Question

Calculate the energy that can be obtained from 1 kg of water through the fusion reaction
2H + 2H → 3H + p.
Assume that 1.5 × 10−2% of natural water is heavy water D2O (by number of molecules) and all the deuterium is used for fusion.

Answer 1

Explanation:

It is given:

$6.023 \times 10^{25}$ molecules are present in 18 g of water.

Therefore, 1000 g of water $=6.023 \times 100018 \times 1023=3.346 \times 1025$ molecules

% of deuterium $=3.346 \times 0.015100 \times 1025=0.05019 \times 10^{23}$

Deuterium’s energy $=30.4486 \times 1025$

$=2 \times m H 2-m H 3-m p c 2$

$=4.542349 \mathrm{MeV}$

$=7.262 \times 10-13$

Total energy = $0.05019 \times 7.262 \times 10 \times 10^{23}-13 J$

Through the reaction of fusion, from 1 kg of water, the energy that can be gained is 3644 MJ.