$10 \ kg$ of ice at $-10^{\circ}C$ is added to $100 \ kg$ of water to lower its temperature from $25^{\circ}C.$ Consider no heat exchange to surroundings. The decrement to the temperature of water is . . . . . . $^{\circ}C.$ (specific heat of ice $= 2100 \ J/kg.^{\circ}C$,specific heat of water $= 4200 \ J/kg.^{\circ}C$,latent heat of fusion of ice $= 3.36 \times 10^{5} \ J/kg$)

$1 \ g$ of ice at $0^\circ C$ is mixed with $1 \ g$ of water at $100^\circ C$. The resulting temperature will be .......... $^\circ C$.

When $100 \ g$ of boiling water at $100^{\circ} C$ is added into a calorimeter containing $300 \ g$ of cold water at $10^{\circ} C$,the temperature of the mixture becomes $20^{\circ} C$. Then,a metallic block of mass $1 \ kg$ at $10^{\circ} C$ is dipped into the mixture in the calorimeter. After reaching thermal equilibrium,the final temperature becomes $19^{\circ} C$. What is the specific heat of the metal in $C$.$G$.$S$. units?

$A$ piece of metal of mass $5 \ kg$ and temperature $-50 \ ^\circ C$ is dropped in $20 \ kg$ water at $52 \ ^\circ C$ temperature. In thermal equilibrium,the temperature of water decreases by $2 \ ^\circ C$. Find the specific heat of the metal in $Cal/g \ ^\circ C$.

When $M_1$ gram of ice at $-10\,^{\circ}C$ (specific heat $= 0.5\, cal\, g^{-1}\,^{\circ}C^{-1}$) is added to $M_2$ gram of water at $50\,^{\circ}C$,finally no ice is left and the water is at $0\,^{\circ}C$. The value of latent heat of ice,in $cal\, g^{-1}$ is

$A$ piece of metal weighing $100 \,g$ is heated to $80^{\circ} C$ and dropped into $1 \,kg$ of cold water in an insulated container at $15^{\circ} C$. If the final temperature of the water in the container is $15.69^{\circ} C$,the specific heat of the metal in $J / g^{\circ} C$ is: