Kettle Efficiency, a Missing Lid and a Claim About Destroyed Energy
A 2.0 kW kettle is filled with 0.60 kg of water at 24 degrees Celsius and switched on. A student times it to reach a steady 100 degrees Celsius: 115 s. She calculates the energy needed using c = 4200 J/kg degrees Celsius and is surprised that the kettle appears to be only about 83% efficient. In a second trial she leaves the lid off; the same quantity of water at the same starting temperature now takes 140 s. She notices that the element continues to bubble briefly after the switch clicks off, and that the kettle's metal base becomes hot even where no water touches it. She writes: 'The missing 17% of the energy has been destroyed by the kettle.' The kettle is rated at 2000 W at 240 V, but the laboratory supply measures 232 V.
1. Calculate the energy needed to heat the water and the efficiency of the kettle, and comment on the figure obtained.
Read the worked answer
Energy to the water:
E = mc(delta theta) = 0.60 x 4200 x 76 = 191 520 J (about 1.9 x 10^5 J)
At the rated 2000 W, energy supplied in 115 s = 2000 x 115 = 230 000 J.
Efficiency = 191 520 / 230 000 = 0.83 = 83%
However, the rating is only nominal. At 232 V the power is approximately
P = 2000 x (232/240)^2 = 1868 W
so the input in 115 s is about 214 800 J and the efficiency is nearer 89%. The true figure is therefore uncertain, and the quoted 17% loss is an overestimate.
The remaining energy heats the element, the metal body and base plate, and is lost by convection and radiation from the sides, and by evaporation at the surface.
2. Explain, using specific heat capacity and latent heat, why the trial with the lid off took 140 s.
Read the worked answer
Water evaporates from the surface even well below 100 degrees Celsius, and each kilogram evaporated carries away about 2.26 x 10^6 J of latent heat. That is roughly 540 times the 4200 J needed to raise 1 kg by 1 degree, so a small loss of vapour carries a large loss of energy.
The extra 25 s at about 1868 W is about 4.7 x 10^4 J, which would evaporate 4.7 x 10^4 / 2.26 x 10^6 = 0.021 kg, about 3.5% of the water. That is a plausible amount to lose in a few minutes.
The lid does two jobs:
- it traps vapour, so some of the latent heat is returned when the vapour condenses on the underside;
- it suppresses convection, keeping a layer of warm air above the water instead of letting cool air circulate in.
With the lid off, steam escapes permanently and cooler air replaces it, so the water must be supplied with more energy to reach the same temperature.
3. The student writes that 'the missing 17% of the energy has been destroyed.' Evaluate this statement.
Read the worked answer
It is wrong. Energy is conserved; it is not destroyed but transferred by heating to the surroundings - the element, the kettle body, the bench and the air - and carried away by escaping vapour. The student is confusing 'not useful' with 'destroyed'.
The useful transfer is the rise in internal energy of the water; the rest is dissipated, and in principle it could be recovered (a lagged kettle with a lid loses less).
Two further points she has missed:
- The bubbling after switch-off is the stored thermal energy of the hot element continuing to boil a little water, not extra electrical input. So the stopwatch timing includes some heat delivered after the circuit opens.
- Some of the electrical energy heats the element itself, which is why the base is hot where no water touches it - an unavoidable loss unless the element is fully immersed and the body well insulated.
A well-lagged, lidded kettle could approach 100%, but never reach it, because a temperature difference between kettle and room always drives some transfer to the surroundings.