8. A copper cube with edge length $l=5 \mathrm{~cm}$ 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 copper $c_{\text {s }}=400$ J $/$ kg $\cdot{ }^{\circ} \mathrm{C}$, the latent heat of fusion of ice $\lambda=3.3 \cdot 10^{5}$ J $/$ kg, the density of copper $\rho_{м}=8900$ kg $/$ m $^{3}$, and the density of ice $\rho_{n}=900$ kg $/ \mathrm{m}^{3}$. (10 points)
Answer: $0.06 . m$

Solution. The mass of the cube: $m_{n}=\rho_{w} V=\rho_{n} l^{3}$ (2 points). The thermal balance equation for the cube and ice: $c_{m} m_{»} \Delta T=\lambda m_{n}$ (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. We obtain: $h=\frac{c_{u} \rho_{l} l^{3} \Delta T}{\lambda \rho_{n} l^{2}}=\frac{c_{m} \rho_{\rho_{l}} l \Delta T}{\lambda \rho_{s_{n}}}=\frac{400 \cdot 8900 \cdot 0.05 \cdot 100}{3.3 \cdot 10^{5} \cdot 900} \approx 0.06, \mu$ (3 points).