Amount of Energy Needed for Boiling a Litre of Water
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!