A refrigerator with coefficient of performance 13releases 200 J of heat to a hot reservoir. Then the work done on the working substance is

The coefficient of performance of a refrigerator is given by β=Q2W=Q2Q1−Q2Substituting the given values, we get 13=Q2200−Q2 or 200−Q2=3Q2or 4Q2=200 or Q2=2004J=50J∴ W=Q1−Q2=200J−50J=150J

A refrigerator with coefficient of performance 13releases 200 J of heat to a hot reservoir. Then the work done on the working substance is

The coefficient of performance of a refrigerator is given by β=Q2W=Q2Q1−Q2Substituting the given values, we get 13=Q2200−Q2 or 200−Q2=3Q2or 4Q2=200 or Q2=2004J=50J∴ W=Q1−Q2=200J−50J=150J

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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

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$\frac{100}{3} J$

$100 J$

$\frac{200}{3} J$

$150 J$

answer is D.

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Detailed Solution

The coefficient of performance of a refrigerator is given by $β = \frac{Q_{2}}{W} = \frac{Q_{2}}{Q_{1} - Q_{2}}$

Substituting the given values, we get

$\begin{array}{l} \frac{1}{3} = \frac{Q_{2}}{200 - Q_{2}} or 200 - Q_{2} = 3 Q_{2} \\ or 4 Q_{2} = 200 or Q_{2} = \frac{200}{4} J = 50 J \\ ∴ W = Q_{1} - Q_{2} = 200 J - 50 J = 150 J \end{array}$