Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 34 - Electromagnetic Fields and Waves - Exercises and Problems - Page 1031: 46

Answer

$\approx 2.7\;\rm min$

Work Step by Step

Recalling that the total heat energy absorbed by the water is given by $$Q=mc_w\Delta T\tag 1$$ The author asks about the time it takes to raise its temperature 50$^\circ$C. So the power of heat transfer is given by $$P=\dfrac{Q}{t}$$ Hence, $$t=\dfrac{Q}{P}$$ Plug from (1), $$t=\dfrac{mc_w\Delta T}{P}\tag 2$$ We know that the energy rate transfer from the beam to the cube is given by $$P=0.80IA$$ where $I=\frac{1}{2}\epsilon_0cE^2$, and $A=L^2$ where $L$ is the side length of the water cube. The $0.80$ since the author told us that the water absorbs only 80% of the incident energy. $$P=\frac{0.80\epsilon_0cE^2L^2}{2}$$ Plug into (2), $$t=\dfrac{2mc_w\Delta T}{0.80\epsilon_0cE^2L^2} $$ Recall that $m=\rho V$, and $V=L^3$ $$t=\dfrac{2\rho L^3c_w\Delta T}{0.80\epsilon_0cE^2L^2} $$ $$t=\dfrac{2\rho L c_w\Delta T}{0.80\epsilon_0cE^2 } $$ Plug the known; $$t=\dfrac{2(1000) (0.1) (4186)(50)}{0.80(8.85\times 10^{-12})(3\times 10^8)(11000)^2 } $$ $$t=\color{red}{\bf 163}\;\rm s$$
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