How much energy is spent on heating water weighing 0.75 kg from 20 to 100C and the subsequent formation of steam weighing 250 g?

To determine the cost of thermal energy for heating and vaporization of the taken water, we will use the formula: Q = L * m2 + Sv * m1 * (tp – t0).

Constants and variables: L – specific heat of vaporization (L = 2.3 * 10 ^ 6 J / kg); m2 – steam mass (m2 = 250 g = 0.25 kg); Sv – specific heat (Sv = 4200 J / (kg * ºС)); m1 is the total mass of water (m1 = 0.75 kg); tpar – temperature of vaporization (tpap = 100 ºС); t0 – initial temperature (t0 = 20 ºС).

Calculation: Q = L * m2 + Sv * m1 * (tpar – t0) = 2.3 * 10 ^ 66 * 0.25 + 4200 * 0.75 * (100 – 20) = 827 * 10 ^ 3 J = 827 kJ.

Answer: 827 kJ of thermal energy was spent.