What is the smallest possible time necessary to freeze $2 \mathrm{~kg}$ of water at $0^{\circ} \mathrm{C}$ if a 50 watt motor is available and the outside air (hot reservoir) is at $27^{\circ} \mathrm{C}$ ?
When $2 \mathrm{~kg}$ of water at $0^{\circ} \mathrm{C}$ becomes ice, the heat released is

$$
Q_{2}=1.44 \times 2 \times 10^{3} / 18=1.6 \times 10^{2} \mathrm{kcal} .
$$

The highest efficiency of the motor is

$$
\varepsilon=\frac{T_{2}}{T_{1}-T_{2}}=\frac{Q_{2}}{W_{\min }} .
$$

Thus,

$$
W_{\min }=Q_{2} \frac{T_{1}-T_{2}}{T_{2}} .
$$

If we use the motor of $P=50 \mathrm{~W}$, the smallest necessary time is

$$
\tau=\frac{W_{\min }}{P}=\frac{Q_{2}}{P} \cdot \frac{T_{1}-T_{2}}{T_{2}} .
$$

With $T_{1}=300 \mathrm{~K}, T_{2}=273 \mathrm{~K}$, we find

$$
\tau=1.3 \times 10^{3} \mathrm{~s} .
$$