Question

A domestic food freezer maintains a temperature of –15°C. The ambient air temperature is 30°C. If heat leaks into the freezer at the continuous rate of 1.75 kJ/s, the least power necessary to pump this heat out continuously will be nearly
1. 0.1 kW 
2. 0.2 kW
3. 0.3 kW
4. 0.4 kW

Answer 1

Correct Answer - Option 3 : 0.3 kW

Concept:

It is asked least work required, the work will be minimum only when refrigerator operates on reversed Carnot cycle.

For reversed Carnot cycle,

\(COP = \frac{{{T_L}}}{{{T_H} - {T_L}}} = \frac{{{Q_L}}}{W}\)

Calculation:

Given T_L = -15°C = 258 K, T_H = 30°C = 303 K, Q_L = 1.75 kJ/s;

\(\frac{{{T_L}}}{{{T_H} - {T_L}}} = \frac{{{Q_L}}}{W} \Rightarrow \frac{{258}}{{303 - 258}} = \frac{{1.75}}{W}\) 

⇒ W = 0.3 kJ/s = 0.3 kW