# Latent heat?

1. Mar 16, 2014

### KiNGGeexD

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

2. Mar 16, 2014

### BvU

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. Mar 17, 2014

### KiNGGeexD

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. Mar 17, 2014

### BvU

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. Mar 18, 2014

### KiNGGeexD

Would N in the top equation not be just Avergadros number?

6. Mar 18, 2014

### BvU

Yes. Since we are talking J/mol, N=NA.

7. Mar 18, 2014

### KiNGGeexD

Ahh ok this calculation must just be rounded up because using avergadros number just now I obtained

1.95E-21 J

Thanks