\(v_{boom}=3.4354*\cfrac{6020}{56+22+8*3}=202.76\,ms^{-1}\)
\(T_{boom}=202.76^2*\cfrac{233.192*10^{-3}}{2*8.3144}=576.50\,K\) or \(303.35\,^oC\)
This is way hot. Temperature is know to decrease conductivity. An impossible hot superconductor.
For \(CaCu_3Ti_4O_{12}\),
\(v_{boom}=3.4354*\cfrac{4700}{20+29*3+22*4+8*12}=55.49\,ms^{-1}\)
\(T_{boom}=55.49^2*\cfrac{614.1789*10^{-3}}{2*8.3144}=43.17\,K\) or \(-229.98\,^oC\)
This is cold. The cell unit is too big to be considered as a quasi-nucleus. Maybe it is possible to parcel out molar mass and arrive at a density value for copper only in the crystal structure.
\(f_{Cu}=\cfrac{total\,molar\,mass\,Cu}{total\,atomic\,molar\,mass}\)
\(M_{m\,Cu}=M_m*f_{Cu}\approx molar\,mass\,Cu\)
where \(M_m\) is molecular mass,
\(\rho_{cu}=\rho_{m}.f_{cu}\)
where \(\rho_{m}\) and \(\rho_{cu}\) are crystal density and density of \(Cu\) in the structure, and so consider only copper in the structure.
So,
\(f_{Cu}=\cfrac{63.546*3}{614.1789}=0.310\)
\(\rho_{cu}=4700*0.310=1458.86\,kgm^{-3}\)
Considering only \(Cu\) in \(CaCu_3Ti_4O_{12}\),
\(v_{boom}=3.4354*\cfrac{1458.86}{29*3}=57.60\,ms^{-1}\)
\(T_{boom}=57.60^2*\cfrac{63.546*3*10^{-3}}{2*8.3144}=38.03\,K\)
It does not seem to work. Furthermore, to consider collisions with \(Cu\) only, the atomic will have to dominate the structure in size. The rest of the molecules is as if does not exist. Even so, three separate \(Cu\) atoms does not form a quasi-nucleus. Maybe,
\(v_{boom}=3.4354*\cfrac{1458.86}{29}=172.82\,ms^{-1}\)
\(T_{boom}=172.82^2*\cfrac{63.546*10^{-3}}{2*8.3144}=114.13\,K\)
What about copper by itself as a metal;
\(v_{boom}=3.4354*\cfrac{8960}{29}=1061.42\,ms^{-1}\)
\(T_{boom}=1061.42^2*\cfrac{63.546*10^{-3}}{2*8.3144}=4305.29\,K\) or \(4032.14\,^oC\)
which is not indicative of its high conductivity of electricity. For this reason a folly...
