What amount of heat should be taken to bring 100 g of water at 10 degrees Celsius to a boil and evaporate 10 g of it?

Given:

m1 = 100 grams = 0.1 kilograms – the mass of water that must be brought to a boil;

t1 = 10 degrees Celsius – initial water temperature;

t2 = 100 degrees Celsius is the boiling point of water;

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

m2 = 10 grams = 0.01 kilograms – the mass of water that needs to be evaporated;

q = 2.26 * 10 ^ 6 J / kg is the specific heat of vaporization of water.

It is required to determine Q (Joule) – how much heat is required to heat and evaporate water.

Let’s find the amount of heat required to heat the water:

Q1 = c * m1 * (t2 – t1) = 4200 * 0.1 * (100 – 10) = 4200 * 0.1 * 90 =

= 420 * 90 = 37800 Joules.

To evaporate water:

Q2 = q * m2 = 2.26 * 10 ^ 6 * 0.01 = 2.26 * 10 ^ 4 = 22600 Joules.

Total heat needed:

Q = Q1 + Q2 = 37800 + 22600 = 60400 Joules = 60.4 kJ.

Answer: you need heat equal to 60.4 kJ.