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 \( m = 50 \) 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 \( \rho_{\mathrm{n}} = 0.9 \mathrm{r} / \mathrm{cm}^{3} \), the density of salt ice \( \rho_{\mathrm{c}} = 0.95 \mathrm{r} / \mathrm{cm}^{3} \), the density of fresh water \( \rho_{\text {fw }} = 1 \mathrm{r} / \mathrm{cm}^{3} \), and the density of saltwater \( \rho_{\mathrm{sw}} = 1.03 \mathrm{r} / \mathrm{cm}^{3} \). Neglect the change in total volume when mixing two liquids.
Answer: $\approx 2.63 \mathrm{~cm}^{3}$.

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

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

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 salt ice of the same mass in salt water:

$\rho_{\text {sw }} g V_{\text {subm }}=m g=\rho_{\text {s}} g V_{2},$,

where $V_{2}$ is the entire volume of salt ice.

When the salt ice melts, it occupies exactly the volume $V_{\text {subm. }}$. 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 overflows:

$\Delta V=\left(\frac{m}{\rho_{\text {f }}}-\frac{m}{\rho_{\text {s}}}\right) \cdot \frac{\rho_{\text {f }}}{\rho_{\text {fw }}} \approx 2.63 \mathrm{~cm}^{3}$.