Question

The temperature of the source and sink of a heat engine are 127°C and 27°C respectively. By how much should the temperature of the source be increased so as to double the efficiency?

• A. 100°C
• B. 200°C
• C. 150°C
• D. 50°C

Hint:

The efficiency of a heat engine increases with the temperature of the source. To double the efficiency, use the Carnot equation and solve for the new source temperature.

Answer 1

The Correct Option is C

The efficiency \( \eta \) of a heat engine is given by the Carnot efficiency formula: \[ \eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}} \] Where: - \( T_{\text{sink}} \) is the temperature of the sink (in Kelvin), - \( T_{\text{source}} \) is the temperature of the source (in Kelvin). Given: - \( T_{\text{source}} = 127^\circ C = 400 \, \text{K} \), - \( T_{\text{sink}} = 27^\circ C = 300 \, \text{K} \). ### Step 1: Initial Efficiency The initial efficiency is: \[ \eta_{\text{initial}} = 1 - \frac{300}{400} = 1 - 0.75 = 0.25 \] ### Step 2: Final Efficiency We want to double the efficiency, so the new efficiency \( \eta_{\text{final}} \) should be: \[ \eta_{\text{final}} = 2 \times \eta_{\text{initial}} = 2 \times 0.25 = 0.50 \] ### Step 3: Solve for New Temperature of Source Using the Carnot efficiency formula for the final efficiency: \[ 0.50 = 1 - \frac{300}{T_{\text{source, final}}} \] Solving for \( T_{\text{source, final}} \): \[ \frac{300}{T_{\text{source, final}}} = 0.50 \] \[ T_{\text{source, final}} = \frac{300}{0.50} = 600 \, \text{K} \] ### Step 4: Increase in Temperature The increase in the temperature of the source is: \[ \Delta T = 600 - 400 = 200 \, \text{K} = 200^\circ C \] Thus, the temperature of the source needs to be increased by: \[ \boxed{(B) \, 200^\circ C} \]