Thank you for reporting, we will resolve it shortly
Q.
A refrigerator with coefficient of performance $\frac{1}{3}$ releases $200 \,J$ of heat to a hot reservoir. Then the work done on the working substance is
Thermodynamics
Solution:
The coefficient of performance of a refrigerator is given by
$\beta = \frac{Q_{2}}{W}= \frac{Q_{2}}{Q_{1} -Q_{2}}$
Substituting the given values, we get
$\frac{1}{3} = \frac{Q_{2}}{200-Q_{2}}$ or
$200-Q_{2} = 3Q_{2} $
or $4Q_{2} = 200$ or $Q_{2}$
$ = \frac{200}{4}J = 50 \,J $
$\therefore W= Q_{1} -Q_{2} $
$= 200 \,J -50 \,J $
$= 150 \,J$