Question

Of all the hydrogen in the oceans, 0.0300 % of the mass is deuterium. The oceans have a volume of 317 million mi³.(a) If nuclear fusion were controlled and all the deuterium in the oceans were fused to ⁴₂He, how many joules of energy would be released?

Answer 1

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×$10^-^1^2$J

How is the energy in joules calculated for a controlled nuclear fusion with all the deuterium in oceans fused to ⁴₂He?

Given the percentage of hydrogen in the ocean = 0.300%

Volume of the ocean = 319 million mi³

Total volume of the ocean, V = 319 × $10^6$ mi³

V = 1.330×$10^1^8$ $m^3$

Density of water = 1000 kg/$m^3$

Avogadro's number =6.022 × $10^2^3$

There are two hydrogen atoms in a water molecule

Molar mass of Hydrogen =2.016×$10^-^3$ kg/mole

Molar mass of deuterium molecule = 4.028204 × $10^-^3$kg/mole

Total mass of sea = Volume × Density

= 1.330 × $10^2^1$ 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 × $10^4^6$

Total mass of hydrogen in the given mass of water =

[Number of water molecules × Molar mass of hydrogen molecule] /  Avogadro's number

= 1.488 × $10^2^0$ kg

Total mass of deuterium in the calculated mass of hydrogen =

Mass of hydrogen × [ density / 100] =

4.464 × $10^1^6$ kg

Total number of deuterium atoms available=

[Total mass of deuterium / Molar mass of deuterium molecule] × Avogadro's number × 2 = 1.335 × $10^4^6$

The combined process consumes 6 deuterium atoms and produces two helium atoms and a total of

E = 43.2×$10^6$ eV of energy.

Therefore the energy release in the consumption of 6 atoms

E=43.2×$10^6$×1.6×$10^1^9$$10^-^1^9$

E=6.912×$10^-^1^2$J

To learn more about nuclear fusion refer

brainly.com/question/23765509

#SPJ4