Question

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:

• A. Decreases by 32.5\%
• B. Decreases by 7.5\%
• C. Increases by 32.5\%
• D. Increases by 7.5\%

Hint:

In a Carnot engine, the efficiency depends on the temperatures of the source and sink. Lowering the sink temperature increases the efficiency.

Answer 1

The Correct Option is D

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.