Specific latent heat of fusion lab

AI Thread Summary
The discussion centers on a lab experiment aimed at determining the specific latent heat of fusion of ice using the method of mixtures. Participants share their equations and calculations, noting discrepancies in results, particularly when substituting specific heat values. It is emphasized that if ice is at zero degrees Celsius, using the specific heat of water instead of ice can yield accurate results. Concerns are raised about potential errors due to heat loss to the surroundings and the need to account for the heat gained by the beaker. The conclusion suggests that significant errors in measurements may require repeating the experiment for more accurate results.
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Homework Statement


It's part of a lab designed to find the specific latent heat of fusion of ice using the method of mixtures. The results are in the 1st attachment..I want to find the specific latent heat of fusion of ice from those nos.

Homework Equations



heat lost by water = heat gained by ice + heat required to melt ice

The Attempt at a Solution


mwcw(\theta<sub>i</sub>) - (\theta<sub>f</sub> - mici(\theta<sub>i</sub> - 0) + milf

lf = mwcw (\theta<sub>i</sub> - \theta<sub>f</sub>) - mici(\theta<sub>i</sub> - 0) / mi

I don't know how to fix this cursed code...anyways this is the equation I come up with. My friend in his attempt, (2nd attachment), got something different. I also found if you used cw instead of ci at the end of the last eqn you got the precise value of the specific latent heat of fusion of ice.
 

Attachments

  • results.jpg
    results.jpg
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  • vir's calc.jpg
    vir's calc.jpg
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I also found if you used cw instead of ci at the end of the last eqn you got the precise value of the specific latent heat of fusion of ice.

If the ice is at zero degree celcius you can use cw in place of ci.
 
If the ice is at zero degree celcius you can use cw in place of ci.

Ok could you explain why that is so, because as far as I know you would get different answers..btw when I worked it out yesterday I got 4.85x105Jkg-1

This is what, like more than 100 000 off the correct answer. Now I didn't use any insulation material during the experiment..can my answer be off by so much because of heat lost to surroundings?
 
You are doing the experiment in a beaker. So you have to take into account the heat gained by the beaker also. The spacific heat of the glass is 840 J/kg .degree. Then the answer is still more. So you have made some error in the observations. Do it again.
 
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