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

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,

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

Here, M is the molar mass of the heavy water, and${\mathit{N}}_{\mathbf{A}}$ is the number of molecules in one mole.

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.

$$\begin{gathered} M=\left(16.0+1.00+2.00\right)g/mol \\ =19g/mol \end{gathered}$$

Substitute all the known values in equation (1).

$$\begin{gathered} N=\frac{0.00015\left(1000g/mol\right)\left(6.022\times {10}^{23}mo{l}^{-1}\right)}{19g/mol} \\ =4.75\times {10}^{21} \end{gathered}$$

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 .

$$\begin{gathered} N_{fu}=\frac{N}{2} \\ =\frac{4.75\times {10}^{21}}{2} \\ =2.375\times {10}^{21} \end{gathered}$$

The expression of power output is given by,

$$\begin{gathered} P=\frac{E}{t} \\ =\frac{N_{fu}Q}{t}..........\left(2\right) \end{gathered}$$

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

Substitute all the known values in equation (2).

$$\begin{gathered} P=\frac{\left(2.375\times {10}^{21}\right)\left(3.27MeV\right)\left(1.602\times {10}^{-13}J/MeV\right)}{24\times 3600s} \\ =1.44\times {10}^{4}W \\ =14.4kW \end{gathered}$$

Therefore, the power output is 14.4kW .