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Ice cube placed in water - find final temperature

  1. Dec 13, 2009 #1
    1. The problem statement, all variables and given/known data
    A 75 gram ice cube at 0 degrees Celcius is placed in 825 grams of water at 25 degrees C. What is the final temperature of the mixture?


    2. Relevant equations
    What is the best formula to use in this question?


    3. The attempt at a solution I have attempted this problem multiple times. The last time I plugged in these variables:
    (75g)(2.108kJ/kg-k)(T2-0)=(825g)(4.184J/g*degrees C)(T2-25 degrees C)
    The answer is supposed to be 16 degrees C. I can't seem to figure this one out. Any help will be greatly appreciated. Am I missing variables?
     
  2. jcsd
  3. Dec 13, 2009 #2

    AEM

    User Avatar

    Re: Thermodynamics

    Well, I can think of two things you might check on the left side of your equation you seem to have mixed units you have 75g of ice but write J/kg.

    Also, don't you have to melt the ice and wouldn't that involve latent heat?
     
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