Answer to Question #281791 in Molecular Physics | Thermodynamics for mekka

Question

A Carnot refrigerator rejects 2500 kJ of heat at 80°C while using 1100 kJ of work

Answer 1

Find the heat absorbed:

$$Q_a=W+Q_r=1100+2500=3600\text{ kJ}.$$

The efficiency of the refrigerator is

$$\eta=\frac{W}{Q_a}=\frac{1100}{3600}=0.31.$$