One gram each of ice, water, and water vapor are in equilibrium together in a closed container. The pressure is $4.58 \mathrm{~mm}$ of $\mathrm{Hg}$, the temperature is $0.01^{\circ} \mathrm{C}$. Sixty calories of heat are added to the system. The total volume is kept constant. Calculate the masses of water vapor now present in the container. 

(Hint: For water at $0.01^{\circ} \mathrm{C}$, the latent heat of fusion is $80 \mathrm{cal} / \mathrm{g}$, the latent heat of vaporization is $596 \mathrm{cal} / \mathrm{g}$, and the latent heat of sublimation is $676 \mathrm{cal} / \mathrm{g}$. Also note that the volume of the vapor is much larger than the volume of the water or the volume of the ice.)
It is assumed that the original volume of water vapor is $V$, it volume is also $V$ after heating, and the masses of ice, water, and water vapor are respectively $x, y$ and $z$ at the new equilibrium. We have

$$
\begin{gathered}
x+y+z=3, \\
(1-x) L_{\mathrm{sub}}+(1-y) L_{\mathrm{vap}}=Q=60, \\
\frac{1-x}{\rho_{\mathrm{ice}}}+\frac{(1-y)}{\rho_{\text {water }}}+V_{0}=V . \\
V_{0}=\frac{R T}{\mu p} . \\
V=\frac{z}{\mu p} R T
\end{gathered}
$$

where $\mu=18 \mathrm{~g} / \mathrm{mole}, p=4.58 \mathrm{mmHg}, T=273.16 \mathrm{~K}, R=8.2 \times 10^{8}$ $\mathrm{m}^{3} \cdot \mathrm{atm} / \mathrm{mol} \cdot \mathrm{K}, \rho_{\text {ice }}=\rho_{\text {water }}=1 \mathrm{~g} / \mathrm{cm}^{3}, L_{\text {sub }}=676 \mathrm{cal} / \mathrm{g}$, and $L_{\text {vap }}=$ $596 \mathrm{cal} / \mathrm{g}$. Solving the equations we find

$$
x=0.25 \mathrm{~g}, \quad y=1.75 \mathrm{~g}, \quad z=1.00 \mathrm{~g} .
$$

That is, the heat of $60 \mathrm{cal}$ is nearly all used to melt the ice.