$200 \,g$ of ice at $-20^{\circ} C$ is mixed with $500 \,g$ of water at $20^{\circ} C$ in an insulating vessel. The final mass of water in the vessel is ........... $g$ (specific heat of ice $= 0.5 \,cal \,g^{-1} {^{\circ}C}^{-1}$,latent heat of fusion of ice $= 80 \,cal \,g^{-1}$).

Objects $A$ and $B$ that are initially separated from each other and well isolated from their surroundings are then brought into thermal contact. Initially $T_A = 0^{\circ}C$ and $T_B = 100^{\circ}C$. The specific heat of $A$ is less than the specific heat of $B$. After some time,the system comes to an equilibrium state. The final temperatures are :

$100 \, g$ 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 $^{\circ}C$?

Two liquids $A$ and $B$ are at $32^{\circ}C$ and $24^{\circ}C$. When mixed in equal masses,the temperature of the mixture is found to be $28^{\circ}C$. Their specific heats are in the ratio of:

Steam at $100^{\circ} C$ is added to $150 \,g$ of water to increase its temperature from $20^{\circ} C$ to $40^{\circ} C$. The total mass of the water at $40^{\circ} C$ is (specific heat capacity of water $= 1 \,cal \,g^{-1} {}^{\circ} C^{-1}$ and latent heat of steam $= 540 \,cal \,g^{-1}$) (in $\,g$)

The water equivalent of a calorimeter is $10 \ g$ and it contains $50 \ g$ of water at $15^{\circ} C$. Some amount of ice, initially at $-10^{\circ} C$, is dropped in it and half of the ice melts till equilibrium is reached. What was the initial amount of ice that was dropped (given specific heat of ice $= 0.5 \ cal \ g^{-1} {}^{\circ} C^{-1}$, specific heat of water $= 1.0 \ cal \ g^{-1} {}^{\circ} C^{-1}$ and latent heat of melting of ice $= 80 \ cal \ g^{-1}$) (in $g$)?