Calculate the heat energy released when 1 liter of ethanol is burned.

The atomic masses are approximately: C = $12.01 \text{ g/mol}$, H = $1.008 \text{ g/mol}$, O = $16.00 \text{ g/mol}$.
The formula for ethanol is $\text{C}_2\text{H}_5\text{OH}$, which can also be written as $\text{C}_2\text{H}_6\text{O}$.

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

Given that 1 liter of ethanol has a mass of $780 \text{ g}$.

$$\text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}}$$ $$\text{Number of moles} = \frac{780 \text{ g}}{46.068 \text{ g/mol}} \approx 16.931 \text{ mol}$$

Given that $1370 \text{ kJ}$ of heat energy is released per mole of ethanol.

$$\text{Total heat energy released} = \text{Number of moles} \times \text{Heat released per mole}$$ $$\text{Total heat energy released} = 16.931 \text{ mol} \times 1370 \text{ kJ/mol}$$ $$\text{Total heat energy released} \approx 23195.67 \text{ kJ}$$

Rounding to three significant figures (due to the $780 \text{ g}$ value):

$$\text{Total heat energy released} \approx 23200 \text{ kJ}$$

The amount of heat energy released when 1 liter of ethanol is completely burned is $\boxed{\text{23200 kJ}}$.