Find the amount of heat required to vaporize 200 grams of water taken at the boiling point.

Given:

m = 200 grams is the mass of water;

T = 100 degrees Celsius – boiling point of water;

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

It is required to determine Q (Joule) – the amount of heat required to evaporate water.

Let’s convert mass units from grams to kilograms:

m = 200 gr = 200 * 10 ^ -3 = 0.2 kilograms.

Then, since, according to the condition of the problem, water is already at the boiling point, then:

Q = q * m = 2.3 * 10 ^ 6 * 0.2 = 2.3 * 10 ^ 5 * 2 = 4.6 * 10 ^ 5 Joules = 460 kJ.

Answer: for the evaporation of water, an energy of 460 kJ is required.