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 :- [ 2015, 2017]
(a) 90 J (b) 99 J
(c) 100 J (d) 1 J
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Answer:
efficiency of carnot cycle is given as
η= Q H/W
where W is work done and Q /H
is amount of heat added to system.
Q H = W/n =100J
Q qC =Q H −W=100−10
=90J
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