How much energy is required to turn water weighing 150 g into steam at a temperature of 100 C.

Problem data: m (mass of water to turn into steam) = 150 g (0.15 kg); t (temperature at which the taken water was located) = 100 ºС (corresponds to the temperature of vaporization).

Constants: L (specific heat of vaporization of water) = 2.3 * 10 ^ 6 J / kg.

The required amount of heat energy is calculated by the formula: Q = L * m.

Let’s make the calculation: Q = 2.3 * 10 ^ 6 * 0.15 = 0.345 * 10 ^ 6 J (345 kJ).

Answer: 345 kJ of thermal energy is required.