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Entropy #1

  1. Oct 19, 2005 #1
    Liquid water can be cooled to a temperature of -10 C and stay a liquid. As it is below its freezing point it is thermodynamically unstable and will eventually turn into solid water. If you put 100 g of supercooled water at -10 C into an insulated thermos (adiabatic) it will eventually freeze into a mixture of solid ice and liquid water at 0 C. How much of the water is solid? What is the entropy change of the universe for this process? (Assume the heat capacities of solid and liquid water are independent of temperature.)

    I have to admit, I have very little clue as to where to even begin with this question. Normally, every time I ask a question, I like to have some answer or idea to let people know I at least thought about it, but I simply cannot decipher this problem. I tried working with absolute entropies of the water in solid and ice forms at 263 and 273 K, but I am getting nowhere. Any advice is appreciated.
  2. jcsd
  3. Oct 22, 2005 #2
    there should be an equation in physics specifically for adiabatic problems.
  4. Oct 23, 2005 #3


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    The first part is simply solved by setting heat lost = heat gained. Can you write down the equation for that ?
  5. Oct 24, 2005 #4


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    here are some questions to consider

    -at constant volume/adiabatic conditions what would be the work for raising the temperature. It's simple calorimetry, except that you might want to use w, instead of q. The energy gained by the liquid phase is approportionated to form the ice. The process occurs due to differences in chemical potential of the liquid and solid phase at -10C

    The change in entropy of the surroudings will be zero right, since no heat is released to the surroundings. The change in entropy of the system can be found by assuming the equivalent changes in state for an adiabatic process as that of constant temperature and constant _______?
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