It takes 40.7 KJ of energy to vaporize 1 mole of water ( heat of vaporization). Calculate the total amount of energy needed to vaporize 25g of water at 100C
When 1 mol of water at 100°C is converted to 1 mol of water vapor at 100°C, 40.7 kJ of heat are absorbed from the surroundings.
Convert the amount of water (H2O) to moles:
(25 g H2O) × (1 mol H2O / 18.0153 g H2O) = 1.3877 mol H2O
Now, use the molar heat of vaporization as a conversion factor to determine how much heat would be needed to boil 1.3877 moles of water at its boiling point:
(1.3877 mol H2O) × (40.7 kJ / 1 mol H2O) = 56.48 kJ
Answer: 56.48 kJ of energy needed.