# Problem No. 8 (15 points)

$100 \, \text{g} \, p$ of ice, taken at a temperature of $t_{n}=-5^{\circ} \text{C}$, was mixed with water taken at a temperature of $t_{B}=10^{\circ} \text{C}$. It is known that the final temperature in the container is $t=0^{\circ} \text{C}$. Determine the mass of the added water. The specific heat capacity of water $c_{s}=4200 \, \text{J/kg} \cdot{ }^{\circ} \text{C}$, the specific heat capacity of ice $c_{n}=2100 \, \text{J/kg} \cdot{ }^{\circ} \text{C}$, and the latent heat of fusion of ice $\lambda=3.3 \cdot 10^{5} \, \text{J/kg}$.

Answer: $0.0028 \, \text{kg} \leq m_{B} \leq 0.811 \, \text{kg}$
# Solution and Evaluation Criteria:

Two extreme situations possible in this problem:

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

The heat balance equation in this case: $c_{J} m_{\pi} 5=c_{B} m_{B} 10+\lambda m_{B}$.

We get: $m_{B}=\frac{2100 \cdot 0.1 \cdot 5}{4200 \cdot 10+330000} \approx 0.0028$ kg

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

The heat balance equation in this case: $c_{\pi} m_{\pi} 5+\lambda m_{J}=c_{B} m_{B} 10$.

We get: $m_{B}=\frac{2100 \cdot 0.1 \cdot 5+330000 \cdot 0.1}{4200 \cdot 10} \approx 0.811$ kg

The mass of the added water 0.0028 kg $\leq m_{b} \leq 0.811$ kg

Multidisciplinary Engineering Olympiad "Star" in Natural Sciences

## Final Stage

2016-2017 academic year

## Problems, Answers, and Evaluation Criteria

![](https://cdn.mathpix.com/cropped/2024_05_06_8b91ba68ee8c87ac0dd6g-5.jpg?height=137&width=531&top_left_y=640&top_left_x=928)