How much heat is needed to convert 100 grams of water into steam, if its initial temperature was 20 degrees.

Data: m (mass of water to be converted to steam) = 100 g (0.1 kg); t0 (initial temperature) = 20 ºС.

Constants: C (specific heat capacity of water) = 4200 J / (kg * K); tpap (temperature of the beginning of vaporization) = 100 ºС; L (specific heat of vaporization of water) = 2.3 * 10^ 6 J / kg.

1) Water heating: Q1 = C * m * (tpap – t) = 4200 * 0.1 * (100 – 20) = 33600 J.

2) Vaporization: Q2 = L * m = 2.3 * 10:^6 * 0.1 = 230,000 J.

3) Required heat: Q = Q1 + Q2 = 33600 + 230,000 = 263600 J (263.6 kJ).