# Problem No. 6 (10 points)

A pot was filled with $2 \pi$ liters of water, taken at a temperature of $t=0{ }^{\circ} C$, and brought to a boil in 10 minutes. After that, without removing the pot from the stove, ice at a temperature of $t=0{ }^{\circ} \mathrm{C}$ was added. The water began to boil again only after 15 minutes. Determine the mass of the added ice. The specific heat capacity of water $c_{B}=4200$ J $/$ kg $\cdot{ }^{\circ} \mathrm{C}$, the latent heat of fusion of ice $\lambda=3.3 \cdot 10^{5}$ J $/$ kg, and the density of water $\rho=1000$ kg $/ \mathbf{m}^{3}$.

## Answer: 1.68 kg

#
# Solution and Evaluation Criteria:

Mass of the initial water: $m_{B}=\rho V=2$ kg

Power of the stove in the first case: $P=\frac{c_{B} m_{B} \Delta T}{t_{1}}$.

And in the second case: $P=\frac{\lambda m_{J}+c_{B} m_{\pi} \Delta T}{t_{2}}$.

We obtain:

$$
\frac{c_{B} m_{B} \Delta T}{t_{1}}=\frac{\lambda m_{J}+c_{B} m_{J} \Delta T}{t_{2}} . \text { From which } m_{J}=\frac{c_{B} m_{B} \Delta T t_{2}}{t_{1}\left(\lambda+c_{B} \Delta T\right)}=\frac{4200 \cdot 2 \cdot 100 \cdot 15}{10(330000+4200 \cdot 100)}=1.68 \text { kg (4 points) }
$$