Kettle Efficiency and the Temperature of Steam
A student checks the specific latent heat of vaporisation of water with a 2.0 kW electric kettle. She puts 1.50 kg of water at 25 °C into the kettle and switches it on. After 5.0 minutes the water has been boiling for a while, the kettle is still on, and a balance shows that 45 g of water has left the kettle. The room is at 25 °C and the pressure is 1 atm. Specific heat capacity of water = 4 200 J kg⁻¹ K⁻¹; specific latent heat of vaporisation = 2.26 × 10⁶ J kg⁻¹; boiling point at 1 atm = 100 °C. The kettle body and element are cold at the start.
Her partner says the experiment 'proves' the kettle is 100 % efficient because the water was boiling at the end. He also says the steam leaving the spout must be hotter than 100 °C because it 'holds the latent heat'.
1. Calculate the electrical energy supplied, the energy needed to heat the water to 100 °C, the energy carried away as steam, and the overall efficiency. Hence judge the partner's 100 % claim.
Read the worked answer
Energy supplied: E = Pt = 2.0 × 10³ × 300 = 6.0 × 10⁵ J.
Sensible heat: Q_h = mcΔθ = 1.50 × 4 200 × 75 = 4.725 × 10⁵ J, which is already 78.8 % of the supply.
Steam: Q_v = mL = 0.045 × 2.26 × 10⁶ = 1.017 × 10⁵ J.
Accounted for: 4.725 × 10⁵ + 1.017 × 10⁵ = 5.74 × 10⁵ J, so efficiency = 5.74/6.00 = 95.7 %, leaving about 2.6 × 10⁴ J (4 %) unaccounted for - consistent with losses through the unlagged walls, heating the element and body, and steam escaping before the balance was read.
Judging the claim: the number is close to 100 %, but the reasoning is wrong. 'The water was boiling at the end' only shows that the heating stage finished inside the 5 min; it says nothing about losses. Note also that 45 g read on a kitchen balance carries ±5 g uncertainty, which alone shifts the efficiency by ±2 %. The honest statement is 'about 96 %, i.e. 100 % within the precision of the measurement', not a proof of perfect efficiency.
2. Use kinetic theory to answer the partner's second claim: is steam at 100 °C hotter than water at 100 °C? Why then does steam burn more severely?
Read the worked answer
Temperature measures the mean translational kinetic energy of the molecules, ½m⟨c²⟩ = (3/2)kT. It is not a measure of total energy.
At 100 °C and 1 atm the molecules in the liquid water and in the saturated steam have the same mean translational KE - otherwise heat would flow between them. The latent heat does not go into speeding the molecules up: it is used to do work against the attractive forces between molecules (raising their potential energy) as they separate, plus the small pΔV work of pushing back the atmosphere. So saturated steam leaving the spout is at 100 °C, not above it. It could only be superheated if the element continued to heat it after evaporation, which is not what happens here.
Steam scalds far worse because it condenses on skin, delivering L = 2.26 × 10⁶ J per kilogram on top of the cooling of the condensed water. The damage comes from the high internal energy of the vapour, not from a higher temperature - a point the partner has conflated.
3. The measured 45 g may differ from the mass the energy budget predicts. Give two effects that would make the measured loss too small and one that would make it too large, and suggest one change that would make the result more trustworthy.
Read the worked answer
The energy budget predicts an ideal m = (6.0 × 10⁵ − 4.725 × 10⁵)/2.26 × 10⁶ = 56 g.
Too small:
- The kettle body, element and base plate have their own heat capacity, so part of the 6.0 × 10⁵ J never reaches the water; less energy is left for boiling, so less than 56 g is vaporised.
- Heat loss through the unlagged wall and around the lid to the 25 °C room reduces the energy available for boiling.
Too large: water evaporates from the open surface and from splashes below 100 °C, and the kettle keeps steaming while it is carried to the balance - these add mass loss that the calculation does not include.
Improvement: lag the kettle and fit the lid, then weigh the kettle (not just the water) immediately before switching on and immediately after switching off, on a balance reading to 1 g. Better still, calibrate the kettle's own heat capacity by a blank run with a known mass of water and no boiling, then compute L = (Pt − mcΔθ − C_kettle Δθ)/m_steam. Stirring before switching off ensures the whole 1.50 kg really is at 100 °C when the timing is stopped.