A carnot engine having an efficiency of 1/10 as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is:- (a) 90 J(b) 99 J(c) 100 J(d) 1 J
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the lower temperature is 1J
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Answer:
A) 90 J
Explanation:
Efficiency of engine, n =1/10 (Given)
Work done on system W = 10J (Given)
Coefficient of performance of refrigerator -
β = Q2/W =1-n/n
= 1-1/10/1/10
= 9/10/1/10
= 9
Energy absorbed from the reservoir -
O = βW
O = 9×10
O = 90 Joules
Thus, if the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is 90J
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