Question

The efficiency of a Carnot heat engine is 25% and the temperature of its source is 127°C. Without changing the temperature of the source, if the absolute temperature of the sink is decreased by 10%, the efficiency of the engine is:

• A. \( 27.5% \)
• B. \( 17.5% \)
• C. \( 32.5% \)
• D. \( 22.5% \)

Hint:

The Carnot efficiency depends on the absolute temperatures of the heat source and sink. A decrease in sink temperature increases the efficiency.

Answer 1

The Correct Option is C

Step 1: Understanding the Carnot Efficiency Formula
The efficiency of a Carnot engine is given by: \[ \eta = 1 - \frac{T_L}{T_H} \] where: - \( \eta \) is the efficiency,
- \( T_H \) is the absolute temperature of the source,
- \( T_L \) is the absolute temperature of the sink.
Step 2: Convert Given Temperatures to Absolute Scale
The given source temperature is 127°C: \[ T_H = 127 + 273 = 400K \] Let the initial sink temperature be \( T_L \). From the efficiency formula: \[ 0.25 = 1 - \frac{T_L}{400} \] Solving for \( T_L \): \[ T_L = 400(1 - 0.25) = 400 \times 0.75 = 300K \] Step 3: New Efficiency After Sink Temperature Decrease
If the absolute temperature of the sink is decreased by 10%, the new sink temperature is: \[ T'_L = 300 - 0.1 \times 300 = 300 - 30 = 270K \] The new efficiency is: \[ \eta' = 1 - \frac{T'_L}{T_H} \] \[ \eta' = 1 - \frac{270}{400} = 1 - 0.675 = 0.325 \] \[ \eta' = 32.5% \] Step 4: Conclusion
Thus, the correct answer is option (C) \( 32.5% \).