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A refrigerator with COP= 1//3 release 20...

A refrigerator with `COP= 1//3` release `200 J` at heat to a reservoir. Then the work done on the working substance is

A

`(100)/(3) J`

B

100 J

C

`(200)/(3) J`

D

150 J of heat has been added to the gas

Text Solution

Verified by Experts

The correct Answer is:
D
The coefficient of performance of a refrigeratior is given by
`beta =(Q_(2))/(W)=(Q_(2))/(Q_(1))-Q_(2))`
substituting the given values we get
`(1)/(3)=(Q_(2))/(200-q_(2))` or `200-Q_(2)=3Q_(2)`
or `4Q_(2) =200 or Q_(2)=(200)/(4)J=50 J`
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