Find the entropy for the process

[chemman218]

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.

 

[Andrew Mason]

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.

To determine the change in entropy, you have to find a reversible path between the initial and final states and then calculate the integral of dQ/T for that path. Supercooled water freezing at -8C is not reversible. Water freezing at 0C is reversible. So you have to get the supercooled water from -8C to 0C reversibly, let it freeze reversibly, and then cool the ice back to -8C. The change in entropy is the intergral of dQ/T along that path.

AM

 

[chemman218]

Thank you so much for the clarification. It makes more sense now to view it in that process.