The variation of density of water with temperature is represented by the

Find the amount of heat supplied to decrease the volume of an ice-water mixture by $1 \, cm^3$ without any change in temperature. (Given: $\rho_{ice} = 0.9 \rho_{water}$,$L_{ice} = 80 \, cal/g$).

How much thermal energy is required to change a $40 \ g$ ice cube from solid at $-10^{\circ} C$ to steam at $110^{\circ} C$ (in $kcal$)?
[Assume, latent heat of fusion for water $= 80 \ kcal/kg$, specific heat of water $= 1 \ kcal/kg^{\circ} C$, specific heat of ice $= 0.5 \ kcal/kg^{\circ} C$, specific heat of steam $= 0.48 \ kcal/kg^{\circ} C$, latent heat of vaporisation of water $= 540 \ kcal/kg$]

Match the following List-$I$ with List-$II$.

List-$I$	List-$II$
$A$. When ice melts into water	$I$. Volume increases
$B$. When water changes into steam	$II$. Volume decreases
$C$. Melting point of ice	$III$. Increases with increase of pressure
$D$. Boiling point of water	$IV$. Decreases with increase in pressure

$500\, g$ of water and $100\, g$ of ice at $0\,^{\circ}C$ are in a calorimeter whose water equivalent is $40\, g$. $10\, g$ of steam at $100\,^{\circ}C$ is added to it. Then the final amount of water in the calorimeter is ....... $g$ (Latent heat of ice $= 80\, cal/g$,Latent heat of steam $= 540\, cal/g$).

$Assertion :$ In the pressure-temperature $(P-T)$ phase diagram of water,the slope of the melting curve is found to be negative.
$Reason :$ Ice contracts on melting to water.