Answer to Question #3368 in Molecular Physics | Thermodynamics for Andrew

Question

A 2kW kettle, 90% energy efficient is used to boil 1 litre of water from 20°C to 100°C.

i) how long does this take?
ii) how long more to completely boil off the water?

Answer 1

The power of kettle is 0.9 * 2x10³ = 1.8 x 10³ W
W = Q/t, t = Q/W
The water is boiled from 20 to 100, therefore, the heat is
Q = C m ΔT = 4.1855 [J/gK]* 1000[g] * 80[K] = 334 840 J
Thus
t₁ = 334840 / 1800 = 186 s = 3.1 minute.
To completely boil Q2 = λ m = 2 270 [kJ/kg] * 1 [kg] = 2 270000 J is required. (λ - Latent heat of evaporation - 2.270 kJ/kg)
t₂ = Q/W = 2270000 /1800 = 1261 s = 21 minute.
Total time = t₁ + t₂ = 186 + 1261 = 1447 s