Calculate the number of `kw-h` of electricity is necessary to produce `1.0` metric ton `(1000 kg)` of aluminium by the Hall process in a cell operating at `15.0 V`.

Explanation

`Al^(3+)+3e rarr Al`
Eq. of `Al, (w)/(E) = (i.t)/(96500)`
`:. i.t = (1000 xx 10^(3) xx 96500)/(27//3) [E_(Al) = 27//3]`
`i.t = 107.22 xx 10^(8)` ampere-sec or coulomb
`E = "Coulomb" xx "volt" = 107.22 xx 10^(8) xx 15.0`
`= 1.61 xx 10^(11) J-sec`
or `E = 1.61 xx 10^(11) "watt-sec"`
`= (1.61 xx 10^(11))/(10^(3)) xx (1)/(3600) kW-hr`
`= 4.47 xx 10^(4) kW-h`