Answer
a) $Q_{\text {cooling }}=36.2 W$
b) $Q_{\text {heating }}=49.7W$
c) $\dot{W}_{\text {in,cooling }} =181 W$
$\dot{W}_{\text {in,heating }} =41.4 W$
Work Step by Step
(a) The average cooling rate of the refrigerator is simply the rate of decrease of the energy content of the canned drinks,$$
\begin{aligned}
& Q_{\text {cooling }}=m c_p \Delta T=(0.771 \mathrm{lbm})\left(1.0 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)(78-38)^{\circ} \mathrm{F}=30.84\ \mathrm{Btu} \\
& \dot{Q}_{\text {cooling }}=\frac{Q_{\text {cooling }}}{\Delta t}=\frac{30.84 \mathrm{Btu}}{15 \times 60 \mathrm{~s}}\left(\frac{1055 \mathrm{~J}}{1 \mathrm{Btu}}\right)=36.2 \mathrm{~W}
\end{aligned}
$$ (b) The average heating rate of the refrigerator is simply the rate of increase of the energy content of the canned drinks, $$
\begin{aligned}
& Q_{\text {heating }}=m c_p \Delta T=(0.771 \mathrm{lbm})\left(1.0 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)(130-75)^{\circ} \mathrm{F}=42.4\ \mathrm{Btu} \\
& \dot{Q}_{\text {heating }}=\frac{Q_{\text {bening }}}{\Delta t}=\frac{42.4 \mathrm{Btu}}{15 \times 60 \mathrm{~s}}\left(\frac{1055 \mathrm{~J}}{1 \mathrm{Btu}}\right)=49.7 \mathrm{~W}
\end{aligned}
$$ (c) The electric power drawn from the car battery during cooling and heating is $$
\begin{aligned}
\dot{W}_{\text {in,cooling }} & =\frac{\dot{Q}_{\text {cooling }}}{\mathrm{COP}_{\text {cooling }}}=\frac{36.2 \mathrm{~W}}{0.2}=181 \mathrm{~W} \\
\mathrm{COP}_{\text {heating }} & =\mathrm{COP}_{\text {cooling }}+1=0.2+1=1.2 \\
\dot{W}_{\text {in,heating }} & =\frac{\dot{Q}_{\text {heating }}}{\mathrm{COP}_{\text {heating }}}=\frac{49.7 \mathrm{~W}}{1.2}\\
&=41.4 \mathrm{~W}
\end{aligned}
$$
