Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 17 - Electric Potential - General Problems - Page 498: 72

Answer

660 volts.

Work Step by Step

The energy stored in the capacitor transforms into the heat energy that raises the water temperature. $$PE=Q$$ $$\frac{1}{2}CV^2=mc\Delta T$$ $$V=\sqrt{\frac{2mc\Delta T }{C}}=\sqrt{\frac{2(2.8kg)(4186J/(kg\cdot ^{\circ}C))74C^{\circ}}{4.0F}}$$ $$\approx 660V$$
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