1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Specific Entropy Of Ice

  1. Apr 7, 2008 #1
    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
     
  2. jcsd
  3. Apr 7, 2008 #2

    Mapes

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  4. Apr 7, 2008 #3
    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?
     
  5. Apr 7, 2008 #4

    Mapes

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  6. Apr 7, 2008 #5
    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
     
  7. Apr 7, 2008 #6

    Mapes

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  8. Apr 7, 2008 #7
    1223/18 = 67.97J/K
    i think that should be the right answer.
    thanks alot, i really appreciate it.
     
  9. Apr 7, 2008 #8

    Mapes

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Specific Entropy Of Ice
  1. Entropy ice-water (Replies: 2)

Loading...