1. The problem statement, all variables and given/known data A domestic kettle is marked 250 V, 2.3 kW and the manufacturer claims that it will heat a pint of water to boiling point in 94 s. (a) Test this claim by calculation and state any simplifying assumptions you make. (b) If the kettle is left switched on after it boils, how long will it take to boil away half a pint of water measured from when it first boils? (c) Estimate the work done against an atmospheric pressure of 100 kPa when 1 cm3 of water evaporates at 100 °C, producing 1600 cm3 of steam. Express this as a percentage of the total energy required to evaporate 1 cm3 of water at 100 °C. Specific heat capacity of water = 4.2 * 103 J kg-1 K-1 Specific latent heat of vaporisation of water = 2.3 * 106 J kg-1 Density of water = 1.0 g cm-3 1 pint = 570 cm3 Answers: (c) 1.6 * 102 J, 7.0 % 2. The attempt at a solution (a) We need to find Δθ in ΔQ = m c Δθ. Q is V I t, where I = P / V = 2300 / 250 = 9.2 A. Then find Q = 250 * 9.2 * 94 = 216 200 J. m is 1 g cm-3 * 570 cm3 = 570 g. 570 g / 1000 = 0.57 kg. Substitute for Δθ = ΔQ / mc = 216 200 / (0.57 * 4.2 * 103) = 90.3 °C. Water boils at 100 °C, so the manufacturer is wrong. In 94 s we'll only get to 90.3 °C. (b) The mass of water that will be in the kettle at 100 °C: m = ΔQ / cΔθ = 216 200 / (100 * 4.2 * 103) = 0.51 kg. Find ΔQ with half of that mass: ΔQ = 0.26 * 4.2 * 103 * 100 = 108 100 J. Find time: t = Q / VI = 108 100 / 250 * 9.2 = 47 seconds. That is half the claimed boiling time. Maybe it was possible to just divide the boiling time (94 s) by two? Since at boiling time half of the water will boil out in half the time (same as a third in a third and 1/4 in 1/4 etc.). I didn't start (c) since I am not sure whether both (a) and (b) are correct. I also worked on this problem some while ago and in (a) I had a different mass = 0.473 kg and I got the temperature to be equal to 109.35 °C (so the manufacturer is right). But I can't remember at all how did I derive that number, maybe it is wrong and this is correct: m is 1 g cm-3 * 570 cm3 = 570 g. 570 g / 1000 = 0.57 kg. Any thoughts on (a) and (b) please?