Physics: Principles with Applications (7th Edition)

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

Chapter 18 - Electric Currents - Problems - Page 523: 45

Answer

See work below.

Work Step by Step

The manufacturer claims that 85 percent of the electrical power used by the heater is sent into the water as heat. $$0.85\;IV=\frac{Q_{absorbed}}{t}=\frac{mc\Delta T}{t}$$ $$I=\frac{mc\Delta T}{0.85Vt}=\frac{(0.120kg)(4186J/kg)(70C)}{0.85(12V)(480s)}\approx 7.2A$$ b. Use Ohm’s law to find the resistance. $$R=\frac{V}{I}=\frac{12V}{7.182A}=1.7\Omega$$
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