\(v_{boom\,water}=3.4354*\cfrac{1000}{10}=343.54\,ms^{-1}\)
\(T_{water}=\cfrac{343.54^2*18.01528*10^{-3}}{3*8.3144}=85.24\,K\) (the boiling point of Fluorine, \(F\) exactly)
which is \(-187.90\,^oC\).
Which make no sense. Water is a solid at this temperature. It can oxidize further to \(H_2O_2\). Does water resist further cooling at this temperature?
If we apply the idea of \(T_{boom}\) to hydrogen gas, \(H_2\),
\(v_{boom\,H_2}=3.4354*\cfrac{0.08988}{2}=0.1544\,ms^{-1}\)
\(T_{H_2}=\cfrac{0.1544^2*2*1.00794*10^{-3}}{3*8.3144}=1.9267*10^{-6}\,K\)
Which is way low. Is this another resistance barrier to cooling?
The biggest assumption is from the use of kinetic theory for ideal gas in the derivation of \(T_{boom}\), where it is assume that the gas particles are free and does not interact with other gas particles other than through collisions. Water is liquid and is hydrogen bonded to approximately 3.4 other water molecules. Ice is a solid and is hydrogen bonded to approximately 4 other water molecules. Both instances invalidate the free particles assumption.
None-the-less, with water at \(100\,^oC\) of a density \(958.4\,kgm^{-3}\),
\(v_{boom\,water}=3.4354*\cfrac{958.4}{10}=329.25\,ms^{-1}\)
\(T_{water}=\cfrac{329.25^2*18.01528*10^{-3}}{3*8.3144}=78.30\,K\)
since water is hydrogen bonded to 3.4 other molecules and we ignore hydrogen bonding beyond the first neighbor. This hydrogen bonded macro-molecule of (\(3.4+1\)) molecules moves as whole and behave like a big gas particle.
\(T_{water,H-Bonded}=\cfrac{329.25^2*(3.4+1)*18.01528*10^{-3}}{3*8.3144}=344.50\,K\) or \(71.35\,^oC\)
Using this as a guide, we choose water at \(80\,^oC\) with a density of \(971.8\,kgm^{-3}\),
\(v_{boom\,water}=3.4354*\cfrac{971.8}{10}=333.85\,ms^{-1}\)
\(T_{water}=\cfrac{333.85^2*18.01528*10^{-3}}{3*8.3144}=80.50\,K\)
and, assuming that water is a hydrogen bonded macro-molecule of 4.4 water molecules,
\(T_{water,H-Bonded}=T_{water}*4.4=354.20\,K\) or \(81.50\,^oC\)
Is this the reason why a bicycle pump goes pop? Or did we have that in the post "Fire Starter" dated 24 Oct 2017. Water definitely does not boil at \(81.50\,^oC\). Does a sealed sphere of pure water have a stable temperature at \(81.50\,^oC\)? Is water nuclear at \(81.50\,^oC\)?
With ice at a density of \(916.7\,kgm^{-3}\),
\(v_{boom\,ice}=3.4354*\cfrac{916.7}{10}=314.92\,ms^{-1}\)
\(T_{ice}=\cfrac{314.92^2*18.01528*10^{-3}}{3*8.3144}=71.63\,K\)
Since ice is hydrogen bonded to 4 other molecules. We ignore hydrogen bonding beyond the first neighbor and apply kinetic theory for ideal gas to this hydrogen bonded macro-molecule, we have,
\(T_{ice,H-bonded}=\cfrac{329.25^2*(4+1)*18.01528*10^{-3}}{3*8.3144}=391.5\,K\) or \(118.35\,^oC\)
And ice does not melt at \(118.35\,^oC\) either?
Water Bombed!
