= Solution
Using $365$ days per year, the required time is $2\times365\times24=17520\,\mathrm h$. Hence
$$
\boxed{Q=(0.010)(17520)=175.2\,\mathrm{Ah}=630720\,\mathrm C}.
$$
<lithium> transfers one <electron> per atom, so
$$
n_{\rm Li}=\frac{Q}{F}=\frac{630720}{96500}=6.53596\,\mathrm{mol},
\qquad\boxed{m_{\rm Li}=6.53596(6.94)=45.4\,\mathrm g}.
$$
This is the stoichiometric minimum for one cell, or one cell-equivalent parallel assembly, with complete utilization and no self-discharge. Practical loading must be larger. If a pack uses several cells in series, each series cell must carry the full delivered charge and the total <lithium> mass increases accordingly; no required pack <voltage> or series count is supplied.
Back to article page
