8. An aluminum cube with edge length $l=10$ cm was heated to a temperature of $t_{1}=100^{\circ} \mathrm{C}$. After that, it was placed on ice, which has a temperature of $t_{2}=0^{\circ} \mathrm{C}$. Determine the maximum depth to which the cube can sink. The specific heat capacity of aluminum $c_{a}=900$ J $/$ kg ${ }^{\circ} \mathrm{C}$, the latent heat of fusion of ice $\lambda=3.3 \cdot 10^{5}$ J $/$ kg, the density of aluminum $\rho_{a}=2700$ kg $/ \mathrm{m}^{3}$, and the density of ice $\rho_{n}=900$ kg $/ \mathrm{m}^{3}$.

## (10 points)
Answer: 0.081 m

Solution. The mass of the cube: $m_{a}=\rho_{a} V=\rho_{a} l^{3}$ (2 points). The thermal balance equation for the cube and ice: $c_{a} m_{a} \Delta T=\lambda m_{n}$ (2 points). The mass of the melted ice: $m_{n}=\rho_{a} V_{n}=\rho_{s} S h=\rho_{a} l^{2} h$ (3 points), where $h$ is the maximum depth to which the cube sinks. As a result, we get: $h=\frac{C_{a} \rho_{a}^{3} \Delta T}{\lambda \rho_{a} l^{2}}=\frac{c_{a} \rho_{a} l \Delta T}{\lambda \rho_{a}}=\frac{900 \cdot 2700 \cdot 0.1 \cdot 100}{3.3 \cdot 10^{5} \cdot 900} \approx 0.081$, (3 points).

## Final Stage

 2017-2018 academic year
## Tasks, answers, and evaluation criteria 8th grade Variant II $\begin{aligned} & \text { half } \text { fəłGr } \\ & \text { r }\end{aligned}$