Physics, asked by Dasrupsa74431, 8 months ago

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

Answers

Answered by aayushi890
4

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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