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Of all the hydrogen in the oceans, 0.0300 % of the mass is deuterium. The oceans have a volume of 319 million mi³. If nuclear fusion were controlled and all the deuterium in the oceans were fused to ⁴₂He, the joules of energy released is E = 6.912×J
Given the percentage of hydrogen in the ocean = 0.300%
Volume of the ocean = 319 million mi³
Total volume of the ocean, V = 319 × mi³
V = 1.330×
Density of water = 1000 kg/
Avogadro's number =6.022 ×
There are two hydrogen atoms in a water molecule
Molar mass of Hydrogen =2.016× kg/mole
Molar mass of deuterium molecule = 4.028204 × kg/mole
Total mass of sea = Volume × Density
= 1.330 × kg
Total number of water molecules available in the given mass of water =
[Mass of sea water / Molar mass of water] × Avogadro's number
= 4.446 ×
Total mass of hydrogen in the given mass of water =
[Number of water molecules × Molar mass of hydrogen molecule] / Avogadro's number
= 1.488 × kg
Total mass of deuterium in the calculated mass of hydrogen =
Mass of hydrogen × [ density / 100] =
4.464 × kg
Total number of deuterium atoms available=
[Total mass of deuterium / Molar mass of deuterium molecule] × Avogadro's number × 2 = 1.335 ×
The combined process consumes 6 deuterium atoms and produces two helium atoms and a total of
E = 43.2× eV of energy.
Therefore the energy release in the consumption of 6 atoms
E=43.2××1.6×
E=6.912×J
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