Answer
(a)
$$\begin{aligned} \text { Chemical energy } &=\Delta P E_{c}=15.12 \mathrm{MJ} \end{aligned}$$
As the battery discharges, the voltage will decrease below the rated voltage. The present chemical energy save in the battery is less useful or not useful.
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(b)
$$\Delta Q=1.26 \mathrm{MC} $$
Work Step by Step
(a)
$\begin{aligned} \Delta V \equiv \frac{\Delta P E_{c}}{\Delta Q} & I=\frac{\Delta Q}{\Delta t} \\ \text { Chemical energy } &=\Delta P E_{c}=\Delta V \cdot \Delta Q=\Delta V \cdot(I \cdot \Delta t) \\ &=12 \mathrm{V} 350 \mathrm{A}-\mathrm{hr} 3600 \frac{\mathrm{s}}{\mathrm{hr}}=15.12 \mathrm{MJ} \end{aligned}$
As the battery discharges, the voltage will decrease below the rated voltage. The present chemical energy save in the battery is less useful or not useful.
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(b)
$\Delta Q$ is the total charge passing through the battery and gaining 12 $\mathrm{J} / \mathrm{C}$ of electrical energy.
$$\Delta Q=I \cdot \Delta t=350 \mathrm{Ahr}=350-\mathrm{hr} \cdot 3600 \frac{\mathrm{s}}{\mathrm{hr}}=1.26 \mathrm{MC} $$
