Question

a) What is the energy released in the fusion reaction \({ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}+Q ?\) b) The oceans have a total mass of water of \(1.50 \cdot 10^{16} \mathrm{~kg}\), and \(0.0300 \%\) of this quantity is deuterium, \({ }_{1}^{2} \mathrm{H} .\) If all the deuterium in the oceans were fused by controlled fusion into \({ }_{2}^{4} \mathrm{He},\) how many joules of energy would be released? c) World power consumption is about \(1.00 \cdot 10^{13} \mathrm{~W}\). If consumption were to stay constant and all problems arising from ocean water consumption (including those of political, meteorological, and ecological nature) could be avoided, how many years would the energy calculated in part (b) last?

Short answer

In summary, the energy released from controlled fusion of all deuterium present in the oceans would last for approximately 1.07 x 10^3 years if the world's power consumption remained constant at 1.00 x 10^13 W. This calculation assumes that all of the deuterium is used for the fusion process and that there are no issues with water consumption from the oceans.

Step-by-step solution

Calculate the energy released (Q) in the fusion reaction

First, we need to find the difference in mass between the initial reactants and the final product. Then, we can use the mass-energy equivalence relation, \(E=mc^2\), to find the released energy Q. $$\Delta m = m_{initial} - m_{final} = (2\cdot m_{H}) - m_{He}$$ Let's substitute the masses of Hydrogen and Helium: $$\Delta m = (2\cdot 2.014102 \mathrm{amu}) - 4.001506 \mathrm{amu} = 0.026698\mathrm{amu}$$ To convert this mass difference to energy, multiply with the conversion factor \(1.6605 \cdot 10^{-27}\mathrm{ kg/amu}\) and the speed of light squared, \(c^2 = (3.00 \cdot 10^8\mathrm{m/s})^2\): $$Q = \Delta m \cdot c^2 = 0.026698 \mathrm{amu} \cdot 1.6605 \cdot 10^{-27}\mathrm{ kg/amu} \cdot (3.00 \cdot 10^8\mathrm{m/s})^2$$ $$Q = 4.032 \cdot 10^{-12}\mathrm{J}$$

Find the mass of deuterium in the oceans

Given that the total mass of water in the oceans is \(1.50 \cdot 10^{16} \mathrm{~kg}\), and the percentage of deuterium is \(0.0300 \%\), we can calculate the mass of deuterium as follows: $$m_{deuterium} = 1.50 \cdot 10^{16}\mathrm{~kg} \cdot (0.0300/100)$$ $$m_{deuterium} = 4.50 \cdot 10^{13}\mathrm{~kg}$$

Find the number of helium nuclei produced from the deuterium

Since it takes two deuterium nuclei to produce one helium nucleus in the given fusion reaction, we can calculate the number of helium nuclei produced as follows: $$n_{He} = \frac{m_{deuterium}}{2 \cdot m_{H}}$$ Replace \(m_{deuterium}\) and \(m_{H}\) with their values: $$n_{He} = \frac{4.50 \cdot 10^{13}\mathrm{~kg}}{2 \cdot 2.014102 \mathrm{amu} \cdot 1.6605 \cdot 10^{-27}\mathrm{ kg/amu}}$$ $$n_{He} = 8.350 \cdot 10^{28}$$

Calculate the total energy produced by controlled fusion of all deuterium in the oceans

We can compute the total energy released by multiplying the energy released per reaction, Q, by the number of helium nuclei produced, \(n_{He}\): $$E_{total} = Q \cdot n_{He}$$ $$E_{total} = 4.032 \cdot 10^{-12}\mathrm{J} \cdot 8.350 \cdot 10^{28}$$ $$E_{total} = 3.369 \cdot 10^{17}\mathrm{J}$$

Calculate how long the energy would last if the world's power consumption remained constant

