= Solution
If the <voltage> were held at $4.57\,\mathrm V$ for the entire ideal discharge, then
$$
\boxed{E=VQ=(4.57)(175.2)=800.664\,\mathrm{Wh}=0.801\,\mathrm{kWh}},
$$
and the electrical <power> would be
$$
\boxed{P=VI=(4.57)(0.010)=0.0457\,\mathrm W=45.7\,\mathrm{mW}}.
$$
These are \b[constant-voltage estimates], not a determination of the actual <energy> from the fresh-cell <open-circuit voltage>. Under load the <voltage> depends on state of discharge, <electric polarization> and internal resistance. The physical result is $E=I\int_0^{t_f}V_{\rm load}(t)\,dt$, and $P(t)=IV_{\rm load}(t)$. The supplied single open-circuit value does not specify that <voltage> curve; practical delivered <energy> will generally be lower than the simple estimate.
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