$200 \,g$ 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 $= 80 \,cal \,g^{-1}$).

$A$ liquid of mass $m$ and specific heat $c$ is at a temperature of $2T$. Another liquid of mass $m/2$ and specific heat $2c$ is at a temperature of $T$. What will be the final temperature of the mixture?

The time required to raise the temperature of $3 \text{ litre}$ of water from $0^{\circ} C$ to $80^{\circ} C$ by a heater operated under $200 \text{ V}$ having resistance of $50 \Omega$ is
[specific heat capacity of water is $4200 \text{ J kg}^{-1} \text{ K}^{-1}$] [density of water $= 1000 \text{ kg/m}^3$] (in $\text{ min}$)

$10\,g$ of ice at $-20\,^{\circ}C$ is dropped into a calorimeter containing $10\,g$ of water at $10\,^{\circ}C$. The specific heat of water is twice that of ice. When equilibrium is reached,the calorimeter will contain

$A$ small quantity of water of mass $m$ at temperature $\theta^{\circ} C$ is mixed with a large mass $M$ of ice which is at its melting point. If $s$ is the specific heat capacity of water and $L$ is the latent heat of fusion of ice,then the mass of ice melted is:

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