How to Calculate Entropy Difference Between Water and Ice?

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SUMMARY

The discussion focuses on calculating the entropy difference between 12 kg of water at 40°C and 12 kg of ice at -10°C using specific heat capacities and the heat of melting. The specific heat capacity of water is 4.184 J/g·K, while that of ice is 2.1 J/g·K. The heat of melting is 336 kJ/kg, and the standard entropy of water at 298K is 69.9 J/K·mol. The solution involves calculating the entropy change in two phases: from -10°C to 0°C and from 0°C to 40°C, incorporating the heat of melting and standard entropy as a reference point.

PREREQUISITES
  • Understanding of thermodynamics and entropy calculations
  • Familiarity with specific heat capacities (Cp) of substances
  • Knowledge of phase changes and their associated heat values
  • Basic proficiency in calculus for integrating entropy equations
NEXT STEPS
  • Learn how to apply the entropy change formula S = ∫(Cp/T)dT in various scenarios
  • Study the implications of phase changes on entropy, particularly for water and ice
  • Explore the concept of standard entropy and its role in thermodynamic calculations
  • Investigate the relationship between temperature, heat capacity, and entropy in different materials
USEFUL FOR

This discussion is beneficial for students studying thermodynamics, particularly those tackling entropy calculations in physical chemistry or physics courses. It is also useful for educators looking for examples of phase change and entropy in real-world applications.

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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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