What is the amount of energy (J or Wh) needed to boil one litre of tap-water in a regular kettle starting at room temperature (20°C)?

The calculation itself assuming ideal and standard prerequisites would be interesting as well as factoring in the non-purity of tap-water and a typical kettle performance factor.


1 Answer 1


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

$Q = c \cdot m \cdot \left( {100{\text{ }^\circ\text{C}} - T} \right) + \Delta {h_{\text{vap}}} \cdot m$


  • $c$ = 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!

  • $\begingroup$ The answer assumes total vaporization of the water. If you only want to heat the water to the boiling point e.g. to make some tea, then the second term should be Δhvap⋅m1, where m1 stands for the water that got vaporized in the process. Typically 0<m1<m. $\endgroup$
    – QT-1
    Aug 10, 2017 at 11:03

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