How much heat is needed to bring 5 kg of water at 0 ° C to a boil and evaporate it?

univerkov | March 25, 2021 | education

Given: m (mass of water) = 5 kg; t0 (temperature at which water was taken) = 0 ºС.

Reference values: С (specific heat capacity of water) = 4200 J / (kg * ºС); tboil (temperature of the beginning of boiling) = 100 ºС; L (specific heat of vaporization) = 2256 * 10 ^ 3 J.

1) Bringing water to a boil: Q1 = C * m * (tboil – t0) = 4200 * 5 * (100 – 0) = 2100000 J.

2) Evaporation of 5 kg of water: Q2 = L * m = 2256 * 10 ^ 3 * 5 = 11280000 J.

3) The expended heat: Q = Q1 + Q2 = 2100000 + 11280000 = 13380000 J (13.38 MJ).

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