College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 22 - Problems - Page 866: 76


(a) Energy is absorbed by the water at a rate of $914~W$ (b) $I = 1.04\times 10^5~W/m^2$ (c) $E_{rms} = 6260~V/m$ $B_{rms} = 2.09\times 10^{-5}~T$

Work Step by Step

(a) We can find the energy required to raise the temperature of the water: $E = m~C~\Delta~T$ $E = (0.350~kg)(4180~J/kg~K)(75.0~K)$ $E = 109,725~J$ We can find the rate at which energy is absorbed: $P = \frac{E}{t} = \frac{109,725~J}{120~s} = 914~W$ Energy is absorbed by the water at a rate of $914~W$ (b) We can find the average intensity: $I = \frac{P}{A} = \frac{914~J}{88.0\times 10^{-4}~m^2} = 1.04\times 10^5~W/m^2$ (c) We can find $E_{rms}$: $E_{rms}^2 = \frac{I}{c~\epsilon_0}$ $E_{rms} = \sqrt{\frac{I}{c~\epsilon_0}}$ $E_{rms} = \sqrt{\frac{1.04\times 10^5~W/m^2}{(3.0\times 10^8~m/s)~(8.85\times 10^{-12}~C^2/N~m^2)}}$ $E_{rms} = 6260~V/m$ We can find $B_{rms}$: $B_{rms} = \frac{E_{rms}}{c}$ $B_{rms} = \frac{6260~V/m}{3.0\times 10^8~m/s}$ $B_{rms} = 2.09\times 10^{-5}~T$
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