# Problem No. 6 (10 points)

A pot was filled with $3 \pi$ liters of water at a temperature of $t=0{ }^{\circ} C$, and it was brought to a boil in 12 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/m$^{3}$.

Answer: 2.1 kg
# Solution and evaluation criteria:

Mass of the initial water: $m_{B}=\rho V=3$ kg

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

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

We get:

$\frac{c_{B} m_{B} \Delta T}{t_{1}}=\frac{\lambda m_{J}+c_{B} m_{J} \Delta T}{t_{2}}$. From which $m_{J}=\frac{c_{B} m_{B} \Delta T t_{2}}{t_{1}\left(\lambda+c_{B} \Delta T\right)}=\frac{4200 \cdot 3 \cdot 100 \cdot 15}{12(330000+4200 \cdot 100)}=2.1$ kg