If we have greater confidence, with ice at a density of \(916.7\,kgm^{-3}\),
\(v_{boom\,ice}=3.4354*\cfrac{916.7}{10}=314.92\,ms^{-1}\)
\(T_{ice}=\cfrac{314.92^2*18.01528*10^{-3}}{3*8.3144}=71.63\,K\)
This is ice melting had it not been hydrogen bonded to 4 other molecules. We are off by,
\(\Delta T=273.15-71.63=201.52\,K\)
This energy load is shared by 4 hydrogen bonds, so per hydrogen bond in ice (\(O--H\)) contributes,
\(\Delta T_{\small{H},ice}=50.38\,K\) per mole
This is non-sense because we are not sharing energy here.
If we propose that water molecules clusters in ice and effectively \(n_{ice}\) molecules move as a free particle with an increased molecular mass by \(n_{ice}\) times, and given that ice melts at \(273.15\,K\).
\(T_{ice}=\cfrac{314.92^2*n_{ice}*18.01528*10^{-3}}{3*8.3144}=273.15\,K\)
\(n_{ice}=3.8\)
which is much lower than when each molecule is hydrogen bonded to 4 other molecules making an expected \(n=5\). However, if ice melting is a surface phenomenon and does not occurs through out the solid, then on the surface, the top most layer of molecules has one less neighbor molecule. This makes three hydrogen bonded neighbors and a expected cluster size of \(n=4\).
With water at \(100\,^oC\) of a density \(958.4\,kgm^{-3}\),
\(v_{boom\,water}=3.4354*\cfrac{958.4}{10}=329.25\,ms^{-1}\)
\(T_{water}=\cfrac{329.25^2*18.01528*10^{-3}}{3*8.3144}=78.30\,K\)
By a similar argument for a cluster of water molecules at 100oC, \(n_{100}\),
\(T_{water}=\cfrac{329.25^2*n_{100}*18.01528*10^{-3}}{3*8.3144}=373.15\,K\)
\(n_{100}=4.76\)
which is higher than the expected \(n=4.4\).
Kinetic theory for ideal gas does not apply to water at 100oC and ice at 0oC. End of this issue.
Goodnight.
