$10 \ kg$ 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 $= 3.36 \times 10^{5} \ J/kg$)

$A$ sphere of $0.047 \; kg$ aluminium is placed for a sufficient time in a vessel containing boiling water,so that the sphere is at $100 \; ^{\circ}C$. It is then immediately transferred to a $0.14 \; kg$ copper calorimeter containing $0.25 \; kg$ water at $20 \; ^{\circ}C$. The temperature of the water rises and attains a steady state at $23 \; ^{\circ}C$. Calculate the specific heat capacity of aluminium in $kJ \; kg^{-1} K^{-1}$.

$A$ piece of metal of $850 \text{ K}$ is dropped into $1 \text{ kg}$ of water at $300 \text{ K}$. If the equilibrium temperature of the mixture is $350 \text{ K}$, then the heat capacity of the metal expressed in $\text{J/K}$ is (Specific heat of water = $4200 \text{ J/kg} \cdot \text{K}$)

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?

$1\, g$ of steam at $100^{\circ}C$ melts ........ $g$ of ice at $0^{\circ}C?$ (Latent heat of ice $= 80\, cal/g$ and latent heat of steam $= 540\, cal/g$)

The specific heat of water is $4200 \, J \, kg^{-1} \, K^{-1}$ and the latent heat of ice is $3.4 \times 10^{5} \, J \, kg^{-1}$. $100 \, g$ 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 the water reaches $0^{\circ} C$ is close to (in grams):