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Question

The temperature inside and outside of the refrigerator is 260 K and 315 K respectively. Assuming that the refrigerator cycle is reversible, calculate the heat delivered to the surroundings for every joule of work done.

A
5 J
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B
6.19 J
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C
5 J
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D
5.73 J
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Solution

The correct option is D 5.73 J
Refrigerator rejects heat to surroundings (high temperature)
Inside temperature; T2=260 K
Outside temperature, T1=315 K
For the reversible cycle (Carnot's refrigerator),
Coefficient of performance (β)=Q2W=T2T1T2
( W=1 Joule )
Q21=260315260
Q2=4.73 J
Hence, for each joule of work done, the heat delivered to the surrounding is,
Q1=Q2+W
Q1=4.73+1=5.73 J

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