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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
What about ice?
i did a search on the internet but couldn't find anything
Please help
Thank you very much MapesHi 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.
1)Latent heat of fusion = 334,000J/kg1) 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.
1223/18 = 67.97J/KTry 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.