The temperature inside a refrigerator is t2°C and the room temperature is t1°C. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be
A
t2+273t1−t2
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B
t1+t2t1+273
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C
t1t1−t2
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D
t1+273t1−t2
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Solution
The correct option is Dt1+273t1−t2 Coefficient of Performance of Refrigerator COP=QcW=t2+273t1−t2
We have to find heat given to room per unit electrical energy.
Room is at higher temperature and act as hot reservoir. And electrical energy is the input energy shown as W.