$10 \, g$ 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 $S_W = 1 \, cal/g \cdot ^oC$)

$10\,g$ 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

$10 \, g$ of ice at $0^{\circ}C$ is mixed with $100 \, g$ of water at $50^{\circ}C$. What is the resultant temperature of the mixture in $^{\circ}C$?

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 $\,g$)

$A$ water heater of power $2000\,W$ is used to heat water. The specific heat capacity of water is $4200\,J\,kg^{-1}\,K^{-1}$. The efficiency of the heater is $70\%$. The time required to heat $2\,kg$ of water from $10^{\circ}C$ to $60^{\circ}C$ is $..........s$. (Assume that the specific heat capacity of water remains constant over the temperature range).

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