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 (in kg) remaining in the container. It is given that the specific heats of water and ice are $1 Kcal/kg / °C$ and  $0.\text{5K}cal/kg/°C$ while the latent heat of fusion of ice is  $\text{8}0Kcal/kg$

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(20\right)+m_i\text{'}\times L=m_w{c}_w\left[20\right]$

$\Rightarrow 2\times 0.5\left(20\right)+m_i\text{'}\times 80=5\times 1\times 20\Rightarrow m_i\text{'}=1kg$
So final mass of water = Initial mass of water + Mass of ice that melts $=\text{ 5 }+\text{ 1}=\text{ 6}kg.$