A Carnot engine, having an efficiency of η=1/10 as heat engine, is used as a refrigerator. If the work done on the system is 10J, the amount of energy absorbed from the reservoir at lower temperature is
A
90J
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B
99J
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C
100J
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D
1J
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Solution
The correct option is A90J Given, η=110 η=Desired effectwork input=WoutQ2=Q2−Q1Q2 Where, Q1 is the heat at higher temperature and Q2 heat at lower temperature
In case of refrigerator Q2 is desired and work input is Q2−Q1=W. Where, W is the work done on the system β=Desired effectwork input=Q1Q2−Q1 1β=Q2Q1−1(1) η=1−Q1Q2=110 Q1Q2=910 put in equation 1 we get 1β=109−1=19⇒β=9 9=Q1W=Q110 Q1=90J