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?

Expert's answer

The power of kettle is 0.9 * 2x10^{3} = 1.8 x 10^{3} 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

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

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

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