Calculate melting point of ice under pressure

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Discussion Overview

The discussion revolves around calculating the melting point of ice under varying pressures, specifically at 6000 kPa. Participants explore relevant equations and concepts related to phase changes, particularly the Clausius–Clapeyron relation, and how to apply them to the problem at hand.

Discussion Character

  • Homework-related
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a problem involving the melting point of ice under pressure and notes a lack of relevant equations in their textbook.
  • Another suggests using the Clausius–Clapeyron relation to approach the problem.
  • Some participants question how density fits into the application of the Clausius–Clapeyron relation.
  • There is a suggestion to calculate volume change using the given densities of ice and liquid water.
  • A participant attempts to apply the Clausius–Clapeyron equation for evaporation but is advised to use the version for melting instead.
  • Confusion arises regarding the calculation of delta T, with participants discussing how to interpret their results and the implications for the melting point.
  • Another participant introduces a related problem involving different pressures and asks about calculating the heat of fusion (delta H).
  • There is an inquiry about the meaning of delta H, indicating a need for clarification on this concept.

Areas of Agreement / Disagreement

Participants express differing views on the application of the Clausius–Clapeyron relation and how to incorporate density into their calculations. There is no consensus on the correct approach or the interpretation of results, particularly regarding delta T and delta H.

Contextual Notes

Participants reference specific densities and the heat of fusion but do not resolve how these values should be used in the context of the Clausius–Clapeyron relation for melting. The discussion includes various assumptions and interpretations that remain unresolved.

Who May Find This Useful

This discussion may be useful for students or individuals interested in thermodynamics, phase transitions, and the application of the Clausius–Clapeyron relation in practical problems.

Pete_01
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Homework Statement


The density of solid water is approximately 920 kg m^-3 and that of liquid water is 1000 kg m^-3. Calculate the melting point of ice under a pressure of 6000 kPa.


Homework Equations



This is my problem. How do I approach this? I don't see any relevant equations in my textbook.


The Attempt at a Solution


See 2.
 
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Clausius–Clapeyron relation perhaps?
 
Last edited by a moderator:
Borek said:
Clausius–Clapeyron relation perhaps?

Not really. That has to do with pressure, but how does the density fit in?
 
Use it to calculate volume change.
 
Last edited by a moderator:
Borek said:
Use it to calculate volume change.

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I'm not following. Use the density to calculate volume change? Or use the Calpeyron equation to calcululate it? The equation I have found so far is:

delta(ln p) = delta_vapH/RT^2 * delta T

Is this correct? Thanks
 
No, this is version for evaporation, you need version for melting. Use given densities to calculate volume change.

See wikipedia article on Clausius–Clapeyron relation.
 
Borek said:
No, this is version for evaporation, you need version for melting. Use given densities to calculate volume change.

See wikipedia article on Clausius–Clapeyron relation.

So, like their example at the bottom of the page...

7ccf0e9ffa7dc258672182cdaa1060e4.png


and then rearrange and solve for deltaT and I get -0.4438K which doesn't mke much sense?
 
What is delta T?
 
Borek said:
What is delta T?

Change in temperature (melting temp).
 
  • #10
And what was delta T you calculated?
 
  • #11
Borek said:
And what was delta T you calculated?

Ok, so it would be: melting point water - (-0.4438K) for a final melting point?
 
  • #12
That would sound logical.
 
  • #13
Borek said:
That would sound logical.

Awesome! Thanks, I appreciate the help!
 
  • #14
i have something like it:

The densities of ice and liquid water at 1atm (101,325KPa) and 0oC are
917Kg/m3 and 999,8Kg/m3, respectively. The heat of fusion of water is 334720J/Kg.
Calculate the melting point of ice at 0,5atm (50,662KPa) and 101atm (10,2338MPa).

i don't know how i find "delta"H
 
  • #15
What IS delta H?
 

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