Question:
Three engines A, B and C take steam at \( 130^\circ\text{C} \) and reject it at \( 20^\circ\text{C} \), \( 40^\circ\text{C} \) and \( 50^\circ\text{C} \) respectively. The most efficient engine will be
Three engines A, B and C take steam at \( 130^\circ\text{C} \) and reject it at \( 20^\circ\text{C} \), \( 40^\circ\text{C} \) and \( 50^\circ\text{C} \) respectively. The most efficient engine will be
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For any heat engine, to maximize its efficiency, you should either increase the temperature of the source (\( T_H \)) or decrease the temperature of the sink (\( T_C \)). Therefore, the engine with the largest temperature difference (\( T_H - T_C \)) is always the most efficient.
Updated On: May 28, 2026
- A
- B
- C
- All the three engines will be equally efficient
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The problem compares the efficiency of three heat engines operating between a common source temperature and different sink temperatures. We need to find which engine is the most efficient.
Step 2: Key Formula or Approach:
The maximum theoretical efficiency (\( \eta \)) of a heat engine is given by the Carnot efficiency formula:
\[ \eta = 1 - \frac{T_C}{T_H} \]
where:
- \( T_H \) is the absolute temperature of the hot reservoir (source) in Kelvin.
- \( T_C \) is the absolute temperature of the cold reservoir (sink) in Kelvin.
Step 3: Detailed Explanation:
1. All three engines take steam at the same source temperature \( T_H \):
\[ T_H = 130^\circ\text{C} = 130 + 273.15 = 403.15\text{ K} \]
2. Convert the sink temperatures of the three engines to Kelvin:
- Engine A: \( T_{C,A} = 20^\circ\text{C} = 20 + 273.15 = 293.15\text{ K} \)
- Engine B: \( T_{C,B} = 40^\circ\text{C} = 40 + 273.15 = 313.15\text{ K} \)
- Engine C: \( T_{C,C} = 50^\circ\text{C} = 50 + 273.15 = 323.15\text{ K} \)
3. Analyze the efficiency relation:
For a constant \( T_H \), the efficiency \( \eta = 1 - \frac{T_C}{T_H} \) increases as the sink temperature \( T_C \) decreases.
- Since Engine A has the lowest sink temperature (\( T_{C,A} = 20^\circ\text{C} \)), the ratio \( \frac{T_C}{T_H} \) is minimized, thereby maximizing the efficiency \( \eta \).
Thus, Engine A is the most efficient engine.
Step 4: Final Answer:
The most efficient engine is A, which corresponds to option (A).
The problem compares the efficiency of three heat engines operating between a common source temperature and different sink temperatures. We need to find which engine is the most efficient.
Step 2: Key Formula or Approach:
The maximum theoretical efficiency (\( \eta \)) of a heat engine is given by the Carnot efficiency formula:
\[ \eta = 1 - \frac{T_C}{T_H} \]
where:
- \( T_H \) is the absolute temperature of the hot reservoir (source) in Kelvin.
- \( T_C \) is the absolute temperature of the cold reservoir (sink) in Kelvin.
Step 3: Detailed Explanation:
1. All three engines take steam at the same source temperature \( T_H \):
\[ T_H = 130^\circ\text{C} = 130 + 273.15 = 403.15\text{ K} \]
2. Convert the sink temperatures of the three engines to Kelvin:
- Engine A: \( T_{C,A} = 20^\circ\text{C} = 20 + 273.15 = 293.15\text{ K} \)
- Engine B: \( T_{C,B} = 40^\circ\text{C} = 40 + 273.15 = 313.15\text{ K} \)
- Engine C: \( T_{C,C} = 50^\circ\text{C} = 50 + 273.15 = 323.15\text{ K} \)
3. Analyze the efficiency relation:
For a constant \( T_H \), the efficiency \( \eta = 1 - \frac{T_C}{T_H} \) increases as the sink temperature \( T_C \) decreases.
- Since Engine A has the lowest sink temperature (\( T_{C,A} = 20^\circ\text{C} \)), the ratio \( \frac{T_C}{T_H} \) is minimized, thereby maximizing the efficiency \( \eta \).
Thus, Engine A is the most efficient engine.
Step 4: Final Answer:
The most efficient engine is A, which corresponds to option (A).
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