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 $g$)

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 .......... $g$.

$A$ small quantity,mass $m$,of water at a temperature $\theta$ (in $^oC$) is poured onto a large mass $M$ of ice which is at its melting point. If $c$ is the specific heat of water and $L$ is the latent heat of fusion of ice,then the mass of ice melted is given by

$A$ vessel contains $110\,g$ of water. The heat capacity of the vessel is equivalent to $10\,g$ of water. The initial temperature of the water in the vessel is $10\,^{\circ}C$. If $220\,g$ of hot water at $70\,^{\circ}C$ is poured into the vessel,the final temperature,neglecting radiation loss,will be nearly equal to ........ $^{\circ}C$.

Water of volume $2 \, L$ in a container is heated with a coil of $1 \, kW$ at $27 \, ^\circ C$. The lid of the container is open and energy dissipates at a rate of $160 \, J/s$. In how much time will the temperature rise from $27 \, ^\circ C$ to $77 \, ^\circ C$? [Given: specific heat of water is $4.2 \, kJ/(kg \cdot K)$]

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 $\,g$)