Physics: Principles with Applications (7th Edition)

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

Chapter 31 - Nuclear Energy; Effects and Uses of Radiation - Search and Learn - Page 914: 3

Answer

$\approx 4 \times10^{30}J$.

Work Step by Step

Find the mass of water in the oceans. The total mass of water is volume multiplied by density. Assume an average ocean depth of 5000m, covering 70 percent of the Earth. Use a density of $1000kg/m^3$. M=0.7(surface area)(depth)(density) $$ M=0.7(4 \pi (6.38\times10^6m)^2)(5000m)( 1000kg/m^3)=1.79\times10^{21}kg$$ Calculate the number of water molecules by dividing M by water’s molar mass of water, then multiplying by Avogadro’s number. A water molecule has 2 hydrogen atoms, so we multiply the number of H atoms by the percentage of deuterium to find the number of deuterium atoms. $$ N=1.377\times10^{43}deuterium\;atoms$$ Consider the reactions of equations 31–8a and 31–8b. Four deuterons produce 7.30 MeV of energy. Find the total amount of energy. $$ E=(1.377\times10^{43}deuterium\;atoms)\frac{7.30MeV}{4\;deuterons}\frac{1.60\times10^{-13}J}{MeV}$$ $$\approx 4 \times10^{30}J$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.