When boiling water, 690 kJ of energy was expended. Find the mass of vaporized water.

When boiling water, 690 kJ of energy was expended. Find the mass of vaporized water. The specific heat of vaporization is 2.3 * 10v ^ 6.

Given:
Q = 690 kJ = 69 * 10 ^ 4 Joules – the amount of energy that was expended on the evaporation of water;
q = 2.3 * 10 ^ 6 Joule / kilogram – specific heat of water vaporization.
It is required to determine m (kilogram) – the amount of evaporated water.
To find the amount of evaporated water, you need to use the following formula:
m = Q / q = 69 * 10 ^ 4 / (2.3 * 10 ^ 6) = 30 * 10 ^ 4/10 ^ 6 = 30 * 10 ^ -2 = 0.3 kilograms.
Answer: the amount of evaporated water is 0.3 kilograms.