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

$5 \ g$ of ice at $-30^{\circ} C$ and $20 \ g$ of water at $35^{\circ} C$ are mixed together in a calorimeter. The final temperature of the mixture is (Neglect heat capacity of the calorimeter,specific heat capacity of ice $= 0.5 \ cal \ g^{-1} {}^{\circ} C^{-1}$,latent heat of fusion of ice $= 80 \ cal \ g^{-1}$,and specific heat capacity of water $= 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$). (in $^{\circ} C$)

Steam at $100^{\circ} C$ is passed into $1 \ kg$ of water contained in a calorimeter of water equivalent $0.2 \ kg$ at $9^{\circ} C$ until the temperature of the calorimeter and water in it increases to $90^{\circ} C$. The mass of steam condensed in $kg$ is nearly (specific heat of water $= 1 \ cal/g^{\circ} C$, latent heat of vaporisation $= 540 \ cal/g$)

$300 \, g$ of water at $25^{\circ}C$ is added to $100 \, g$ of ice at $0^{\circ}C$. The final temperature of the mixture is ........ $^{\circ}C$.

$20 \, g$ of boiling water is poured into an ice-cold brass vessel (specific heat $0.1 \, cal/g-^{\circ}C$) of mass $100 \, g$. The resulting temperature is ........ $^{\circ}C$.

How many grams of ice at $0 \, ^\circ \text{C}$ will be melted by $1 \, \text{g}$ of steam at $100 \, ^\circ \text{C}$? (Latent heat of fusion of ice $L = 80 \, \text{cal/g}$ and latent heat of vaporization of water $L' = 540 \, \text{cal/g}$)