How much heat must be spent to turn 2 kg of water at a temperature of 10 s into steam at 100 s?

The amount of energy required for vaporization:
Q = Q1 + Q2.
Q1 = C * m * (tк – tн), C – specific heat capacity of water (С = 4187 J / (K * kg)), m – mass of water (m = 2 kg), tк – final temperature (tк = 100 ºС ), tн – initial temperature (tн = 10 ºС).
Q2 = L * m, L is the specific heat of vaporization of water (L = 2256 kJ / kg = 2256 * 10 ^ 3 J / kg).
Q = Q1 + Q2 = C * m * (tк – tн) + L * m = 4187 * 2 * (100 – 10) + (2256 * 10 ^ 3) * 2 = 753660 + 4512000 = 5265660 J ≈ 5.27 MJ.
Answer: 5.27 MJ of heat is needed for vaporization.