Latent heat?

  • Thread starter KiNGGeexD
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  • #1
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ImageUploadedByPhysics Forums1395008862.765744.jpg


I was looking through my notes and came across this (above) I understand the calculation etc but I don't see where the number of atoms N was calculated?

I don't see where it has come from


Any help would be great thanks in advanced
 

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  • #2
BvU
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Neat handwriting you are using to produce your notes....not nice you (they?) break the line between the minus sign and the 23, though....

28 gram is one mole
density is 0.81 g/cm3
so 28 gram is 28/0.81 = 35 cm3, right ? i.e. 35 cm3/mole

one mole is N_avogadro molecules and 35/(6 e23) = 5.8 e-23 cm3/molecule.

Take the ##\root 3 \of \ ## and you get something like 3.9 e-8 cm or 0.39 nm
 
  • #3
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It's the top calculation for latent heat I'm not sure about I get the surface tension one, but the one above makes no mention of N but must use it:)
 
  • #4
BvU
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Why don't you type the part of "your" notes that you have a question about ? I can hardly read what is to the left of ##{1\over 2} nN\epsilon##. Anyway, since we are talking J/mol, N=NA.
 
  • #5
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Would N in the top equation not be just Avergadros number?
 
  • #6
BvU
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Yes. Since we are talking J/mol, N=NA.
 
  • #7
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Ahh ok this calculation must just be rounded up because using avergadros number just now I obtained

1.95E-21 J

Thanks
 

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