JEE Main & Advanced Physics Thermodynamical Processes Question Bank Topic Test - Thermodynamical Process (2021)

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

Correct Answer: C

Solution :

[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 \[{{T}_{1}}=312K\to 39{}^\circ C\]