2. A Carnot heat engine receives heat at the rate of 74,000kJ/min from a high-temperature source at 510^circ C If the maximum possible power output of the engine is 653 kW, find the temperature of the low.temperature sink in which the engine rejects heat (15pts) Assumptions: The system operates steadily.

To find the temperature of the low-temperature sink, we can use the Carnot efficiency equation:<br /><br />\[ \eta = 1 - \frac{T_c}{T_h} \]<br /><br />where:<br />- \( \eta \) is the Carnot efficiency,<br />- \( T_c \) is the temperature of the cold reservoir (sink),<br />- \( T_h \) is the temperature of the hot reservoir (source).<br /><br />Given:<br />- The heat input rate (\( \dot{Q}_h \)) is \( 74,000 \, kJ/min \),<br />- The power output (\( P \)) is \( 653 \, kW \),<br />- The temperature of the hot reservoir (\( T_h \)) is \( 510^{\circ}C \).<br /><br />First, convert the heat input rate to power:<br /><br />\[ \dot{Q}_h = 74,000 \, kJ/min \times \frac{1 \, kJ}{1 \, s} \times \frac{1 \, min}{60 \, s} = 12,333.33 \, W \]<br /><br />The Carnot efficiency is also given by:<br /><br />\[ \eta = \frac{P}{\dot{Q}_h} \]<br /><br />Substitute the given values:<br /><br />\[ \eta = \frac{653 \, kW}{12,333.33 \, W} = \frac{653,000 \, W}{12,333.33 \, W} \approx 0.527 \]<br /><br />Now, use the Carnot efficiency equation to find \( T_c \):<br /><br />\[ 0.527 = 1 - \frac{T_c}{T_h} \]<br /><br />Rearrange to solve for \( T_c \):<br /><br />\[ \frac{T_c}{T_h} = 1 - 0.527 = 0.473 \]<br /><br />\[ T_c = 0.473 \times T_h \]<br /><br />Convert \( T_h \) to Kelvin:<br /><br />\[ T_h = 510^{\circ}C + 273.15 = 783.15 \, K \]<br /><br />Now, calculate \( T_c \):<br /><br />\[ T_c = 0.473 \times 783.15 \, K \approx 370.5 \, K \]<br /><br />Convert \( T_c \) back to Celsius:<br /><br />\[ T_c = 370.5 \, K - 273.15 = 97.35^{\circ}C \]<br /><br />Therefore, the temperature of the low-temperature sink is approximately \( 97.35^{\circ}C \).