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 :

[a]

\[\frac{\Delta {{Q}_{1}}-\Delta {{Q}_{2}}}{\Delta {{Q}_{1}}}=\frac{1}{10}=\frac{\Delta w}{\Delta {{Q}_{1}}}\]

\[\Rightarrow \frac{1}{10}=\frac{\Delta w}{\Delta {{Q}_{1}}}\]

\[\therefore \,\,\,\,\,\,\,\,\Delta {{Q}_{1}}=\Delta w\times 10=100\,\,J\]

So, \[\Delta {{Q}_{1}}-\Delta {{Q}_{2}}=\Delta w\]

\[\Rightarrow \,\,\,\,\,\,100-10=\Delta {{Q}_{2}}=90\,\,J\]