The temperature of water of mass $100 \ g$ is raised from $24^{\circ} C$ to $90^{\circ} C$ by adding steam to it. The mass of the steam added is (Latent heat of steam $= 540 \ cal \ g^{-1}$ and specific heat capacity of water $= 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$) (in $g$)

$2 \; g$ of steam condenses when passed through $40 \; g$ of water initially at $25 \; ^\circ C$. The condensation of steam raises the temperature of water to $54.3 \; ^\circ C$. What is the latent heat of steam in $cal/g$?

$1\, kg$ of ice at $-10\,^{\circ}C$ is mixed with $4.4\, kg$ of water at $30\,^{\circ}C$. The final temperature of the mixture is ........ $^{\circ}C$. (Specific heat of ice = $2100\, J/kg\cdot K$,specific heat of water = $4200\, J/kg\cdot K$,latent heat of fusion of ice = $3.36 \times 10^5\, J/kg$)

$A$ $2 \, kg$ block of ice at $-20^{\circ}C$ is added to $5 \, kg$ of water at $20^{\circ}C$. What will be the total mass of water in $kg$? (Specific heat of water = $1 \, kcal/kg/^{\circ}C$,specific heat of ice = $0.5 \, kcal/kg/^{\circ}C$,latent heat of fusion of ice = $80 \, kcal/kg$)

If $540 \, g$ of ice at $0^{\circ}C$ is mixed with $540 \, g$ of water at $80^{\circ}C$,what will be the final temperature of the mixture in $^{\circ}C$?

$20 \, g$ 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 ........ $^{\circ}C$.