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Finding equilibrium temperature of water and a solid

  1. Apr 9, 2012 #1
    1. The problem statement, all variables and given/known data

    If 100 grams of glass (specific heat .2 cal/g*°C) at a temperature of 500 °C is added to 20 g of 0°C water, what is the final equilibrium temperature?

    2. Relevant equations

    Q= mcΔT


    3. The attempt at a solution

    I know that I have to set the two equation against each other:

    (100 g)*(.2 cal/g*°C)*(500°C -T) = (20 g)*(0°C+T)

    20*(500-T)= 20T
    10,000 - 20T= 20T
    10,000 = 40T
    250 = T

    I have an answer of 250°C, but my answer key says that the correct answer is 100°C. Any help?? Thank you!
     
  2. jcsd
  3. Apr 9, 2012 #2

    gneill

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    Staff: Mentor

    What happens when water reaches 100 C and there's still heat energy available?
     
  4. Apr 9, 2012 #3
    Ah, it vaporizes! So whenever the temperature rises enough to go past 100°C, the equilibrium will be at 100, because the rest of the water vaporizes, correct? Thank you! It is this sort of thing that causes me to miss points on our exams.
     
  5. Apr 9, 2012 #4

    gneill

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    Staff: Mentor

    Well, at least some of the water will evaporate. You could end up with a mixture of liquid water and steam at 100C. If there's enough heat available to evaporate all of the water then the steam temperature could also rise. You should check whether or not any water will remain liquid.
    Glad to help :smile:
     
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