Water weighing 2 kg taken at a temperature of 20 ° C was poured into a kettle and boiled at a temperature of 100 ° C?

Water weighing 2 kg taken at a temperature of 20 ° C was poured into a kettle and boiled at a temperature of 100 ° C? (The specific heat of water is 4200 J / kg * ° C, the specific heat of vaporization of water is 2.3 * 10 ^ 6 J / kg).

Given:

m = 2 kilograms is the mass of water;

T1 = 20 degrees Celsius – initial water temperature;

T2 = 100 degrees Celsius – boiling point of water;

c = 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 determine Q (Joule) – how much energy must be expended to evaporate water.

To heat water to the boiling point, you need to expend energy:

Q1 = c * m * (T2 – T1) = 4200 * 2 * (100 – 20) = 8400 * 80 = 672000 Joules.

To evaporate water heated to the boiling point, you need to expend energy:

Q2 = q * m = 2.3 * 10 ^ 6 * 2 = 4,600,000 Joules.

Total energy:

Q = Q1 + Q2 = 672000 + 4600000 = 5272000 Joules = 5.272 MJ.

Answer: it is necessary to expend energy equal to 5.272 MJ.