question_answer

A Carnot engine having an efficiency of \[\frac{1}{10}\] is being used as a refrigerator. If the work done on the refrigerator is 10 J, the amount of heat absorbed from the reservoir at lower temperature is: [JEE MAIN Held on 08-01-2020 Evening]

A) 90 J     

B) 1 J

C) 99 J                 

D) 100 J

Correct Answer: A

Solution :

\[\frac{\Delta {{Q}_{1}}-\Delta {{Q}_{2}}}{\Delta {{Q}_{1}}}=\frac{1}{10}=\frac{\Delta w}{\Delta {{Q}_{1}}}\] \[\Rightarrow \frac{1}{10}=\frac{\Delta w}{\Delta {{Q}_{1}}}\] \[\therefore \,\,\,\,\,\,\,\,\Delta {{Q}_{1}}=\Delta w\times 10=100\,\,J\] So, \[\Delta {{Q}_{1}}-\Delta {{Q}_{2}}=\Delta w\] \[\Rightarrow \,\,\,\,\,\,100-10=\Delta {{Q}_{2}}=90\,\,J\]