$2$ kg of ice at $-20$°C is mixed with $5$ kg of water at $20$°C in an insulating vessel having a negligible heat capacity. Calculate the final mass of water remaining in the container. It is given that the specific heats of water and ice are $1$ kcal/kg per °C and $0.5$ kcal/kg/°C while the latent heat of fusion of ice is $80$ kcal/kg:

 Initially ice will absorb heat to raise it's temperature to $0^{\circ }C$ then it's melting takes place
If $m$= Initial mass of ice, $m\text{'}$ = Mass of ice that melts and $m_w$ = Initial mass of water, 

By Law of mixture, Heat gain by ice = Heat loss by water

$\begin{array}{l}m\times c\times \left(20\right)+m^{\text{'}}\times L=m_w{c}_w\left[20\right]\\ 2\times 0.5\left(20\right)+m^{\text{'}}\times 80=5\times 1\times 20\\ m^{\text{'}}=1 \mathrm{kg}\end{array}$

So, final mass of water = Initial mass of water + Mass of ice that melts $=5+1=6kg$