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Temperature Changes

  1. Apr 20, 2009 #1
    1. The problem statement, all variables and given/known data
    If 25g of water at a temperature of 10 degrees Celsius is mixed with 40 grams of water at 80 degrees Celsius what is the final temperature of the mixture?


    2. Relevant equations
    Qlost + Qgained = 0
    Q=mcΔt


    3. The attempt at a solution
    Attempt 2:
    0.025(4186)(x-10) + (0.04)(4186)(x-40) = 0
    x = 28.46

    I don't know if I'm doing this right...
     
  2. jcsd
  3. Apr 20, 2009 #2
    You have the right idea, just remember that delta T = Tf - Ti

    So the water at 10 degrees is going to heat to some final temperature T, and the water at 80 degrees is going to cool to the same temperature T

    we don't need the specific heat of water since it's common in both terms

    (0.025)(10 - x) = (0.04)(x-80)

    x = 53 degrees

    Also note in your equation - your heat is 40 degrees rather than what the question states - 80 degrees. Perhaps you were thinking of 40g

    hope this helps,

    cheers
     
  4. Apr 20, 2009 #3
    Thank you so much!
     
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