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

$100 \, g$ of water is supercooled to $-10 \, ^\circ C$. At this point,due to some disturbance,some of it suddenly freezes to ice. What will be the temperature of the resultant mixture and how much mass would freeze? $[S_W = 1 \, cal \, g^{-1} \, ^\circ C^{-1}$ and $L_{fusion} = 80 \, cal \, g^{-1}]$

Steam at $100\,^{\circ}C$ is passed into $20\,g$ of water at $10\,^{\circ}C$. When water acquires a temperature of $80\,^{\circ}C$,the mass of water present will be ........ $g$. [Take specific heat of water $= 1\,cal\,g^{-1}\,^{\circ}C^{-1}$ and latent heat of steam $= 540\,cal\,g^{-1}$]

$100 \text{ g}$ of ice at $0^{\circ}C$ is mixed with $100 \text{ g}$ of water at $100^{\circ}C$. The final temperature of the mixture is. [Take,$L_f = 3.36 \times 10^5 \text{ J kg}^{-1}$ and $S_w = 4.2 \times 10^3 \text{ J kg}^{-1} \text{ K}^{-1}$] (in $^{\circ}C$)

$1\, g$ 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 $= 540\, cal/g$)

$A$ calorimeter of water equivalent $20 \, g$ contains $180 \, g$ of water at $25^{\circ} C$. '$m$' grams of steam at $100^{\circ} C$ is mixed in it until the temperature of the mixture is $31^{\circ} C$. The value of '$m$' is close to (Latent heat of water $= 540 \, \text{cal} \, g^{-1}$,specific heat of water $= 1 \, \text{cal} \, g^{-1} {}^{\circ} C^{-1}$)