• # question_answer 2 kg of ice at $\text{ }20{}^\circ C$ is mixed with 5kg of water at $20{}^\circ C$ 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 $1\text{ }kcal/kg\text{ }/{}^\circ C$ and $0.5\text{ }kcal/kg/{}^\circ C$ while the latent heat of fusion of ice is 80 kcal/kg A) 7kg                   B) 6kg     C) 4kg                   D) 2kg

 Initially ice will absorb heat to raise its temperature to $0{}^\circ C$ then it's melting takes place If ${{m}_{1}}$= Initial mass of ice, ${{m}_{2}}$' = Mass of ice that melts and ${{m}_{W}}$= Initial mass of water By Law of mixture Heat gained by ice = Heat lost by water $\Rightarrow {{m}_{1}}\times \left( 20 \right)+{{m}_{1}}'\times L={{m}_{W}}{{c}_{W}}\left[ 20 \right]$ $\Rightarrow 2\times 0.5\left( 20 \right)+{{m}_{1}}'\times 80=5\times I\times 20$ $\Rightarrow {{\text{m}}_{\text{1}}}\text{ }\!\!'\!\!\text{ =1kg}$ So, final mass of water = Initial mass of water + Mass of Ice that melts = 5 +1 = 6kg