Water weighing 500 g was taken at a temperature of 20 degrees, then heated and turned into steam.

Water weighing 500 g was taken at a temperature of 20 degrees, then heated and turned into steam. How much heat was required for this.

Given:

m = 500 grams = 0.5 kilograms;

T = 20 degrees Celsius – initial water temperature;

T1 = 100 degrees Celsius – boiling point of water;

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

q = 2.3 * 10 ^ 6 J / kg is the specific heat of vaporization of water.

It is required to find the amount of heat Q (Joule) that will be required to heat and evaporate water.

It takes energy to heat water to boiling point:

Q1 = c * m * (T1 – T) = 4200 * 0.5 * (100 – 20) = 4200 * 0.5 * 80 = 168000 Joules.

It takes energy to evaporate heated water:

Q2 = q * m = 2.3 * 10 ^ 6 * 0.5 = 1,150,000 Joules.

Total energy required:

Q = Q1 + Q2 = 168000 + 1150000 = 131800 Joules = 1.3 MJ.

Answer: you need an energy equal to 1.3 MJ.