Of all the hydrogen in the oceans, 0.0300% of the mass is...

Of all the hydrogen in the oceans, 0.0300$\%$ of the mass is deuterium. The oceans have a volume of 317 million mi $^{3}$ . (a) If nuclear fusion were controlled and all the deuterium in the oceans were fused to $\frac{4}{2}$ He, how many joules of energy would be released? (b) What If? World power consumption is approximately $7.00 \times 10^{12} \mathrm{W}$ . If consumption were 100 times greater, how many years would the energy calculated in part (a) last?

Key Concepts

Energy Yield Calculation

Calculating the energy yield from a nuclear fusion reaction involves determining the mass difference between the reactants and the products, which is then converted into energy according to Einstein's mass-energy equivalence principle (E = mc²). This process requires knowledge of the reaction’s specific energy output per fusion event and scaling that value to the total amount of fuel available.

Dimensional Analysis and Unit Conversion

In complex energy calculations, converting units such as volume to mass, or converting energy units (for instance from joules to kiloton equivalents), is essential. This includes understanding the density of materials, the volume of a substance, and working with various physical constants to accurately scale laboratory measurements to real-world quantities.

Nuclear Fusion

Nuclear fusion is the process in which two light atomic nuclei combine to form a heavier nucleus, releasing a considerable amount of energy in the process. It is the fundamental reaction that powers stars, and controlled fusion is being studied as a potential source of power on Earth. The process relies on overcoming the electrostatic repulsion between positively charged nuclei, typically requiring high temperatures and pressures to initiate the reaction.

Deuterium as a Fusion Fuel

Deuterium is an isotope of hydrogen that contains one proton and one neutron in its nucleus. It is considered an ideal fuel for nuclear fusion because it is relatively abundant and has a lower nuclear charge compared to heavier nuclei, which makes fusion reactions more attainable under controlled conditions. The energy released from fusing deuterium is significant, making it a key candidate in the study of fusion-based energy production.

Power Consumption and Energy Duration

Assessing how long a given amount of energy will last involves comparing it to the rate at which energy is consumed, typically measured in watts (joules per second). This concept is crucial in energy policy and resource management, as it ties together the total energy available with the demands of real-world power consumption over time.

Transcript

00:01 Hi everyone, this is the problem based on nuclear fusion reaction.

00:05 Here it is given volume of the water in ocean is 317 million of miles cube, that is 1 .32 into 10 to the power 18 meter cube.

00:32 Here it is given the mass of deuterium present is 0 .003 % of the percent of, of 10 ,000, of, sorry, 0 .03 % of mass of h2 in the ocean.

00:56 We have to calculate amount of energy released if all deuterium in ocean fused to helium.

01:19 In second part we have to calculate the time duration for which this energy can meet the energy of the world requirement if world power consumption is 100 times of 7 into 10 to the power 12 watt, let us start solving it.

02:04 Mass of water we will find volume into density.

02:11 Volume density is 1000.

02:16 Volume is 1 .32 into 10 to the power 80.

02:22 So it is to be 1 .32 into 10.

02:28 10 to the power 18 kg.

02:31 So mass of h2 will be molecular mass of hydrogen gas upon molecular mass of water into total mass of water present in the ocean.

02:45 So it is to be 2 .016 upon 18 .015 into 1 .3 to 10 to the power 18 kg.

02:59 So on solving it it is to be 1 .48 10 to the power 20 kg.

03:09 Mass of deuterium in the ocean you can find that is 0 .03 percentage of mass of that is 0 .003 into 1 .48 10 to the power 20 so you will get 4 .43 10 to the power 60 so you will get 4 .43 10 to the power 60 now number of deuterium atoms will be mass of deuterium upon molecular weight of deuterium into abogadro number, that is 4 .43, 10 to the power 16, 6 .023, 10 to the power 23, 10 to the power 23 upon mass of deuterium...

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