The temperatures of the source and sink of a Carnot’s heat engine are 27$^\circ$C and 127$^\circ$C respectively. If the absolute temperature of the sink is decreased by 10\%, the efficiency of the engine:
Show Hint
- Decreases by 32.5\%
- Decreases by 7.5\%
- Increases by 32.5\%
- Increases by 7.5\%
The Correct Option is D
Solution and Explanation
Step 1: The efficiency \( \eta \) of a Carnot engine is given by: \[ \eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}} \] where \( T_{\text{source}} \) and \( T_{\text{sink}} \) are the temperatures of the source and sink, respectively.
Step 2: The initial temperatures are \( T_{\text{source}} = 27 + 273 = 300 \, \text{K} \) and \( T_{\text{sink}} = 127 + 273 = 400 \, \text{K} \). The initial efficiency is: \[ \eta_{\text{initial}} = 1 - \frac{400}{300} = 1 - \frac{4}{3} = -\frac{1}{3} \]
Step 3: After the sink temperature decreases by 10\%, the new temperature of the sink is \( T_{\text{sink}}' = 0.9 \times 400 = 360 \, \text{K} \). The new efficiency is: \[ \eta_{\text{new}} = 1 - \frac{360}{300} = 1 - \frac{6}{5} = \frac{1}{5} \]
Step 4: The change in efficiency is: \[ \Delta \eta = \frac{1}{5} - (-\frac{1}{3}) = 0.2 + 0.3333 = 0.5333 \]
Step 5: The efficiency increases by 7.5\%, so Option (4) is correct.
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