# Problem No. 8 (15 points)

50 g of ice, taken at a temperature of $t_{\pi}=-10{ }^{\circ} \mathrm{C}$, was mixed with water taken at a temperature of $t_{B}=10{ }^{\circ} \mathrm{C}$. It is known that the final temperature in the container is $t=0{ }^{\circ} \mathrm{C}$. Determine the mass of the added water. The specific heat capacity of water $c_{B}=4200$ J $/$ kg $\cdot{ }^{\circ} \mathrm{C}$, the specific heat capacity of ice $c_{л}=2100$ J $/$ kg $\cdot{ }^{\circ} \mathrm{C}$, the latent heat of fusion of ice $\lambda=3,3 \cdot 10^{2}$ J/kg

Answer: $0,0028 \kappa g \leq m_{B} \leq 0,418$ kg
# Solution and Evaluation Criteria:

Two extreme situations possible in this problem:

First situation - only ice remains in the vessel at a temperature of $t=0{ }^{\circ} \mathrm{C}$

(2 points)

The heat balance equation in this case: $c_{L} m_{L} 10=c_{B} m_{B} 10+\lambda m_{B}$.

We get: $m_{B}=\frac{2100 \cdot 0.05 \cdot 10}{4200 \cdot 10+330000} \approx 0.0028$ kg

Second situation - only water remains in the vessel at a temperature of $t=0{ }^{\circ} \mathrm{C}$

The heat balance equation in this case: $c_{L} m_{L} 10+\lambda m_{L}=c_{B} m_{B} 10$.

We get: $m_{B}=\frac{2100 \cdot 0.05 \cdot 10+330000 \cdot 0.05}{4200 \cdot 10} \approx 0.418$ kg

The mass of the added water $0.0028 \text{ kg} \leq m_{B} \leq 0.418 \text{ kg}$