5. (15 points) A glass is filled to the brim with saltwater. On the surface, there is a piece of fresh ice with a mass of $m=100$ g. What volume $\Delta V$ of water will spill out of the glass by the time the ice melts? Neglect surface tension. The density of fresh ice $\rho_{\mathrm{n}}=0.9 \text{ g/cm}^{3}$, the density of salt ice $\rho_{\mathrm{c}}=0.95 \text{ g/cm}^{3}$, the density of fresh water $\rho_{\text {fw }}=1 \text{ g/cm}^{3}$, the density of salt water $\rho_{\text {sw }}=1.03 \text{ g/cm}^{3}$. Neglect the change in total volume when mixing two liquids.


Answer: $\approx 5.26 \mathrm{~cm}^{3}$.

Solution. The condition for the floating of fresh ice in salt water:

$\rho_{\text {sw }} g V_{\text {sub }}=m g=\rho_{\text {f }} g V_{1}$,

where $V_{1}$ is the entire volume of fresh ice, $\rho_{\text {sw }}$ is the density of salt water.

The condition for the floating of the same mass of salt ice in salt water:

$\rho_{\text {sw }} g V_{\text {sub }}=m g=\rho_{\text {s}} g V_{2}$,
where $V_{2}$ is the entire volume of salt ice. When the salt ice melts, it will occupy exactly the volume $V_{\text {sub }}$.

The excess volume of ice that will end up outside the glass after melting:

$$
V_{1}-V_{2}=\frac{m}{\rho_{\text {f}}}-\frac{m}{\rho_{\text {s}}}
$$

Considering that it will melt, the volume of water that will spill out:

$\Delta V=\left(\frac{m}{\rho_{\text {f}}}-\frac{m}{\rho_{\text {s}}}\right) \cdot \frac{\rho_{\text {f}}}{\rho_{\text {fw }}} \approx 5.26 \mathrm{~cm}^{3}$.