How much heat is needed to convert 100 grams of water into steam if its initial temperature was 29 degrees.

Initial data: m (mass of taken water) = 100 g (0.1 kg); t (initial water temperature) = 29 ºС.

Reference data: C (specific heat) = 4200 J / (kg * ºС); tp (temperature of the beginning of water vaporization) = 100 ºС; L (specific heat of vaporization) = 23 * 10 ^ 5 J / kg.

The required amount of heat is calculated by the formula: Q = Q1 + Q2 = C * m * (tp – t) + L * m = (C * (tp – t) + L) * m.

Calculation: Q = (4200 * (100 – 29) + 23 * 10 ^ 5) * 0.1 = 259 820 J.

Answer: You need 259 820 J of heat.