2 kg of water is brought to a boil and 100 g of it turns into steam.

2 kg of water is brought to a boil and 100 g of it turns into steam. Determine how much heat is required for this. The initial water temperature is 15 degrees.

Given: m = 2 kg, mp = 0.1 kg; t = 15 oC. find Q.

The amount of heat will be determined as Q = Qn + Qp, where Qn is the amount of heat required to heat water up to 100 oC; Qп – the amount of heat required for vaporization. Qн = C * m * (100 – t), where C is the heat capacity of water, J / (kg * oC); m is the mass of water, kg; t – initial water temperature, oC. Qp = L * mp, where L is the specific heat of vaporization of water, J / kg; mп – mass of evaporated water, kg.

According to the reference book, we find C = 4180 J / (kg * oC); L = 2260000 J / kg.

Substituting the data, we consider: Qн = C * m * (100 – t) = 4180 * 2 * (100 – 15) = 710600J; Qp = L * mp = 2260000 * 0.1 = 226000 J; Q = Qн + Qп = 710600 + 226000 = 936600 J = 936.6 kJ.

Answer: Q = 936.6 kJ.