Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

Roughly 0.0150% of the mass of ordinary water is due to “heavy water,” in which one of the two hydrogens in an $H_{2} O$ molecule is replaced with deuterium, ${}^{2}H$ . How much average fusion power could be obtained if we “burned” all the ${}^{2}H$ in 1.00 litre of water in 1.00 day by somehow causing the deuterium to fuse via the reaction ${}^{2}H + {}^{2}H \to {}^{3}H e + n$ ?

The power output is 14.4 kW .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Describe the expression for average kinetic energy

Let m be the mass of 1 litter of water, then the mass of the heavy water in 1 litter is 0.000150m .

The expression for number of molecules is given by,

$N = \frac{0.00015 m N_{A}}{M} . . . . . . . . . . (1)$

Here, M is the molar mass of the heavy water, and $N_{A}$ is the number of molecules in one mole.

Step 2: Find the average fusion power

A heavy water molecule contains one oxygen atom, one hydrogen atom and one deuterium atom so the molar mass of the heavy water will be as follows.

$M = (16.0 + 1.00 + 2.00) g / m o l = 19 g / m o l$

Substitute all the known values in equation (1).

$N = \frac{0.00015 (1000 g / m o l) (6.022 \times 10^{23} m o l^{- 1})}{19 g / m o l} = 4.75 \times 10^{21}$

This is the number of heavy water molecules (or deuterium), the deuterium fuse with another deuterium atom, according to the following reaction.

${}^{2}H + {}^{2}H \to^{3} H e + n$

So, the number of fusion events will be half of N .

$N_{f u} = \frac{N}{2} = \frac{4.75 \times 10^{21}}{2} = 2.375 \times 10^{21}$

The expression of power output is given by,

$P = \frac{E}{t} = \frac{N_{f u} Q}{t} . . . . . . . . . . (2)$

Here, E is total energy, and t is time.

Substitute all the known values in equation (2).

$P = \frac{(2.375 \times 10^{21}) (3.27 M e V) (1.602 \times 10^{- 13} J / M e V)}{24 \times 3600 s} = 1.44 \times 10^{4} W = 14.4 k W$

Therefore, the power output is 14.4kW .

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Fundamentals Of Physics

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Short Answer

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Step 1: Describe the expression for average kinetic energy

Step 2: Find the average fusion power

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