Clausius-Clapeyron equation and water

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SUMMARY

The discussion focuses on applying the Clausius-Clapeyron equation to determine the slope of dp/dT for the equilibrium between water and ice I. Key calculations involve the change in entropy (ΔS) and change in volume (ΔV) for 1 gram of H2O, where ΔS is derived from the heat required to melt ice (79.7 cal or 333 J) divided by the temperature at equilibrium. The densities of ice (0.917 g/cm³) and water (1.000 g/cm³) are essential for calculating ΔV. The conversation also touches on the practical implications of skating on ice versus solid argon.

PREREQUISITES
  • Understanding of the Clausius-Clapeyron equation
  • Knowledge of thermodynamic concepts such as entropy and volume changes
  • Familiarity with the properties of water and ice, including their densities
  • Basic skills in unit conversion and heat calculations
NEXT STEPS
  • Research the application of the Clausius-Clapeyron equation in phase transitions
  • Study the thermodynamic properties of water and ice, focusing on melting and freezing points
  • Explore the concept of latent heat and its role in phase changes
  • Learn about the physical properties of solid argon and its phase behavior
USEFUL FOR

This discussion is beneficial for students in physics or chemistry, particularly those studying thermodynamics, as well as educators looking for practical examples of phase transitions and their implications.

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Homework Statement



Using the Clausius-Clapeyron equation, determine the slope of dp/dT for water-ice I equilibrium and explain why you can skate on ice but not on solid argon.

Homework Equations



Clausius-Clapeyron equation

The Attempt at a Solution



I know what the clausius-clapeyron equation is but I have no idea how I'm supposed to determine the slope for water - there is no information in the book so far on water so I really have no idea how I could write the change in entropy and volume as functions of temperature..

I'm completely at a loss here (and most likely overlooking something simple) :confused:

(I wasn't sure if this should be in introductory or advanced physics, the textbook I'm using doesn't really mention what level it's geared at)
 
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Consider the melting of 1 gram of H2O.

Note ΔS= Q/T where Q is the heat required to melt 1 gram of ice (79.7 cal = 333 J) and T is the temperature corresponding to equilibrium between ice and water.

ΔV = Vf-Vi where Vi is the volume of 1 gram of ice and Vf is the volume of 1 gram of water. You can calculate these volumes from the known densities of ice and water at the melting point temperature (ice: 0.917 g/cm3, water: 1.000 g/cm3).

Watch your units.
 
Ah, thank you! :biggrin:
I knew it'd turn out to be something simple
 

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