Given, mass of water $m_{w}=300\, g$
Temperature of water, $T_{w}=25^{\circ} C$
Mass of ice, $m_{i}=100\, g$
$T_{i}=0^{\circ} C$
Heat required to convert the ice at $0^{\circ} C$ to water at $0^{\circ} C$ $Q_{1}=m_{t} L_{i}=100 \times 80\left[L_{i}=80\, cal / g \right]$
$=8000\, cal \ldots$. (i)
Heat given by water from $25^{\circ} C$ water to $0^{\circ} C$ water $Q_{2}=m_{w} c \Delta T=300 \times 1 \times(25-0)$
$\Rightarrow =7500\, cal \ldots \ldots .$ (ii)
From Eqs. (i) and (ii), we observed that
$\$ Q_{-}\{2\}$ Hence, total ice will not be melt,
so final temperature of mixture will be $0^{\circ} C$ with some unmelted ice in mixture.