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Thermodynamics question

  1. Jan 11, 2005 #1
    I figured that this was a thermodynamics post, so here's my question:

    0.835 kJ is added to 10g ice @ zero-Celsius. The enthalpy of fusion for ice is 6 kJ/mol. What is the final temperature??

    I do: (10g/1)*(1mol/18g)*(6.0kJ/mol) and get 3.33kJ to change the zero-

    celsius ice to zero-celsius water; thus, my answer is zero-degrees Celsius.

    Something MUST be wrong---What is it??
    Perhaps I should use specific heat instead of enthalpy of fusion?....
    Specific heat of ice = 1.8 J/C = 0.0018 kJ/C

    if so, I find the temperature change as q/(m*Cp), and get
    (835 J)/((10g/1)*(1mol/18g)*(1.8 J/C)) = 835 Change in Celsius??
    (where 0.835 kJ = 835 J)
    (so i convert everything to joules)
  2. jcsd
  3. Jan 11, 2005 #2

    Andrew Mason

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    Why do you think your first answer is wrong? (Hint: have you changed the ice to liquid water?)

  4. Jan 11, 2005 #3
    Well, today i asked my teacher, and she said the first answer was right!

    It's not that...but, i just get concerned about questions where I might expect a change to occur....but it doesn't!! I mean, it's just that having not enough energy to raise the temperature....in this case, for example...doesn't seem right to put on a problem (well, if it changed zero degrees celsius, what's the point of the problem??)
  5. Jan 11, 2005 #4
    Maybe this is just a trick question, and not all the ice is supposed to melt.
  6. Jan 11, 2005 #5


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    Bomba's initial answer is correct. The final temperature is zero degrees C and the result is a water-ice mix at that temperature.

    The specific heat of ice only comes in when you're heating up ice at subzero temperatures to the melting point. After that point is reached, it's all about enthalpy of fusion, and the temperature remains constant during phase transition. If you don't put enough energy into the system to melt all the ice, the temperature is going to remain at zero deg C.
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