2 kg of ice at – 10°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 $0^{o}C$ then it's melting takes place 
If $m_i$ = Initial mass of ice, $m_i\text{'}$ = Mass of ice that melts and $m_W$ = Initial mass of water 
By Law of mixture Heat gained by ice = Heat lost by water  $\Rightarrow m_i\times c\times \left(10\right)+m_i\text{'}\times L=m_W{c}_W\left[20\right]$

$\Rightarrow 2\times 0.5\left(10\right)+m_i\text{'}\times 80=5\times 1\times 20\Rightarrow m_i\text{'}=\frac{9}{8}kg$

So final mass of water = Initial mass of water + Mass of ice that melts  $=5+\frac{9}{8}=\frac{49}{8}kg$