Determine the energy that can be extracted from 1 kg of water via...

Number of water molecules in $1 \text{ kg}$ of water is:

$N = \dfrac{1000}{18} \times 6.02 \times 10^{23} \approx 3.34 \times 10^{25}$

Number of $\text{D}_2\text{O}$ molecules is:

$N_{\text{D}_2\text{O}} = \dfrac{1.5 \times 10^{-2}}{100} \times 3.34 \times 10^{25} = 5.01 \times 10^{21}$

Since each $\text{D}_2\text{O}$ molecule contains two deuterium atoms, the total number of deuterium atoms is:

$N_{\text{D}} = 2 \times 5.01 \times 10^{21} = 1.002 \times 10^{22}$

In the given fusion reaction, two deuterium atoms fuse to release approximately $4.0 \text{ MeV}$ of energy. The number of fusion reactions is:

$n = \dfrac{N_{\text{D}}}{2} = 5.01 \times 10^{21}$

Total energy released is:

$E = n \times 4.0 \text{ MeV} = 5.01 \times 10^{21} \times 4.0 \times 10^{6} \times 1.6 \times 10^{-19} \text{ J}$

$E \approx 32 \times 10^{8} \text{ J} = 3200 \text{ MJ}$