calculate the amount of heat involved in the conversion of 25.00 g of water at 100 *C to steam at 100 *C. ( delta H vaporization = 40.7 kJ/mol)
q = mass water x heat vap
I usually change delta Hvap to J/g since water is in g.
24k/j
To calculate the amount of heat involved in the conversion of water to steam, we can use the formula:
Heat = Mass × Specific Heat × Temperature Change + Heat of Vaporization
Where:
- Mass is the mass of the water in grams.
- Specific Heat is the specific heat capacity of water, which is 4.18 J/g°C.
- Temperature Change is the change in temperature from the initial temperature to the boiling point of water, which is 100°C.
- Heat of Vaporization is the heat required to convert water at its boiling point to steam, which is given as 40.7 kJ/mol.
First, let's convert the mass of water from grams to moles:
Molar mass of water (H₂O) = 18.02 g/mol
Moles = Mass / Molar mass
Moles = 25.00 g / 18.02 g/mol
Moles ≈ 1.386 mol
Now, we can calculate the amount of heat required for the temperature change:
Heat for temperature change = Mass × Specific Heat × Temperature Change
Heat for temperature change = 25.00 g × 4.18 J/g°C × 100°C
Heat for temperature change = 10450 J = 10.45 kJ
Finally, we can calculate the amount of heat required for the phase change (vaporization):
Heat of Vaporization = 40.7 kJ/mol
Total Heat = Heat for temperature change + Heat of Vaporization × Moles
Total Heat = 10.45 kJ + 40.7 kJ/mol × 1.386 mol
Total Heat ≈ 67.615 kJ
Therefore, the amount of heat involved in the conversion of 25.00 g of water at 100°C to steam at 100°C is approximately 67.615 kJ.