{T_2} - \frac{Q_1}{T_1} = 0 \implies \Delta S_{\text{system}} = 0$$		$$\frac{Q_2}{T_2} - \frac{Q_1}{T_1} = 0 \implies \Delta S_{\text{system}} = 0$$
	Thus, the efficiency of a heat engine is maximal if the process is completely reversible and the total entropy of the system does not change.		Thus, the efficiency of a heat engine is maximal if the process is completely reversible and the total entropy of the system does not change.

	#### Answer

	[Insert a concise answer or boxed result]