One kilogram of ice at O°C is mixed with one kilogram of water at 80°C. The final temperature of the mixture is (Take: specific heat of water = 4200 J ${\mathrm{kg}}^{-1} {\mathrm{K}}^{-1}$, latent heat of ice = 80 kJ/${\mathrm{kg}}^{-1}$)

Heat rejected by water to reduce its its temperature to $0^0$=1000$\times 1\times (80-0)=80000 cal$

Heat absorbed by ice to convert itself into water at $0^0=1000\times 80 cal=80000 cal$

So finally the mixture contains only water at $0^0$c .