How to Calculate Entropy Difference Between Water and Ice?

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Homework Statement


Compute the entropy difference between 12 kg of water at 40°C and 12kg of ice at -10°C.
Cp,water=4.184 J/g.K
Cp,ice=2.1 J/g.K
Heat of melting of ice=336 kJ/kg
Standard entropy of water at 298K = 69.9 J/K.mol


Homework Equations



S=the integral of cp/TdT

The Attempt at a Solution



I broke it down into different phases, the first one being delta S from -10 to 0, then from 0 to 40 degrees. Then I will add the heat of melting to the total entropy. Mass should be irrelevant since it remains constant. My problem is what to do with the standard entropy of water. Am I supposed to add it to my entropy equaiton as well?
 
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"My problem is what to do with the standard entropy of water. Am I supposed to add it to my entropy equaiton as well?"

Why not use it as the basic datum point for the whole problem? Compute the change in entropy to go from water at 298K or 25C to 0, then phase change, then to -10C. Then compute the change in entropy to go from 25C to 40C. Now you will have the difference when you compare the two values.
 

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