Answer to Question #3368 in Molecular Physics | Thermodynamics for Andrew
i) how long does this take?
ii) how long more to completely boil off the water?
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
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
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