Thermal expansion due to an increase in temperature...

The heat energy required to convert $10 \,kg$ of ice at $-10^{\circ} C$ into water at $0^{\circ} C$ is (specific heat capacity of ice $=0.5 \,cal \,g^{-1} \,^{\circ}C^{-1}$ and latent heat of fusion of ice $=80 \,cal \,g^{-1}$)

$22320 \ cal$ of heat is supplied to $100 \ g$ of ice at $0^{\circ} C$. If the latent heat of fusion of ice is $80 \ cal \ g^{-1}$ and latent heat of vaporization of water is $540 \ cal \ g^{-1}$, the final amount of water thus obtained and its temperature respectively are:

$2 \ kg$ of ice at $-20^{\circ} C$ is mixed with $5 \ kg$ of water at $20^{\circ} C$. What is the final amount of water in the mixture (in $kg$)? (Given: specific heat of ice $= 0.5 \ cal/g^{\circ} C$,specific heat of water $= 1 \ cal/g^{\circ} C$,latent heat of fusion of ice $= 80 \ cal/g$)

The heat extracted out of $x \ \text{g}$ of water initially at $50^\circ \text{C}$ to cool it down to $0^\circ \text{C}$ is sufficient to evaporate $(1000 - x) \ \text{g}$ of water also initially at $50^\circ \text{C}$. The value of $x$ (closest integer) is . . . . . . . (Take latent heat of water $L = 2256 \ \text{kJ/kg}$, specific heat capacity of water $c = 4200 \ \text{J/kg}\cdot \text{K}$)

$A$ flask of volume $10^3 \ cc$ is completely filled with mercury at $0 \, ^oC$. The coefficient of cubical expansion of mercury is $180 \times 10^{-6} / ^oC$ and that of glass is $40 \times 10^{-6} / ^oC$. If the flask is now placed in boiling water at $100 \, ^oC$,how much mercury (in $cc$) will overflow?