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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

100J

C

`(200)/(3)` J

D

150 J

Text Solution

Verified by Experts

The correct Answer is:
D
`beta=(Q_2)/(W) = > W=(Q_2)/(beta)`
`implies Q_1-Q_2=(Q_2)/(1//3)=3Q_2 implies 4Q_2=Q_1 implies Q_2 = 50J`
Now, `W=Q_1-Q_2=200-50=150 J`
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Refrigerator is an apparatus which takes heat from a cold body, work is done on it and the work done together with the heat absorbed is rejected to the source. An ideal refrigerator can be regarded as Carnot's ideal heat engine working in the reverse direction. The coefficient of performance of refrigerator is defined as beta = ("Heat extracted from cold reservoir")/("work done on working substance") = (Q_(2))/(W) = (Q_(2))/(Q_(1)- Q_(2)) = (T_(2))/(T_(1) - T_(2)) A Carnot's refrigerator takes heat from water at 0^(@)C and discards it to a room temperature at 27^(@)C . 1 Kg of water at 0^(@)C is to be changed into ice at 0^(@)C. (L_(ice) = 80"kcal//kg") What is the work done by the refrigerator in this process ( 1 "cal" = 4.2 "joule" )

Refrigerator is an apparatus which takes heat from a cold body, work is done on it and the work done together with the heat absorbed is rejected to the source. An ideal refrigerator can be regarded as Carnot's ideal heat engine working in the reverse direction. The coefficient of performance of refrigerator is defined as beta = ("Heat extracted from cold reservoir")/("work done on working substance") = (Q_(2))/(W) = (Q_(2))/(Q_(1)- Q_(2)) = (T_(2))/(T_(1) - T_(2)) A Carnot's refrigerator takes heat from water at 0^(@)C and discards it to a room temperature at 27^(@)C . 1 Kg of water at 0^(@)C is to be changed into ice at 0^(@)C. (L_(ice) = 80"kcal//kg") How many calories of heat are discarded to the room?

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