• # question_answer An ideal refrigerator has a freezer at a temperature of $-13{}^\circ C.$ The coefficient of performance of the engine is 5. The temperature of the air (to which heat is rejected) will be A) 325°C B) 325K C) 39°C     D) 320°C

 [c]Coefficient of performance $K=\frac{{{T}_{2}}}{{{T}_{1}}-{{T}_{2}}}$Þ$5=\frac{(273-13)}{{{T}_{1}}-(273-13)}=\frac{260}{{{T}_{1}}-260}$ Þ $5{{T}_{1}}-1300=260$Þ $5{{T}_{1}}=1560$ Þ ${{T}_{1}}=312K\to 39{}^\circ C$