Question:
A Carnot engine operating between 430 K and 330 K does a work of 60 kJ. The amount of ice that can melt from its exhaust is (Latent heat of fusion of ice = 330 J g\(^{-1}\)).
A Carnot engine operating between 430 K and 330 K does a work of 60 kJ. The amount of ice that can melt from its exhaust is (Latent heat of fusion of ice = 330 J g\(^{-1}\)).
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In Carnot engines, heat rejected is used for phase change calculations like melting ice.
Updated On: Jun 20, 2026
- 0.6 kg
- 0.75 kg
- 1.2 kg
- 0.4 kg
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The Correct Option is A
Solution and Explanation
Step 1: Carnot efficiency.
\[ \eta = 1 - \frac{T_2}{T_1} = 1 - \frac{330}{430} \] \[ \eta = \frac{100}{430} \]
Step 2: Relation between work and heat input.
\[ \eta = \frac{W}{Q_1} \Rightarrow Q_1 = \frac{W}{\eta} \] \[ Q_1 = \frac{60 \times 10^3}{100/430} \] \[ Q_1 = 60 \times 10^3 \times \frac{430}{100} \] \[ Q_1 = 258000 \, \text{J} \]
Step 3: Heat rejected.
\[ Q_2 = Q_1 - W = 258000 - 60000 = 198000 \, \text{J} \]
Step 4: Ice melting.
\[ m = \frac{Q_2}{L} \] \[ m = \frac{198000}{330 \times 10^3} \] \[ m = 0.6 \, \text{kg} \]
Step 5: Final conclusion.
\[ \boxed{0.6 \, \text{kg}} \]
\[ \eta = 1 - \frac{T_2}{T_1} = 1 - \frac{330}{430} \] \[ \eta = \frac{100}{430} \]
Step 2: Relation between work and heat input.
\[ \eta = \frac{W}{Q_1} \Rightarrow Q_1 = \frac{W}{\eta} \] \[ Q_1 = \frac{60 \times 10^3}{100/430} \] \[ Q_1 = 60 \times 10^3 \times \frac{430}{100} \] \[ Q_1 = 258000 \, \text{J} \]
Step 3: Heat rejected.
\[ Q_2 = Q_1 - W = 258000 - 60000 = 198000 \, \text{J} \]
Step 4: Ice melting.
\[ m = \frac{Q_2}{L} \] \[ m = \frac{198000}{330 \times 10^3} \] \[ m = 0.6 \, \text{kg} \]
Step 5: Final conclusion.
\[ \boxed{0.6 \, \text{kg}} \]
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