Latent heat of fusion for water question

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SUMMARY

The discussion centers on the calculations involving the latent heat of fusion for water and the specific heat of water in the context of supercooled water flash freezing into ice. The initial calculation erroneously suggested a temperature increase of 79 degrees Celsius, which was corrected to reflect that the temperature rise should only be from the supercooled state to 0 degrees Celsius. Participants emphasized the importance of correctly applying the specific heat and latent heat values, specifically 4.2 J/g°C for specific heat and 334 J/g for latent heat of fusion, to avoid dimensional inconsistencies in calculations.

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  • Understanding of latent heat of fusion (334 J/g for water)
  • Knowledge of specific heat capacity (4.2 J/g°C for water)
  • Familiarity with supercooling phenomena in water
  • Basic principles of thermodynamics, including Hess's Law
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  • Research Hess's Law and its applications in thermodynamic cycles
  • Explore the concept of supercooling and its effects on phase changes
  • Study detailed thermodynamic calculations involving latent heat and specific heat
  • Review experimental studies on supercooled water, such as those found in academic journals
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Researchers, physicists, and students studying thermodynamics, particularly those interested in phase transitions and the behavior of supercooled liquids.

soronemus
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Hello,

I am doing some research involving supercooled water flash freezing into ice.

I am doing some calculations and I think that I must be wrong judging by my result.

Using the latent heat of fusion of water, and the specific heat of water, I can calculate a temperature value which should be the amount that the water heats up due to the energy released by the exothermic reaction of water turning into ice. This value was 79 degrees Celsius. This seems very unrealistic to me... According to that the water would jump from freezing almost to boiling. Am I doing something wrong?

Calculations: (4.2J/gC)/(334J/g) = 1/0.0125C ~79C
(latent heat of fusion / specific heat)
 
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soronemus said:
(4.2J/gC)/(334J/g) = 1/0.0125C ~79C
You might try this again.
 
I guess I don't see what I did wrong. Would you elaborate?
 
soronemus said:
Would you elaborate?
4.21/334 = 79? Think about it a moment.
 
4.21/334 is not 79 you are correct, it is .0125. The problem is that I still have the C unit left in the denominator when everything else cancels. I would think that you would have to take the inverse of .0125 (79) to get the temperature unit into the numerator and get my answer. My notation on my original post was kind of bad.
 
Specific heat will be in J/gC for one thing. You are supercooling liquid water; the only heat you have to account for is that removed to reduce the temperature below the freezing point. The only temperature rise in the system is from the supercooled temperature to the freezing temperature. That temperature difference times specific heat divided by latent heat will give you the total mass of solid you can form.
 
I did use J/gC for specific heat. The operation you are proposing cancels all units and leaves you with a dimensionless number if i am looking at it correctly. Thank you for trying to help, I think we might be on different pages. I found something more geared towards what I'm looking for here. http://www.phy.mtu.edu/~cantrell/agu2009-PosterSzedlakV5.pdf
Would you suggest any other resources on the subject?
 
soronemus said:
Would you suggest any other resources on the subject?
Do a Hess's Law cycle, or pair of cycles in parallel, summing heats for equilibrium freezing, and comparing that sum to a supercooling cycle to see where you're losing/gaining a latent heat term to "boil" water.
 
soronemus said:
I did use J/gC for specific heat. The operation you are proposing cancels all units and leaves you with a dimensionless number if i am looking at it correctly. Thank you for trying to help, I think we might be on different pages. I found something more geared towards what I'm looking for here. http://www.phy.mtu.edu/~cantrell/agu2009-PosterSzedlakV5.pdf
Would you suggest any other resources on the subject?
Some of the water will become a solid, and some will remain as a liquid. The liquid water will have a temperature increase no more than to 0C as heat is released from the freezing part of water, which also ends up at 0C.

You can read this lab example to get the idea.
http://www.colorado.edu/MCEN/flowvis/galleries/2011/Team-1/Reports/Schollenberger_Scott.pdf
 

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