Question

if a refrigerator discards $q_h = 80 j$ of heat per cycle and its coefficient of performance is $k = 6.0$, what are (a) the quantity of heat it removes per cycle from a cold reservoir and (b) the amount of work per cycle required for its operation?
a) $q_c=?$ $k=\frac{q_c}{w}=\frac{q_c}{q_h - q_c}$ $q_h=q_c + w$ $kq_h - kq_c=q_c$ $q_c=\frac{kq_h}{1 + k}=\frac{(6.0\times80)}{1 + 6.0} j=?$ $japprox?$ $j$
b) $w =?$ $k=\frac{q_c}{w}$ $w=\frac{q_c}{k}=\frac{?}{6.0}j$ $q_c,q_h
ightarrow heat$ $w
ightarrow work done.$ $q
ightarrow heat.$

Explanation:

Step1: Recall coefficient - of - performance formula

The coefficient of performance of a refrigerator is $K=\frac{Q_c}{W}$, and also $Q_h = Q_c+W$. From $K=\frac{Q_c}{W}$ and $Q_h = Q_c+W$, we can express $W=\frac{Q_c}{K}$ and substitute into $Q_h = Q_c+W$ to get $Q_h=Q_c+\frac{Q_c}{K}=Q_c(1 + \frac{1}{K})$. Then $Q_c=\frac{KQ_h}{1 + K}$.

Step2: Calculate $Q_c$

Given $K = 6.0$ and $Q_h=80\ J$, we substitute into the formula $Q_c=\frac{KQ_h}{1 + K}$. So $Q_c=\frac{6.0\times80}{1 + 6.0}=\frac{480}{7}\approx68.6\ J$.

Step3: Calculate $W$

Since $K=\frac{Q_c}{W}$, then $W=\frac{Q_c}{K}$. Substituting $Q_c=\frac{480}{7}\ J$ and $K = 6.0$ into the formula, we have $W=\frac{\frac{480}{7}}{6.0}=\frac{480}{7\times6}=\frac{80}{7}\approx11.4\ J$.

Answer:

(a) $Q_c\approx68.6\ J$
(b) $W\approx11.4\ J$