To calculate the heat absorbed/released when the steam condenses to liquid water, we need to use the equation:
q = n * ΔHvap
Where:
q = heat absorbed/released (in kJ)
n = number of moles of steam
ΔHvap = heat of vaporization (in kJ/mol)
First, we need to calculate the number of moles of water using the formula:
n = m / M
Where:
m = mass of water (in g)
M = molar mass of water (18.01528 g/mol)
m = 8.60 g
M = 18.01528 g/mol
n = 8.60 g / 18.01528 g/mol
n ≈ 0.4777 mol
Now we can plug the values into the equation:
q = 0.4777 mol * 40.66 kJ/mol
q ≈ 19.436 kJ
Therefore, approximately 19.436 kJ of heat is absorbed/released when 8.60 g of steam condenses to liquid water at 100 degrees Celsius.
At 100 degrees Celsius. Heat of vaporization is 40.66 kj/mol for water. Calculate the quantity of heat that is absorbed/released when 8.60g of steam condenses to liquid water at 100 degrees Celsius
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