A refrigerator with coefficient of performance $\frac{1}{3}$ releases $200 \mathrm{~J}$ of heat to a hot…

A refrigerator with coefficient of performance $\frac{1}{3}$ releases $200 \mathrm{~J}$ of heat to a hot reservoir. Then the work done on the working substance is
  1. $\frac{100}{3} \mathrm{~J}$
  2. $100 \mathrm{~J}$
  3. $\frac{200}{3} \mathrm{~J}$
  4. $150 \mathrm{~J}$

Solution

The coefficient of performance of a refrigerator is given by
$$
\alpha=\frac{\mathrm{Q}_{2}}{\mathrm{~W}}=\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}-\mathrm{Q}_{2}}
$$
Substituting the given values, we get
$$
\begin{aligned}
& \frac{1}{3}=\frac{\mathrm{Q}_{2}}{200-\mathrm{Q}_{2}} \\
\Rightarrow & 200-\mathrm{Q}_{2}=3 \mathrm{Q}_{2} \Rightarrow 4 \mathrm{Q}_{2}=200 \\
\text { or } & \mathrm{Q}_{2}=\frac{200}{4} \mathrm{~J}=50 \mathrm{~J} \\
\therefore \quad \mathrm{W}=\mathrm{Q}_{1}-\mathrm{Q}_{2}=200 \mathrm{~J}-50 \mathrm{~J}=150 \mathrm{~J}
\end{aligned}
$$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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