Determine the energy that can be extracted from 1 kg of water via the fusion reaction ² H + ² H ³ H + p . Assume that 1.5 10⁻² % of natural water consists of heavy water D ₂ O (by number of molecules) and that all the available deuterium undergoes fusion.

Step-by-step solution

Number of water molecules in 1 kg of water is: N = 1000 18 6.02 10²³ 3.34 10²⁵ Number of D ₂ O molecules is: N_ D ₂ O = 1.5 10⁻² 100 3.34 10²⁵ = 5.01 10²¹ Since each D ₂ O molecule contains two deuterium atoms, the total number of deuterium atoms is: N_ D = 2 5.01 10²¹ = 1.002 10²² In the given fusion reaction, two deuterium atoms fuse to release approximately 4.0 MeV of energy. The number of fusion reactions is: n = N_ D 2 = 5.01 10²¹ Total energy released is: E = n 4.0 MeV = 5.01 10²¹ 4.0 10⁶ 1.6 10⁻¹⁹ J E 32 10⁸