Answer
a. 1.3 million volts.
b. 2.0 kg.
Work Step by Step
a. The energy is the charge multiplied by the potential difference, Equation 17–3.
$$\Delta PE=q \Delta V$$
$$\Delta V=\frac{\Delta PE}{q}=\frac{5.2\times10^6 J}{4.0C}=1.3\times10^6 V$$
b. The energy raises the temperature of the water and then vaporizes it. Assume that the water started at $20^{\circ}C$. Use Equations 14–2 and 14–4.
$$Q=mc\Delta T+mL_V$$
$$m=\frac{Q}{c\Delta T+L_V}$$
$$m=\frac{5.2\times10^6 J }{(4186\frac{J}{kg \cdot C^{\circ}})(80 C^{\circ})+2.26\times10^6\frac{J}{kg}}=2.0kg$$
