How much heat is necessary to vaporize 100 g of water at 100 °C to form steam at 100 °C?
To determine the amount of heat required to vaporize a substance, we need to use the formula:
Q = m * ΔHv
where Q is the heat required, m is the mass of the substance, and ΔHv is the heat of vaporization.
In this case, we know the mass of water (m = 100 g) and that we want to vaporize it at 100 °C, which is the boiling point of water at atmospheric pressure. Therefore, we need to calculate the heat of vaporization for water at 100 °C.
The heat of vaporization for water can vary slightly with temperature, but for simplification, we can consider it as a constant value of 40.7 kJ/mol.
To calculate the heat of vaporization for water at 100 °C, we need to convert grams to moles:
1 mole of water (H2O) = 18.01528 g
moles = mass (g) / molar mass (g/mol)
moles = 100 g / 18.01528 g/mol ≈ 5.548 mol
Now we can calculate the heat required:
Q = m * ΔHv
Q = 5.548 mol * 40.7 kJ/mol
Q ≈ 222.44 kJ
Therefore, approximately 222.44 kJ of heat is required to vaporize 100 g of water at 100 °C to form steam at 100 °C.