Problem 5. An electric kettle heats water from room temperature $T_{0}=20^{\circ} \mathrm{C}$ to $T_{m}=100^{\circ} \mathrm{C}$ in $t=10$ minutes. How long will it take $t_{1}$ for all the water to boil away if the kettle is not turned off and the automatic shut-off system is faulty? The specific heat capacity of water $c=4200$ J/kg $\cdot$ K. The specific latent heat of vaporization of water $L=2.3$ MJ/kg. Round the answer to the nearest whole number of minutes.
Solution. The power $P$ of the kettle is fixed and equal to $P=Q / t$. From the heat transfer law $Q=c m\left(T_{m}-T_{0}\right)$ we get $P t=c m\left(T_{m}-T_{0}\right)$.

To evaporate the water, the amount of heat required is $Q_{1}=L m \Rightarrow P t_{1}=L m$.

By comparing these relations, we obtain $\frac{t_{1}}{t}=\frac{L m}{c m\left(T_{m}-T_{0}\right)}=\frac{L}{c\left(T_{m}-T_{0}\right)} \Rightarrow$

$$
\frac{t_{1}}{t}=\frac{2.3 \cdot 10^{6}}{4200 \cdot 80}=\frac{2300}{336} \approx 6.845
$$

Thus, the water will boil away in $t_{1}=6.845 \cdot t=68.45$ minutes.

Answer: 68 minutes.