One kilogram of ice at $0^{\circ}C$ is mixed with one kilogram of water at $80^{\circ}C$. The final temperature of the mixture is ........ $^{\circ}C$. (Take: specific heat of water $= 4200 \ J \ kg^{-1} \ K^{-1}$,latent heat of ice $= 336 \ kJ \ kg^{-1}$)

The heat required to raise the temperature of $50 \, g$ of copper by $10^\circ C$ is given to $10 \, g$ of water. What is the rise in temperature of the water in $^\circ C$? (Specific heat of copper $= 420 \, J \cdot kg^{-1} \cdot ^\circ C^{-1}$,Specific heat of water $= 4200 \, J \cdot kg^{-1} \cdot ^\circ C^{-1}$)

$15 \text{ g}$ of ice at $0^{\circ}\text{C}$ is added to a vessel containing water at $40^{\circ}\text{C}$. The mass of water and water equivalent of the vessel is $60 \text{ g}$. Assuming that negligible heat is taken from the surroundings, the final temperature of the mixture will be $[L_{\text{ice}} = 80 \text{ cal/g}, S_{\text{water}} = 1 \text{ cal/g}]$ (in $^{\circ}\text{C}$)

Steam at $100\,^{\circ}C$ is passed into $22\,g$ of water at $20\,^{\circ}C$. The mass of water that will be present when the water acquires a temperature of $90\,^{\circ}C$ (Latent heat of steam is $540\,cal/g$) is ......... $g$.

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

$10 \, g$ 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 $^{\circ}C$?