$37 \ g$ of ice at $0^{\circ} C$ temperature is mixed with $74 \ g$ of water at $70^{\circ} C$ temperature. The resultant temperature is (Specific heat capacity of water $= 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$ and latent heat of fusion of ice $= 80 \ cal \ g^{-1}$) (in $^{\circ} C$)

When $50 \ g$ of water at $10^{\circ} C$ is mixed with $50 \ g$ of water at $100^{\circ} C$,the resultant temperature is: (in $^{\circ} C$)

$10\,g$ of ice at $-20\,^{\circ}C$ is dropped into a calorimeter containing $10\,g$ of water at $10\,^{\circ}C$. The specific heat of water is twice that of ice. When equilibrium is reached,the calorimeter will contain

$10 \, g$ of ice at $0^{\circ}C$ is mixed with $100 \, g$ of water at $50^{\circ}C$. What is the resultant temperature of the mixture in $^{\circ}C$?

The time taken for a calorimeter containing $75 \ g$ of water at $62^{\circ} C$ to cool to $58^{\circ} C$ is $9 \ minutes$. When the calorimeter contains $105 \ g$ of water,it takes $12 \ minutes$ to cool from $62^{\circ} C$ to $58^{\circ} C$. The water equivalent of the calorimeter is $.........$ (in $g$)

$1 \,kg$ 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.)