Iron Boiled At $T_{boom\,p}$

For Iron $Fe$, $Z=26$, molar mass of $0.055845\,kgmol^{-1}$, density of $7874\,kgm^{-3}$

$v_{boom}=3.4354*\cfrac{7874}{26}=1040.40\,ms^{-1}$

$T_{boom\,p}=\cfrac{1040.40^2*0.055845}{2*8.3144}=3635.16\,K$  or $3362.01\,^oC$

Iron has a melting point of  $1811\,K$ and a boiling point of $3134 K$

And the most nuclear $T_{boom\,p}$ bring iron beyond its boiling point.