# Specific Entropy Of Ice

1. Apr 7, 2008

### physics_newbi

The Specific Entropy of Water is 69.9J/K.
i did a search on the internet but couldn't find anything

2. Apr 7, 2008

### Mapes

Hi physics_newbi, welcome to PF. The decrease in entropy when a liquid freezes is $\frac{L}{T}$, where L is the latent heat (J/kg) and T is the freezing temperature (K). This will help you approximate the entropy of ice near the freezing point.

3. Apr 7, 2008

### physics_newbi

Thank you very much Mapes
here is a little calculation i did, can you please check if it's right
latent heat = 334(J/kg)
Temperature =273(K)
L/T = 1.22
and so the entropy of ice would be 69.91/1.22?
or would it be 69.91X1.22?

4. Apr 7, 2008

### Mapes

1) Your latent heat is off by several orders of magnitude.
2) You need to subtract $\frac{L}{T}$ from the entropy of water at 0°C, which is actually 63.3 J/K-mol (you had the value at STP).
3) Some values are in kilograms and some are in moles, so you will want to convert everything to one or another.

5. Apr 7, 2008

### physics_newbi

1)Latent heat of fusion = 334,000J/kg
334000/273 = 1223
2)63.3-1223 = -1160J/K? is it possible? did i make a mistake somewhere?
3)how do you convert moles to kg?
thanks

6. Apr 7, 2008

### Mapes

Try to carry the units along with every number. If the units don't match up, then the calculation is invalid. The calculation

$$63.3\,\frac{\mathrm{J}}{\mathrm{K}\cdot \mathrm{mol}} - 1223\,\frac{\mathrm{J}}{\mathrm{K}\cdot \mathrm{kg}} = \dots$$

The number of grams in a mole of a substance is its molecular weight.

7. Apr 7, 2008

### physics_newbi

1223/18 = 67.97J/K
i think that should be the right answer.
thanks alot, i really appreciate it.

8. Apr 7, 2008

### Mapes

That's not the right answer, because you didn't carry the units along with the numbers. Your answer has units (J-mol)/(K-kg-g); does that seem right to you?