The specific heat of water $=4200\, J\, kg ^{-1}\, K ^{-1}$ and the latent heat of ice $=3.4 \times 10^{5}\, J\, kg ^{-1}.$ $100$ grams of ice at $0^{\circ} C$ is placed in $200\, g$ of water at $25^{\circ} C$. The amount of ice that will melt as the temperature of water reaches $0^{\circ} C$ is close to (in $grams$) 

$100\,g$ of water is supercooled to $-\,10\,^oC$. At this point, due to some disturbance mechanised or otherwise some of it suddenly freezes to ice. What will be the temperature of the resultant mixture and how much mass would freeze ? $[S_W = 1\,cal\,g^{-1}\,^oC^{-1}$ and ${L^W}_{{\text{fussion}}}$ $= 80\,cal\,g^{-1}]$

$2\  kg$ ice at $-20^o\ C$ is mixed with $5\  kg$ water at $20^o\ C$. Then final amount of water in the mixture would be ; Given specific heat of ice $= 0.5\ cal/g^o\ C$, specific heat of water $= 1\  cal/g^o\ C$, Latent heat of fusion of ice $= 80\  cal/g$ …….. $kg$

$1 \,kg$ of ice at $-20^{\circ} C$ is mixed with $2 \,kg$ of water at $90^{\circ} C$. Assuming that there is no loss of energy to the environment, the final temperature of the mixture is ………… $^{\circ} C$ (Assume, latent heat of ice $=334.4 \,kJ / kg$, specific heat of water and ice are $4.18 \,kJ kg ^{-1} K ^{-1}$ and $2.09 \,kJ kg ^{-1}- K ^{-1}$, respectively.)

A $2\,kg$ copper block is heated to $500^o\,C$ and then it is placed on a large block of ice  at $0^o\,C$. If the specific heat capacity of copper is $400\, J/kg/ ^o\,C$ and latent heat of  fusion of water is $3.5 \times 10^5\, J/kg$, the amount of ice, that can melt is :-