= Negative melting-point slope of water
For a local illustrative <ice>–<water> <melting> calculation, stipulate $T_m=273.15\,\mathrm K$, <latent heat of fusion> $L=334\,\mathrm{kJ\,kg^{-1}}$, <solid> <mass density> $\rho_s=917\,\mathrm{kg\,m^{-3}}$, and <liquid> <mass density> $\rho_l=1000\,\mathrm{kg\,m^{-3}}$. These are explicitly stipulated rounded inputs, not measurements made here. The <specific volume> change on <melting> is
$$
\Delta v=\frac{1}{\rho_l}-\frac{1}{\rho_s}=-9.0513\times10^{-5}\,\mathrm{m^3\,kg^{-1}}.
$$
The <common-pressure melting-point slope> therefore gives
$$
\frac{dT_m}{dp}=\frac{T_m\Delta v}{L}=-7.4022\times10^{-8}\,\mathrm{K\,Pa^{-1}}\simeq-0.07402\,\mathrm{K\,MPa^{-1}}.
$$
A small $1\,\mathrm{MPa}$ increase in the common <pressure> thus lowers the <melting> <temperature> by about $0.074\,\mathrm K$ in this local model. Positive <latent heat> does not force a positive <phase coexistence curve> slope: <ice> contracts on conversion to <liquid> <water>. Increasing <pressure> favors the smaller <specific volume>. This calculation compares equilibrium states of the same two <thermodynamic phases>; extending its tangent to large <pressures> would require updated <mass densities>, <latent heat of fusion>, and the relevant <ice> <phase diagram>.
Back to article page
