Chapter 31, Problem 56GP

To determine

Estimation of total deuterium in the Earth’s oceans and the total energy released.

Answer to Problem 56GP

The estimated value of total deuterium in the Earth’s oceans is $5\times {10}^{16}\text{kg}$ and the total energy released is $\mathrm{E}=4.61\times {10}^{30}\text{J}$

Explanation of Solution

Given info:

Deuterium makes up 0.0115% of natural hydrogen.

Known data:

Density of water, $\mathrm{\delta }={10}^3{\text{kg/m}}^3$

Depth of the sea, $\mathrm{d}=4000\text{ m}$

Radius of the Earth, $\mathrm{R}=6.371\times {10}^6\text{m}$

Formula used:

Mass of water can be calculated as:

  $\mathrm{m}=\mathrm{A}\mathrm{d}\mathrm{\delta }$

Calculation:

Firstly, we need to find the mass of water as follows:

  $\begin{array}{l}\mathrm{m}=\left(4\mathrm{\pi }{\mathrm{R}}^2\right)\mathrm{d}\mathrm{\delta }\\ \mathrm{m}=4\left(3.14\right){\left(6.371\times {10}^6\text{m}\right)}^2\left(4000\text{ m}\right)\left({10}^3{\text{kg/m}}^3\right)\\ \mathrm{m}=2.05\times {10}^{21}\text{kg}\end{array}$

Now, the number of atoms can be calculated using the mass of water as follows:

  $\begin{array}{l}\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}\\ \mathrm{n}=\frac{\left(2.05\times {10}^{21}\text{kg}\right)\left(6.022\times {10}^{23}\text{mol}\right)\left(0.0115\text{d atoms}\right)}{0.018\text{ kg}}\\ \mathrm{n}=2\times {10}^{43}\end{array}$

Hence, the mass of total deuterium can be given by:

  $\begin{array}{l}=\left(2\times {10}^{43}\right)\left(\frac{2\times {10}^{-3}\text{kg}}{6.022\times {10}^{23}}\right)\\ =5\times {10}^{16}\text{kg}\end{array}$

Lastly, the total energy released will be given as:

  $\begin{array}{l}\mathrm{E}=\frac{\left(2\times {10}^{43}\right)\left(7.3\text{ MeV}\right)}{4}\\ \mathrm{E}=\frac{\left(2\times {10}^{43}\right)\left(7.3\times {10}^6\right)\left(1.602\times {10}^{-19}\right)\text{J}}{4}\\ \mathrm{E}=4.61\times {10}^{30}\text{J}\end{array}$

Conclusion:

Thus, the estimated value of total deuterium in the Earth’s oceans is $5\times {10}^{16}\text{kg}$ and the total energy released is $\mathrm{E}=4.61\times {10}^{30}\text{J}$