Speed of Heating Water in the Electric Kettle

Task number: 1792

How much time will an electric kettle with a power input of 1.2 kW and efficiency of 90 % need to boil 2.0 kg of water, whose initial temperature is 10 °C?

Note: Heat loss as the result of heating the kettle itself is included in the efficiency of the kettle, therefore there is no need to take into account the heat capacity of the kettle.

  • Notation

    P = 1.2 kW = 1.2×103 W  power input of the kettle
    η = 90 % = 0.9efficiency of the kettle
    m = 2.0 kgmass of the heated water
    ti = 10 °Cinitial water temperature
    τ = ? time of heating

    Other necessary values

    cw = 4180 Jkg−1K−1specific heat capacity of water
    tw = 100 °C  boiling point of water
  • Analysis

    First of all, we need to determine the amount of heat required for the water to reach the desired temperature. This amount of energy has to be supplied by the electric kettle, whose power output we can compute from its power input and efficiency.

  • Solution

    The heat needs to be supplied to the water \(Q=c_w m (t_w - t_i)\).

    Kettle’s power is equal to \(\eta P\), which means that during the time \(\tau\) the kettle will provide the energy \(E= \eta P \tau\).

    The kettle has to supply the energy required to heat the water. Thus it holds:


    \[\eta P \tau = c_w m (t_w - t_i)\,. \]

    From the last equation we can express the desired time of heating:

    \[\tau = \frac {c_w m (t_w - t_i)}{\eta P} = \frac {4180 \cdot{2} \cdot{(100-10)}}{0.9 \cdot{ 1200}} \,\mathrm s = 697 \,\mathrm s \dot= 12 \,\mathrm {min}\,.\]

  • Answer

    The kettle will bring the water to the boil approximately in 12 minutes.

Difficulty level: Level 1 – Lower secondary level
Task requires extra constants
Tasks requiring comparison and contradistinction
Cs translation
Pl translation
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