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Found in: Page 1333

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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

The power output is 14.4 kW .

See the step by step solution

## 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,

${\mathbit{N}}{\mathbf{=}}\frac{\mathbf{0}\mathbf{.}\mathbf{00015}\mathbf{m}{\mathbf{N}}_{\mathbf{A}}}{\mathbf{M}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}\left(1\right)$

Here, M is the molar mass of the heavy water, and ${{\mathbit{N}}}_{{\mathbf{A}}}$ 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.

$M=\left(16.0+1.00+2.00\right)g/mol\phantom{\rule{0ex}{0ex}}=19g/mol$

Substitute all the known values in equation (1).

$N=\frac{0.00015\left(1000g/mol\right)\left(6.022×{10}^{23}mo{l}^{-1}\right)}{19g/mol}\phantom{\rule{0ex}{0ex}}=4.75×{10}^{21}$

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

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

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

${N}_{fu}=\frac{N}{2}\phantom{\rule{0ex}{0ex}}=\frac{4.75×{10}^{21}}{2}\phantom{\rule{0ex}{0ex}}=2.375×{10}^{21}$

The expression of power output is given by,

$P=\frac{E}{t}\phantom{\rule{0ex}{0ex}}=\frac{{N}_{fu}Q}{t}..........\left(2\right)$

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

Substitute all the known values in equation (2).

$P=\frac{\left(2.375×{10}^{21}\right)\left(3.27MeV\right)\left(1.602×{10}^{-13}J/MeV\right)}{24×3600s}\phantom{\rule{0ex}{0ex}}=1.44×{10}^{4}W\phantom{\rule{0ex}{0ex}}=14.4kW$

Therefore, the power output is 14.4kW .