Question

How much energy (J) is contained in 1.2 kg of water if its natural deuterium is used in the fusion reaction of 21H+21H?31H+ 11H? Compare to the energy obtained from the burning of 1.2 kg of gasoline, about 6.0×107 J .

Answer 1

The given equation is ₁H² + ₁H² -> ₁H³ + ₁H¹.

To find out the energy (in joule) contained in 1.2 kg of water.

m(₁H²) = 2.014 amu; m(₁H³) = 3.0160 amu; m(₁H¹) = 1.0072 amu

If the sum of products and reactants is considered, then there is a mass defect of 0.004875 amu. This defect in mass is converted into energy for every nucleus of Deuterium or tritium undergoing fusion reaction. The energy is

E = m c²

Thus E = 0.004875 amu X (1.6605402 X 10^-27 kg / 1 amu) X (3 X 10⁸ m / s)²

= (0.0702 X 10^-11) J.

This is for every nucleus of Deuterium or tritium undergoing fusion reaction.

When 1.2 kg of water is used, the energy contained in it is  

18 g of water contains 6.023 x 10²³ atoms

thus 1200 g of water contains 401.53 X 10²³ atoms

Energy released from 1.2 kg of water is = 401.53 X 10²³ x 0.0702 X 10^-11 J

= 28.18 x 10¹² J

The same energy produced when 1.2 kg of gasoline is burned is approximately 1/5 of the amount prouduced.