Please be reminded of \(E=mc^2\). Small amount of mass is equivalent to huge amount of energy.
For Iron \(Fe\), \(Z=26\), molar mass of \(0.055845\,kgmol^{-1}\), density of \(7874\,kgm^{-3}\)
\(v_{rms}=3.4354*\cfrac{7874}{26}=1040.40\,ms^{-1}\)
\(T_{boom}=\cfrac{1040.40^2*0.055845}{3*8.3144}=2423.44\,K\) or \(2150.29\,^oC\)
Iron has a melting point of \(1811\,K\) and a boiling point of \(3134 K\)
What happens to iron at \(2423.44\,K\)? Red Hot? No, red hot in all shades are below the melting point.
The density of iron at \(2150.29\,^oC\) is expected to be different from \(7874\,kgm^{-3}\). So, what happens to iron around \(2150.29\,^oC\)? If at this temperature iron is completely sealed off from oxidizing does it remain at \(T_{boom}\)?
