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chemman218
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Homework Statement
Supercooled water is water that is liquid and yet BENEATH the freezing point.
a) A sample of 131 g of supercooled liquid water freezes to solid ice at a temperature of -8.00 ° C. Using the following,
Cp,ice = 38.09 J/molK
Cp,liquid = 74.539 J/molK
fusH° (at T=0 ° C)=6.01 kJ/mol,
calculate S° for the process. Hint: you must set this up in steps! Report your answer to 3 significant figures.
Homework Equations
delta s = CpLn(T2/T1)
delta Sfus = delta Hfus/T
The Attempt at a Solution
I converted the fusH 6.01kj/mol(131g/18g/mol)=.043 joules
I know Delta S has to be an addition of the steps of freezing the water but am not sure how to get there.