Question

The minimum amount of work required to operate a refrigerator which removes 1000 Cal heat at $0^\circ$C and rejects at $50^\circ$C will be:

• A. 170 Cal
• B. 120 Cal
• C. 150 Cal
• D. 183.15 Cal

Hint:

Use $\text{COP}_{\text{ref}} = \frac{T_L}{T_H - T_L}$ for Carnot refrigerators. Always convert Celsius to Kelvin before substitution.

Answer 1

The Correct Option is D

To find the minimum work input, we use the Carnot refrigerator formula:
\[ \text{COP}_{\text{refrigerator}} = \frac{T_L}{T_H - T_L} \]
Where:
$T_L = 0^\circ C = 273 \text{ K}$
$T_H = 50^\circ C = 323 \text{ K}$
Substitute the values:
\[ \text{COP} = \frac{273}{323 - 273} = \frac{273}{50} = 5.46 \]
The work input $W$ is related to the heat extracted $Q_L$ and COP as:
\[ W = \frac{Q_L}{\text{COP}} = \frac{1000}{5.46} \approx 183.15 \text{ Cal} \]