Calculate the heat energy released when 1 litre of ethanol burns, given its density is 789 g/L and the heat of combustion is 1370 kJ/mol.

The atomic masses are approximately: C = $12.01 \text{ g/mol}$, H = $1.008 \text{ g/mol}$, O = $16.00 \text{ g/mol}$.

$$\text{Molar mass of } \text{C}_2\text{H}_5\text{OH} = (2 \times 12.01) + (6 \times 1.008) + (1 \times 16.00)$$ $$= 24.02 + 6.048 + 16.00$$ $$= 46.068 \text{ g/mol}$$

Given the density of ethanol is $789 \text{ g/L}$ and the volume is $1 \text{ L}$.

$$\text{Mass} = \text{Density} \times \text{Volume}$$ $$\text{Mass of ethanol} = 789 \text{ g/L} \times 1 \text{ L}$$ $$= 789 \text{ g}$$

$$\text{Moles} = \frac{\text{Mass}}{\text{Molar mass}}$$ $$\text{Moles of ethanol} = \frac{789 \text{ g}}{46.068 \text{ g/mol}}$$ $$\approx 17.126 \text{ mol}$$

Given that $1 \text{ mole}$ of ethanol releases $1370 \text{ kJ}$ of heat energy.

$$\text{Total heat energy released} = \text{Moles} \times \text{Heat released per mole}$$ $$= 17.126 \text{ mol} \times 1370 \text{ kJ/mol}$$ $$\approx 23462.62 \text{ kJ}$$

Rounding to three significant figures (based on the given density $789 \text{ g/L}$), the heat energy released is:

$$\approx 23500 \text{ kJ}$$

The amount of heat energy released when 1 litre of ethanol is completely burnt is $\boxed{\text{23500 kJ}}$.