Question

A Carnot heat engine has an efficiency of 10%. If the same engine is worked backward to obtain a refrigerator, then the coefficient of performance of the refrigerator is

• A. 8
• B. 9
• C. 5
• D. 6

Hint:

For refrigerators working on a Carnot cycle, use the inverse of the efficiency to calculate the coefficient of performance (COP).

Answer 1

The Correct Option is B

The coefficient of performance of a refrigerator is given by the formula: \[ \text{COP} = \frac{T_{\text{cold}}}{T_{\text{hot}} - T_{\text{cold}}} \] Given that the efficiency \( \eta \) of the Carnot engine is 10\%, we can calculate the temperatures. The efficiency is related to the temperatures by: \[ \eta = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}} \] For a Carnot engine, \( \eta = 0.1 \), so: \[ 0.1 = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}} \] Solving for \( T_{\text{cold}} \): \[ T_{\text{cold}} = 0.9 T_{\text{hot}} \] Now, using the COP formula for a refrigerator: \[ \text{COP} = \frac{T_{\text{cold}}}{T_{\text{hot}} - T_{\text{cold}}} = \frac{0.9 T_{\text{hot}}}{T_{\text{hot}} - 0.9 T_{\text{hot}}} = \frac{0.9}{0.1} = 9 \] Thus, the coefficient of performance of the refrigerator is 9.