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Valter edited Aug 22, 2026, 6:22 PM

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		+	### Statement
		+
		+	$5.9.16.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	### Statement
		+
		+	$5.9.16.$ In the ocean, there is a boat with a $1\text{ kg}$ block of ice at $0^\circ\text{C}$ on board. Determine the maximum work that can be obtained using the ice melting process. The water temperature is $27^\circ\text{C}$.
		+
		+	### Solution
		+
		+	The previous author made a fundamental mistake. They multiplied the efficiency ($\eta$) by the heat of melting ($Q_C$), assuming that $Q_C$ is the heat received from the hot reservoir. However, in this case, the ice acts as the <b>cold reservoir (sink)</b>, and the ocean is the <b>hot reservoir (source)</b>.
		+
		+	The heat engine takes heat $Q_H$ from the ocean, performs useful work $A$, and rejects the remaining heat $Q_C$ to the ice, causing it to melt.
		+	The heat required to melt the ice is:
		+	$$Q_C = \lambda m$$
		+	where $\lambda = 334 \text{ kJ/kg}$, $m = 1 \text{ kg}$, $T_C = 273 \text{ K}$.
		+	The temperature of the ocean (hot reservoir) is $T_H = 300 \text{ K}$.
		+
		+	For a reversible Carnot cycle (which yields maximum work), the following relation holds:
		+	$$\frac{Q_H}{T_H} = \frac{Q_C}{T_C}$$
		+	From this, the heat the engine takes from the ocean is:
		+	$$Q_H = Q_C \frac{T_H}{T_C}$$
		+	The maximum work is the difference between the heats:
		+	$$A_{\max} = Q_H - Q_C = Q_C \left(\frac{T_H}{T_C} - 1\right) = \lambda m \frac{T_H - T_C}{T_C}$$
		+
		+	Substituting the numerical values:
		+	$$A_{\max} = 334000 \cdot \frac{300 - 273}{273} = 334000 \cdot \frac{27}{273} \approx 33033 \text{ J} \approx 33 \text{ kJ}$$
		+
		+	Regarding the heating of the resulting water: the problem explicitly states <i>"using the ice melting process"</i>, so the work is calculated only for the phase transition.
		+
		+	#### Answer
		+	$33 \text{ kJ}$.
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### Statement

			$5.9.16.$ [Insert the problem statement]

			### Solution

			### Statement

			$5.9.16.$ In the ocean, there is a boat with a $1\text{ kg}$ block of ice at $0^\circ\text{C}$ on board. Determine the maximum work that can be obtained using the ice melting process. The water temperature is $27^\circ\text{C}$.

			### Solution

			The previous author made a fundamental mistake. They multiplied the efficiency ($\eta$) by the heat of melting ($Q_C$), assuming that $Q_C$ is the heat received from the hot reservoir. However, in this case, the ice acts as the <b>cold reservoir (sink)</b>, and the ocean is the <b>hot reservoir (source)</b>.

			The heat engine takes heat $Q_H$ from the ocean, performs useful work $A$, and rejects the remaining heat $Q_C$ to the ice, causing it to melt.
			The heat required to melt the ice is:
			$$Q_C = \lambda m$$
			where $\lambda = 334 \text{ kJ/kg}$, $m = 1 \text{ kg}$, $T_C = 273 \text{ K}$.
			The temperature of the ocean (hot reservoir) is $T_H = 300 \text{ K}$.

			For a reversible Carnot cycle (which yields maximum work), the following relation holds:
			$$\frac{Q_H}{T_H} = \frac{Q_C}{T_C}$$
			From this, the heat the engine takes from the ocean is:
			$$Q_H = Q_C \frac{T_H}{T_C}$$
			The maximum work is the difference between the heats:
			$$A_{\max} = Q_H - Q_C = Q_C \left(\frac{T_H}{T_C} - 1\right) = \lambda m \frac{T_H - T_C}{T_C}$$

			Substituting the numerical values:
			$$A_{\max} = 334000 \cdot \frac{300 - 273}{273} = 334000 \cdot \frac{27}{273} \approx 33033 \text{ J} \approx 33 \text{ kJ}$$

			Regarding the heating of the resulting water: the problem explicitly states <i>"using the ice melting process"</i>, so the work is calculated only for the phase transition.

			#### Answer
			$33 \text{ kJ}$.

			#### Answer

			[Insert a concise answer or boxed result]