= Solution
At coexistence, the two phases have equal <chemical potential>. For a single component, $d\mu=-s\,dT+v\,dp$, with <entropy> and volume measured on the same molar or <mass> basis. Equating the phase <derivatives> along coexistence gives $\Delta v\,dp=\Delta s\,dT$. Reversible <latent heat> satisfies $L=T\Delta s$, hence
$$
\boxed{\frac{dp}{dT}=\frac{L}{T\Delta v}.}
$$
Here $p$ and $T$ are coexistence <pressure> and absolute <temperature>, $L$ is <latent heat> for transforming phase 1 to phase 2, and $\Delta v=v_2-v_1$ uses the same amount of material as $L$.
For vaporization, neglect liquid <molar volume> compared with ideal-vapour volume $RT/p$. The ideal-vapour Clausius-Clapeyron approximation becomes
$$
\frac{dp}{dT}=\frac{pL_e}{RT^2},\qquad
\ln\frac p{p_b}=-\frac{L_e}{R}\left(\frac1T-\frac1{T_b}\right).
$$
Using the normal boiling reference $T_b=373.15\,\mathrm K$, $p_b=1.01325\times10^5\,\mathrm{Pa}$ and the supplied constant $L_e$ gives
$$
\boxed{p(273\,\mathrm K)\simeq8.2\times10^2\,\mathrm{Pa}.}
$$
This deliberately constant-latent-heat extrapolation does not reproduce the supplied triple-point <pressure> exactly. It predicts the liquid-vapour branch, metastable against ice below the triple-point <temperature>; it is not a precision sublimation-pressure prediction.
The <phase diagram> has solid, liquid and vapour regions meeting at the <triple point>. The <sublimation> and vaporization boundaries have positive slopes, whereas the ordinary ice–water boundary slopes negatively because ice has larger specific volume. The liquid–vapour line terminates at a <critical point>; the original schematic below emphasizes topology and marks the supplied <triple point>.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2012/paper-2-water-phase.png]
{title=Schematic pressure–temperature phase topology of ordinary water, with negative melting slope and the supplied triple point}
For the melting-pressure estimate, <Archimedes' principle> gives $\rho_i=0.90\rho_w=900\,\mathrm{kg\,m^{-3}}$, so
$$
\Delta v=v_w-v_i=\frac1{1000}-\frac1{900}
=-1.111\times10^{-4}\,\mathrm{m^3\,kg^{-1}}.
$$
Convert the molar evaporation <heat> using water <molar mass> $M\simeq0.0180\,\mathrm{kg\,mol^{-1}}$, rounded from the tabulated value $18.0153\,\mathrm{g\,mol^{-1}}$.<NIST Chemistry WebBook, SRD 69>[Water entry, CAS 7732-18-5, molecular weight 18.0153]{ref} An <enthalpy> cycle through vapour then gives
$$
L_f=L_s-\frac{L_e}{M}
\simeq2.8\times10^6-\frac{40700}{0.0180}
\simeq5.39\times10^5\,\mathrm{J\,kg^{-1}}.
$$
This uses the question's temperature-independent evaporation <heat> all the way to the <triple point>; no replacement experimental fusion <heat> is inserted. The <melting pressure inferred from flotation and latent heats>, on holding $L_f$ and $\Delta v$ constant, is
$$
\boxed{p-p_t=\frac{L_f}{\Delta v}\ln\frac T{T_t}
\simeq5.37\times10^7\,\mathrm{Pa},\qquad
p\simeq54\,\mathrm{MPa},}
$$
using $T_t=273.16\,\mathrm K$ and $T=270.15\,\mathrm K$. Linearizing the logarithm about the <triple point> gives $53.4\,\mathrm{MPa}$ instead of $53.7\,\mathrm{MPa}$. Both are estimates within the stated constant-property approximation; the negative melting slope makes a positive <pressure> necessary to lower the melting <temperature>.
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