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The heat capacity of liquid water is 4.18 J/g·°C and the heat of vaporization is 40.7 kJ/mol. How many kilojoules of heat must be provided to convert 1.00 g of liquid water at 67°C into 1.00 g of steam at 100°C?
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First find the energy from 67 to 100\[q= {4.18J \over g ^\circ C}(1.00g)(100.0^\circ C - 67.0^\circ C)\] Then find the amount of joules it would take to convert all that liquid water at 100 to steam at 100\[q= {40.7KJ \over mol}\times{1.00g(mol) \over 18.02 g}\] Add those to together and that's your answer. Remeber that the last one is in kilojoules and the first is in joules.
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