Final answer:
To vaporize 600 grams of water at 100°C and 101.3 kPa, approximately 1360 kJ of heat is required.
Step-by-step explanation:
The quantity of heat required to vaporize water can be calculated using the equation:
q = m * ΔHvap
where q is the quantity of heat, m is the mass of water, and ΔHvap is the enthalpy of vaporization. The enthalpy of vaporization for water is approximately 40.7 kJ/mol. To determine the quantity of heat required to vaporize 600 grams of water, we need to convert grams to moles:
moles=mass/molar mass
The molar mass of water is approximately 18 g/mol, so:
moles = 600 g / 18 g/mol = 33.33 mol
Now, we can calculate the quantity of heat:
q = 33.33 mol * 40.7 kJ/mol = 1360 kJ
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