My kettle is rated 2000 W and holds 1.5 L. That’s all the data you need for the problem I use to open the electricity unit: how long does a full kettle take to boil, what does it cost, and where does the energy actually go? The arithmetic is short. The physics underneath it took James Joule years.
Start with the water. One liter of water weighs 1 kg, so 1.5 L is m = 1.5 kg. My tap runs at 20 °C and the kettle stops at 100 °C, so ΔT = 80 °C. Water needs 4186 J to heat one kilogram by one degree, a number worth memorizing because it comes up constantly. Total energy: Q = mcΔT = 1.5 × 4186 × 80 = 502,320 J, call it 5.0 × 10⁵ J.
The arithmetic
Power is energy per second, so time is energy divided by power: t = 502,320 J / 2000 W = 251 s. That’s 4 minutes 11 seconds. I ran the experiment before writing this: 4 min 30 s. The difference means about 90% of the electrical energy landed in the water, and the rest heated the kettle body, the counter, and my patience. That 90% is typical for a kettle, which is one reason I boil water electrically instead of on a gas ring.
A caution that trips up half the class: once the water hits 100 °C, the temperature stops climbing. The next 2.26 × 10⁶ J per kilogram goes into turning water into steam while the thermometer sits still. That flat line is water’s built-in thermostat.
Where the heat comes from
The heating element works by Joule heating: push current through a resistance and the electrical energy turns into thermal energy. The relations are P = IV and P = I²R. At 230 V, my kettle draws I = 2000/230 = 8.7 A, and its element has resistance R = V/I = 230/8.7 = 26 Ω. Every electron that squeezes through those 26 ohms pays a toll, and the toll is heat.
Joule established the exchange rate in 1843 with a falling weight turning paddle wheels in water: mechanical work becomes heat at 4.186 J per calorie. The old site had his experiment on the front page for years. Same physics, better kettle.
The same I²R explains two household facts worth knowing. Handles are plastic because they conduct heat poorly and cost little; the element is metal because it must resist current and survive 300 °C. And a worn socket with 0.5 Ω of contact resistance at 8.7 A dissipates I²R = 38 W, which is a small nightlight hidden in your wall. That’s why I check plugs before I check anything else.
The bill
Energy is power times time, so a boil is 2000 W × 251 s = 5.0 × 10⁵ J. Divide by 3.6 × 10⁶ J per kilowatt-hour and you get 0.14 kWh. At 30 cents per kWh, a full kettle costs about 4 cents. Four boils a day, every day, is under 50 cents a week. Tea remains one of the cheaper thermodynamic pleasures available.
Your numbers will differ: a 2200 W kettle, colder tap water, a lid left open. Redo the multiplication with your own measurements and see what your boil costs. If the clock disagrees with your calculation by more than 10%, something in your house is losing heat faster than mine.