A refrigerator with coefficient of performance (1/3) releases 200 J of heat to a hot reservoir. Then the work done on the working substance is

Question

A refrigerator with coefficient of performance (1/3) releases 200 J of heat to a hot reservoir. Then the work done on the working substance is (A) (100/3) J (B) 100 J (C) (200/3) J (D) 150 J

Tardigrade Admin · Accepted Answer

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

\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

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

$\frac{100}{3} J$

$100 J$

$\frac{200}{3} J$

$150 J$

Q. A refrigerator with coefficient of performance 31​ releases 200J of heat to a hot reservoir. Then the work done on the working substance is

3100​J

100J

3200​J

150J

The coefficient of performance of a refrigerator is given by β=WQ2​​=Q1​−Q2​Q2​​ Substituting the given values, we get 31​=200−Q2​Q2​​ or 200−Q2​=3Q2​ or 4Q2​=200 or Q2​ =4200​J=50J ∴W=Q1​−Q2​ =200J−50J =150J

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

$\frac{100}{3} J$

$100 J$

$\frac{200}{3} J$

$150 J$