It is simple. Density of water is $1\;\frac{\text{g}}{\text{ml}}$ as we have 1 kg water.
Now according to the laws of calorimetry the amount of heat required
$Q = c \cdot m \cdot \left( {100{\text{ }^\circ\text{C}} - T} \right) + \Delta {h_{\text{vap}}} \cdot m$
Here
- $c$ = specific heat capacity of water, which is 1 cal/(g °C)
- $m$ = mass of water; i.e. 1 kg
- $T$ = the room temperature in °C
- $\Delta {h_{\text{vap}}}$ = specific enthalpy of vaporization of water; 540 cal/g
So you will get the amount of energy needed (in cal) to boil a litre of water
To get answer in Joule multiply the answer in cal by 4.184 J/cal. And be careful with units!