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

β=Q2W=Q2Q1−Q213=Q2200−Q2 , 200−Q2=3Q2Q2=50J , W=Q1−Q2=150J

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

β=Q2W=Q2Q1−Q213=Q2200−Q2 , 200−Q2=3Q2Q2=50J , W=Q1−Q2=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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$\frac{200}{3} J$

$100 J$

$150 J$

$\frac{100}{3} J$

answer is D.

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Detailed Solution

$β = \frac{Q_{2}}{W} = \frac{Q_{2}}{Q_{1} - Q_{2}}$
$\frac{1}{3} = \frac{Q_{2}}{200 - Q_{2}}, 200 - Q_{2} = 3 Q_{2}$
$Q_{2} = 50 J, W = Q_{1} - Q_{2} = 150 J$

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

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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