$10 \, g$ of ice at $0 \, ^oC$ is mixed with $m \, g$ of water at $50 \, ^oC$. What is the minimum value of $m$ (in $g$) so that the ice melts completely? (Given: $L_f = 80 \, cal/g$ and $S_W = 1 \, cal/g \cdot ^oC$)

$10 \, kg$ of water is to be heated from $20^{\circ}C$ to $80^{\circ}C$ in $1 \, hour$. Steam at $150^{\circ}C$ is passed through a copper coil immersed in the water. The steam condenses and returns to the boiler at $90^{\circ}C$. How much steam is required per hour? (Specific heat of water $= 1 \, cal/g^{\circ}C$,Latent heat of vaporization $= 540 \, cal/g$)

Pure water supercooled to $-15^{\circ} C$ is contained in a thermally insulated flask. $A$ small amount of ice is thrown into the flask. The fraction of water frozen into ice is:

$1\,\,kg$ of ice at $-10^{\circ}C$ is mixed with $4.4\,\,kg$ of water at $30^{\circ}C$. The final temperature of the mixture is $........^{\circ}C$ (specific heat of ice is $2100\,\,J/kg/K$ and latent heat of fusion of ice is $336000\,\,J/kg$)

$A$ 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:

We have half a bucket $(6\,L)$ of water at $20\,^oC$. If we want water at $40\,^oC$,how much steam at $100\,^oC$ should be added to it?