Question

Heat is being removed from a refrigerator at a rate of 300 kJ/min to maintain its inside temperature at \( 2^\circ \text{C} \). If the input power to the refrigerator is 2 kW, the coefficient of performance of the refrigerator is _________.
\text{[round off to one decimal place]}

Hint:

To calculate the coefficient of performance for a refrigerator, use the formula \( COP = \frac{\text{Heat removed}}{\text{Work input}} \).

Answer 1

Correct Answer: 2.4

The coefficient of performance (COP) of a refrigerator is given by: \[ COP = \frac{\text{Heat removed}}{\text{Work input}}. \] Convert the heat removed to the same units as the work input: \[ \text{Heat removed} = 300 \, \text{kJ/min} = \frac{300}{60} \, \text{kJ/s} = 5 \, \text{kW}. \] Now, calculate the COP: \[ COP = \frac{5}{2} = 2.5. \] Thus, the coefficient of performance is approximately \( 2.6 \).