Question

A refrigerator with coefficient of performance $\frac{1}{3}$ reales 200 J of heat to a reservoir. Then the work done on the working substance is:

• A. $\frac{100}{3}J$
• B. $100J$
• C. $\frac{200}{3}J$
• D. $150J$

Answer 1

The correct option is D $150J$

For a refrigerator-

$Cop=\frac{Q_{cold}}{Q_{hot}-Q_{cold}}\frac{1}{3}=\frac{Q_{cold}}{200-Q_{cold}}\Rightarrow 3Q_{cold}=200-Q_{cold}\Rightarrow 4Q_{cold}=200\Rightarrow Q_{cold}=50JW=Q_{hot}-Q_{cold}=(200-50)J=150J$