Question

# A refrigerator whose coefficient of performance $K=5$, extracts heat from the cooling compartment at the rate of $250J/cycle$. What is the work done per cycle to operate the refrigerator? How much heat is discharged per cycle to the room which acts as the high-temperature reservoir?

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Solution

## Step 1: Given dataThe coefficient of performance of the refrigerator is $\eta =5.$The cooling rate of the cold reservoir is ${Q}_{c}=250J/cycle.$Step 2: Formula UsedWe know that the performance of a refrigerator is the ratio of the amount of heat extracted from the cold body to the work done on it.The performance or efficiency of a refrigerator is defined by the form, $\eta =\frac{{Q}_{C}}{W}=\frac{heatextractedfromthecoldbody}{workdoneonthemachine}.$Heat discharged by a refrigerator is the sum of total heat taken by the machine and the work done by the machine, i.e, ${Q}_{d}={Q}_{c}+W$Step 3: DiagramStep 4: CalculationAs we know, the work done by a refrigerator is $W=\frac{{Q}_{C}}{\eta }$So, $W=\frac{250}{5}=50\phantom{\rule{0ex}{0ex}}orW=50joule.$Therefore, the work done per cycle to operate the refrigerator is $50joule.$Now heat discharged by the refrigerator is ${Q}_{d}={Q}_{c}+W$.So, ${Q}_{d}=250+50=300\phantom{\rule{0ex}{0ex}}or{Q}_{d}=300joules$Therefore, heat discharged per cycle to the room is $300joules$.

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