7. (15 points) In a cylindrical vessel, ice has frozen at the bottom. Its temperature is $0^{\circ} \mathrm{C}$. Above it, water is poured at the same temperature. The vessel is brought into a warm room. The dependence of the water level in the vessel on time is shown in the graph. Determine the initial masses of ice and water. The area of the base of the vessel $S=15 \mathrm{~cm}^{2}$, the density of water $\rho_{6}=1 \mathrm{~g} /$ cm $^{3}$, the density of ice $\rho_{n}=0.9 \mathrm{~g} / \mathrm{cm}^{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9c0cded10981f14efe49g-09.jpg?height=542&width=1037&top_left_y=1999&top_left_x=538)
Answer: mass of ice -675 g, mass of water - 1050 g

Solution. The change in the volume of water in the vessel $\Delta V=S \Delta h=15 \cdot 5=75 \mathrm{~cm}^{3}$. (2 points)

This change is the difference between the volumes of the initial ice and the water into which it has transformed. $\Delta V=V_{I}-V_{W}=\frac{m_{I}}{\rho_{I}}-\frac{m_{I}}{\rho_{W}}$.

We obtain that the mass of the initial ice: $m_{I}=\frac{\Delta V \cdot \rho_{I} \cdot \rho_{W}}{\rho_{W}-\rho_{I}}=6752$.

The final volume of water in the vessel: $V_{F}=115 \cdot 15=1725 \mathrm{~cm}^{3}$.

(2 points)

Therefore, the initial mass of water $m_{W}=\rho_{W} V_{F}-m_{I}=10502$.

(3 points)