Question

(a) Estimate the years that the deuterium fuel in the oceans could supply the energy needs of the world. Assume world energy consumption to be ten times that of the United States which is \(8 \times {10^9}J/y\) and that the deuterium in the oceans could be converted to energy with an efficiency of \(32\% \). You must estimate or look up the amount of water in the oceans and take the deuterium content to be \(0.015\% \) of natural hydrogen to find the mass of deuterium available. Note that approximate energy yield of deuterium is \(3.37 \times {10^{14}}J/kg\).

(b) Comment on how much time this is by any human measure. (It is not an unreasonable result, only an impressive one.)

Short answer

(a) The years that the deuterium fuel in the oceans could supply the energy needs of the world is\(t = 2.975 \times {10^8}\).

(b) By human measure it would take millions of years to consume all the energy from deuterium.

Step-by-step solution

Concept Introduction

The power kept in the centre of an atom is known as nuclear energy. All stuff in the cosmos is made up of tiny particles called atoms. The centre of an atom's nucleus is typically where most of its mass is concentrated. Neutrons and protons are the two subatomic particles that make up the nucleus. Atomic bonds that hold them together carry a lot of energy.

Mass of deuterium

First, calculate the total amount of water in the oceans which is\(1.33 \times {10^{18}}\;c{m^3}\). So, to get the mass, simply use the standard density formula –

\(m = \rho V\)

Putting the values –

\(\begin{array}{c}m = 1.33 \times {10^{18}}\;c{m^3} \times 1\;g/c{m^3}\\m = 1.33 \times {10^{18}}\;g\end{array}\)

Now the mass of all water is known, so calculate the mass of one deuterium molecule, it has a mass number of\(2\), and diving it with Avogadro–

\({m_d} = \frac{M}{{{N_A}}}\)

Putting the values –

\(\begin{array}{c}{m_d} = \frac{{2\;g/mol}}{{6.022 \times {{10}^{23}}\;mo{l^{ - 1}}}}\\{m_d} = 3.32 \times {10^{ - 24}}\;g\end{array}\)

Since the ratio of deuterium within the hydrogen inside the water is known, calculate the water molecules. One water molecule has a molar mass of\(18\;g/mol\)–

\(\begin{array}{c}n = \frac{m}{M} = \frac{{1.33 \times {{10}^{18}}\;g/g}}{{18\;g/mol}}\\n = 7.28 \times {10^{16}}\;mol\end{array}\)

To get the absolute number of water molecules, simply multiply the number of moles with Avogadro number –

\(\begin{array}{c}N = 4.44 \times {10^6}\;mol \times 6.022 \times {10^{23}}\;mo{l^{ - 1}}\\N = 4.38 \times {10^{40}}{\rm{ }}molecules\end{array}\)

Only\(0.015\% \)of calculated water molecules is possibly having deuterium inside, so a number of deuterium will be –

\(\begin{array}{c}{N_d} = 0.00015 \times 4.38 \times {10^{40}}molecules\\{N_d} = 6.57 \times {10^{36}}\end{array}\)

And knowing the mass of the deuterium molecule we know that the total mass of all deuterium in the whole ocean will be –

\(\begin{array}{c}{m_{all}} = 6.57 \times {10^{36}} \times 3.32 \times {10^{ - 24}}\;g\\{m_{all}} = 2.21 \times {10^{16}}\;kg\end{array}\)

Calculation for time

To get approximate total energy with \(32\% \) efficiency multiply yield with total mass and efficiency –

\(\begin{array}{c}E = eY{m_{all}}\\E = 0.32 \times 2.21 \times {10^{16}}\;kg \times 3.37 \times {10^{14}}\;J/K\\E = 2.38 \times {10^{29}}\;J\end{array}\)

To get the times of world consumption divide this energy by the yearly consumption –

\(\begin{array}{c}t = \frac{{{E_t}}}{{{E_y}}}\\t = \frac{{2.38 \times {{10}^{29}}\;J}}{{80 \times {{10}^{19}}\;J/y}}\\t = 2.975 \times {10^8}\end{array}\)

Therefore, the value for time is obtained as \(t = 2.975 \times {10^8}\).

Time in human measure

It would take a million years to consume all this energy from deuterium. We know that Earth is old more than \(4\) billion, so these years one could consume deuterium energy, that would mean impressive achievement and everyone would be able to live long without energyworries. Energy used from this deuterium would be green and without pollution.

Therefore, by human measure it is million years.