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

Question #3368

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?

i) how long does this take?

ii) how long more to completely boil off the water?

Expert's answer

The power of kettle is 0.9 * 2x10

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

t

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)

t

Total time = t

^{3}= 1.8 x 10^{3}WW = 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

t

_{1}= 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)

t

_{2}= Q/W = 2270000 /1800 = 1261 s = 21 minute.Total time = t

_{1}+ t_{2}= 186 + 1261 = 1447 s
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