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Specific Entropy Of Ice

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

Answers and Replies

Mapes
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Hi physics_newbi, welcome to PF. The decrease in entropy when a liquid freezes is [itex]\frac{L}{T}[/itex], 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.
 
Hi physics_newbi, welcome to PF. The decrease in entropy when a liquid freezes is [itex]\frac{L}{T}[/itex], 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.
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?
 
Mapes
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1) Your latent heat is off by several orders of magnitude.
2) You need to subtract [itex]\frac{L}{T}[/itex] 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.
 
1) Your latent heat is off by several orders of magnitude.
2) You need to subtract [itex]\frac{L}{T}[/itex] 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.
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
 
Mapes
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Try to carry the units along with every number. If the units don't match up, then the calculation is invalid. The calculation

[tex]63.3\,\frac{\mathrm{J}}{\mathrm{K}\cdot \mathrm{mol}} - 1223\,\frac{\mathrm{J}}{\mathrm{K}\cdot \mathrm{kg}} = \dots[/tex]

cannot give any meaningful answer.

The number of grams in a mole of a substance is its molecular weight.
 
Try to carry the units along with every number. If the units don't match up, then the calculation is invalid. The calculation

[tex]63.3\,\frac{\mathrm{J}}{\mathrm{K}\cdot \mathrm{mol}} - 1223\,\frac{\mathrm{J}}{\mathrm{K}\cdot \mathrm{kg}} = \dots[/tex]

cannot give any meaningful answer.

The number of grams in a mole of a substance is its molecular weight.
1223/18 = 67.97J/K
i think that should be the right answer.
thanks alot, i really appreciate it.
 
Mapes
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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?
 

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