Q21 / 155 · WBJEE 2026 · Chemistry
Three engines A, B and C take steam at $130°$C and reject it at $20°$C, $40°$C and $50°$C respectively. The most efficient engine will be
- Correct answerAA
- BB
- CC
- DAll the three engines will be equally efficient
Solution
The efficiency of a heat engine is given by $\eta = 1 - \dfrac{T_2}{T_1}$, where $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink.
For all three engines, the source temperature $T_1 = 130^\circ\text{C} = 403\text{ K}$.
The sink temperatures for engines A, B, and C are $T_{2A} = 20^\circ\text{C} = 293\text{ K}$, $T_{2B} = 40^\circ\text{C} = 313\text{ K}$, and $T_{2C} = 50^\circ\text{C} = 323\text{ K}$ respectively.
Since $\eta$ is inversely related to $T_2$ for a given $T_1$, the efficiency will be maximum when the sink temperature $T_2$ is minimum.
Engine A has the lowest sink temperature ($20^\circ\text{C}$), so it will be the most efficient.
Answer: A
For all three engines, the source temperature $T_1 = 130^\circ\text{C} = 403\text{ K}$.
The sink temperatures for engines A, B, and C are $T_{2A} = 20^\circ\text{C} = 293\text{ K}$, $T_{2B} = 40^\circ\text{C} = 313\text{ K}$, and $T_{2C} = 50^\circ\text{C} = 323\text{ K}$ respectively.
Since $\eta$ is inversely related to $T_2$ for a given $T_1$, the efficiency will be maximum when the sink temperature $T_2$ is minimum.
Engine A has the lowest sink temperature ($20^\circ\text{C}$), so it will be the most efficient.
Answer: A
Hence, the correct answer is option: 'A'.