Problem — boiling time of a kettle

Problem

Boiling time (Povey 2015, §9.7). An electric kettle draws power $P$ from the mains and takes a time $\tau$ to bring a litre of water to the boil, starting from $20\ ^\circ\text{C}$. Another model, of a different brand, has internal resistance $R'$ half that of the first, but is connected to the same mains. Without doing numerical calculations, predict the time $\tau'$ of the second kettle (as a function of $\tau$), neglecting losses to the environment. Discuss what changes if the losses to the environment are not negligible.

Solution

Setup. Both kettles are connected to the same mains voltage $V$. The power dissipated by Joule heating at fixed voltage is $P = \frac{V^2}{R}$ so halving the resistance doubles the power: $P' = V^2/R' = V^2/(R/2) = 2P$.

Ideal case (negligible losses). The energy needed to heat the same litre of water by the same $\Delta T$ is the same: $E = m c \,\Delta T$. The time is energy divided by power: $\tau' = \frac{E}{P'} = \frac{E}{2P} = \frac{1}{2}\,\frac{E}{P} = \frac{\tau}{2}$ The second, more powerful kettle takes half the time.

With non-negligible losses. While the water heats, some of the heat is dissipated to the environment at a rate that depends mainly on the temperature difference, not on the kettle’s power. The more powerful kettle heats faster, so it stays on for less time: there is less time for the heat to be lost, and the fraction of energy “wasted” to the environment is smaller. As a result the second kettle is somewhat more efficient and $\tau' > \frac{\tau}{2}$ (the ratio remains, in any case, better for the more powerful model). In short: the “useful” fast-heating time wins out over the “wasted” time.

$\ev{\tau' = \tau/2 \text{ (without losses)}\;;\quad \tau' > \tau/2 \text{ (with losses)}}$

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Topics: Circuiti elettrici Concepts: Effetto Joule Skills: Analisi di casi limite e fantafisica