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Chemistry question? Vaporization?
This is the specific question:
How much heat in kilojoules is absorbed when 63.7g H2O (l) at STP are converted to steam at 100C?
So I've done:
x = 63.7g H2O x (1mol/18.0g) x (6.01kJ/1mol)
x = 21.3 kJ
My question is, do I use the heat of vaporization of water or heat of fusion?
So, would I multiply the number of moles of H2O by 6.01 kJ/mol or 40.7 kJ/mol?
How can I tell which to use?
1 Answer
- Roger the MoleLv 77 years agoFavorite Answer
Never mind kJ/mol. Use 4.184 J/g·°C :
(4.184 J/g·°C) x (63.7 g) x (100 - 0)°C = 26652 J to heat the water to 100°C
The heat of fusion has no bearing on this problem. (The heat of fusion has to do with making or melting ice.) You need the heat of vaporization. Again, forget about kJ/mol and use the value already given in grams:
(2257 J/g) x (63.7 g) = 143770 J to vaporize the water
26652 J + 143770 J = 170422 J = 170 kJ total
