Find the amount of energy released when heating water from t = 20 ° to boiling and evaporation

Find the amount of energy released when heating water from t = 20 ° to boiling and evaporation of 1/2 water Vwater = 6 liters

Given:

V = 6 liters = 0.006 m ^ 3 – volume of heated water;

V1 = V / 2 – volume of evaporated water;

ro = 1000 kg / m ^ 3 – water density;

T1 = 20 degrees Celsius – initial water temperature;

T2 = 100 degrees Celsius – boiling point of water;

s = 4200 J / (kg * C) – specific heat capacity of water;

q = 2258 kJ / kg = 2258000 J / kg is the specific heat of vaporization of water.

It is required to determine Q (Joule) – the amount of heat for heating and evaporation of water.

Let’s find the mass of water:

m = ro * V = 1000 * 0.006 = 6 kilograms.

Qheating = m * c * dT = m * c * (T2 – T1) = 6 * 4200 * (100 – 20) = 6 * 4200 * 80 = 2016000 Joules.

Qevaporation = m * q / 2 = 6 * 2258000/2 = 3 * 2258000 = 6774000 Joules.

Q = Qheat + Qevaporation = 2016000 + 6774000 = 8790000 Joules = 8.79 MJ.

Answer: it is necessary to expend energy equal to 8.79 MJ.