Правка разделов «Statement», «Answer»
en/5.9.16.md
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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: | |||
| $$A_{\max} = 334000 \cdot \frac{300 - 273}{273} = 334000 \cdot \frac{27}{273} \approx 33033 \text{ J} \approx 33 \text{ kJ}$$ | |||
| @@ -31,8 +25,4 @@Solution | |||
| 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] | ||
| + | $$A_{\max} \approx 33 \text{ kJ}$$. | ||
| @@ -1,11 +1,5 @@ | |||
| ### Statement | ### 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 $27^\circ\text{C}$. | $5.9.16.$ 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 | ### 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 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 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: | The heat required to melt the ice is: | ||
| $$Q_C = \lambda m$$ | $$Q_C = \lambda m$$ | ||
| where $\lambda = 334 \text{ kJ/kg}$, $m = 1 \text{ kg}$, $T_C = 273 \text{ K}$. | where $\lambda = 334 \text{ kJ/kg}$, $m = 1 \text{ kg}$, $T_C = 273 \text{ K}$. | ||
| The temperature of the ocean (hot reservoir) is $T_H = 300 \text{ K}$. | 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: | For a reversible Carnot cycle (which yields maximum work), the following relation holds: | ||
| $$\frac{Q_H}{T_H} = \frac{Q_C}{T_C}$$ | $$\frac{Q_H}{T_H} = \frac{Q_C}{T_C}$$ | ||
| From this, the heat the engine takes from the ocean is: | From this, the heat the engine takes from the ocean is: | ||
| $$Q_H = Q_C \frac{T_H}{T_C}$$ | $$Q_H = Q_C \frac{T_H}{T_C}$$ | ||
| The maximum work is the difference between the heats: | The maximum work is the difference between the heats: | ||
| $$A_{\max} = Q_H - Q_C = Q_C \left(\frac{T_H}{T_C} - 1\right) = \lambda m \frac{T_H - T_C}{T_C}$$ | $$A_{\max} = Q_H - Q_C = Q_C \left(\frac{T_H}{T_C} - 1\right) = \lambda m \frac{T_H - T_C}{T_C}$$ | ||
| Substituting the numerical values: | Substituting the numerical values: | ||
| $$A_{\max} = 334000 \cdot \frac{300 - 273}{273} = 334000 \cdot \frac{27}{273} \approx 33033 \text{ J} \approx 33 \text{ kJ}$$ | $$A_{\max} = 334000 \cdot \frac{300 - 273}{273} = 334000 \cdot \frac{27}{273} \approx 33033 \text{ J} \approx 33 \text{ kJ}$$ | ||
| @@ -31,8 +25,4 @@Solution | |||
| 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. | 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 | #### Answer | ||
| $33 \text{ kJ}$. | $$A_{\max} \approx 33 \text{ kJ}$$. | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||