We have half a bucket $(6 \text{ litre})$ of water at $20^{\circ}C$. If we want water at $40^{\circ}C$,how much steam at $100^{\circ}C$ should be added to it?

The temperature of $100 \,g$ of water is to be raised from $24^{\circ} C$ to $90^{\circ} C$ by adding steam at $100^{\circ} C$ to it. The mass of the steam required in this process is (latent heat of steam is $540 \,cal \,g^{-1}$). (in $\,g$)

$150 \,g$ of ice is mixed with $100 \,g$ of water at temperature $80^{\circ} C$. The latent heat of ice is $80 \,cal/g$ and the specific heat of water is $1 \,cal/g^{\circ} C$. Assuming no heat loss to the environment,the amount of ice which does not melt is ........... $g$.

$2\, kg$ of ice at $-20^{\circ}C$ is mixed with $5\, kg$ of water at $20^{\circ}C$ in an insulating vessel having a negligible heat capacity. Calculate the final mass of water remaining in the container. It is given that the specific heats of water and ice are $1\, kcal/kg/^{\circ}C$ and $0.5\, kcal/kg/^{\circ}C$ respectively,while the latent heat of fusion of ice is $80\, kcal/kg$.

Ice in a freezer is at $-7^{\circ} C$. $100 \, g$ of this ice is mixed with $200 \, g$ of water at $15^{\circ} C$. Take the freezing temperature of water to be $0^{\circ} C$,the specific heat of ice equal to $2.2 \, J/g^{\circ} C$,specific heat of water equal to $4.2 \, J/g^{\circ} C$,and the latent heat of ice equal to $335 \, J/g$. Assuming no loss of heat to the environment,the mass of ice in the final mixture is closest to .......... $g$.

$A$ calorimeter contains $0.5 \,kg$ of water at $30^{\circ} C$. When $0.3 \,kg$ of water at $60^{\circ} C$ is added to it, the resulting temperature is found to be $40^{\circ} C$. The water equivalent of the calorimeter is (in $\,kg$)