These tasks: m1 (mass of heated water) = 200 g (0.2 kg); m2 (mass of water to be converted into steam) = 10 g (0.01 kg); t (water temperature before heating) = 20 ºС.
Constants: C (specific heat capacity of water) = 4200 J / (kg * K); tpap (vaporization temperature) = 100 ºС; L (specific heat of vaporization) = 23 * 10 ^ 5 J / kg.
The required amount of heat: Q = Q1 + Q2 = C * m1 * (tp – t) + L * m2.
Calculation: Q = 4200 * 0.2 * (100 – 20) + 23 * 10 ^ 5 * 0.01 = 90200 J (90.2 kJ).
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