Given that the world power consumption is \(1.00 \cdot 10^{13} \mathrm{~W}\), we can find how long the energy from the controlled fusion of deuterium would last by dividing the total energy by the power consumption: $$t = \frac{E_{total}}{P_{consumption}}$$ $$t = \frac{3.369 \cdot 10^{17}\mathrm{J}}{1.00 \cdot 10^{13}\mathrm{~W}}$$ $$t = 3.369 \cdot 10^4\mathrm{s}$$ To convert this time to years, divide by the number of seconds in a year: $$t = \frac{3.369 \cdot 10^4\mathrm{s}}{3.1536 \cdot 10^7\mathrm{s/year}}$$ $$t \approx 1.07 \cdot 10^3\mathrm{~years}$$ So, if the world's power consumption remained constant and all problems arising from ocean water consumption could be avoided, the energy released from controlled fusion of all deuterium in the oceans would last for approximately \(1.07 \cdot 10^3\) years.

Key concepts

Mass-Energy Equivalence

The concept of mass-energy equivalence is central to understanding nuclear fusion energy calculations. It is based on the famous equation by Albert Einstein, denoted as \( E=mc^2 \). This equation tells us that mass \((m)\) and energy \((E)\) are interchangeable; they are different forms of the same thing. In simpler terms, a certain amount of mass can be converted into energy.

In the context of nuclear fusion, tiny differences in mass can result in large amounts of energy being released. This happens because the speed of light \((c)\), which is squared in the equation, is a very large number \((3.00 \times 10^8\ \text{m/s})\). For example, when two deuterium nuclei (a form of hydrogen) fuse to form helium, there is a small loss in mass because the mass of the helium nucleus is slightly less than the combined mass of the original deuterium nuclei.

This lost mass, calculated through the mass-energy equivalence formula, results in the release of a substantial amount of energy, which is the core principle behind harnessing fusion power. Understanding this concept is essential because it shows how a small amount of mass, usually in atomic scales, contributes to enormous energy outputs.

Nuclear Fusion

Nuclear fusion is the process where two light atomic nuclei combine to form a heavier nucleus, releasing energy in the process. It is the opposite of nuclear fission, where heavy atoms split apart. In a fusion reaction, like the one involving deuterium - \({}_{1}^{2} \text{H} + {}_{1}^{2} \text{H} \rightarrow {}_{2}^{4} \text{He} + Q\), deuterium nuclei (isotopes of hydrogen) collide and fuse to form helium.

This reaction does not occur easily; it requires extremely high temperatures and pressures to overcome the electric force of repulsion between the positively charged nuclei. Once the conditions are met, however, the strong nuclear force takes over, allowing the two nuclei to bond and form a new element. The energy released \((Q)\) in this reaction comes from the small amount of mass that is lost in the process, which, as previously mentioned, is converted into energy according to the mass-energy equivalence principle.

Fusion reactions like this are the power source of stars, including our Sun. They hold great potential for being a sustainable and clean energy source here on Earth, as they produce negligible amounts of radioactive waste compared to nuclear fission. The challenge lies in creating and maintaining the extreme conditions necessary for controlled fusion, which researchers continue to work on.

World Power Consumption

World power consumption refers to the total amount of energy used by all activities across the globe. It includes the energy required for electricity, heating, industrial processing, and transportation, among many others. As of the current figures, this is approximately \(1.00 \times 10^{13} \text{W}\), which is a significant demand on global energy resources.

The exercise presented above estimates how long the energy from the fusion of all deuterium in the oceans could meet this demand. With a potential energy release of around \(3.369 \times 10^{17}\ \text{J}\) from oceanic deuterium reserves, this covers roughly \(1.07 \times 10^3\) years of this level of consumption, provided that the consumption remains constant and assuming all technological and environmental barriers can be overcome.

This hypothetical scenario highlights the immense energy potential held in fusion, suggesting a nearly inexhaustible energy source. Transitioning to such a source could greatly alleviate reliance on fossil fuels and reduce the environmental impact of energy production. It emphasizes the need for ongoing research and development in fusion technology to make such a future feasible.