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Specific Heat Capacity question

  1. Jul 10, 2008 #1
    1. The problem statement, all variables and given/known data
    If 100 mL test of water at 60°C is added to 100 mL of water at 20°C, then the final temperature of the water will be?


    2. Relevant equations
    c=Specific Heat Capacity

    Specific Heat Capacity of water = 4186

    Quantity of Heat Energy=mass * Specific Heat Capacity of substance * temperature difference
    -or-
    Q-=m*c*temp diff

    100mL=0.1 kg

    3. The attempt at a solution

    Water quantity at 60 degrees + Water quantity at 20 degrees = sum water quantity at new temperature

    0.1 kg * 4186 * (60 - final temp) + 0.1 kg * 4186 * (final temp - 20) = 0.2 kg * 4186 * final temp

    418.6 (60-final temp) + 418.6 (final temp-20) = 837.2 * final temp
    25116 - 418.6final temp + 418.6final temp - 8372 = 837.2final temp
    16744=837.2final temp
    final temp= 20 degrees celsius


    So, the final temp is 20°C ??? that doesnt make sense?
     
  2. jcsd
  3. Jul 10, 2008 #2
    Oh hold on now. I think I see my mistake. I didn't account for the Law of Conservation of Energy.
     
  4. Aug 10, 2008 #3
    Final Heat Change of 60°C Water + Final Heat Change of 20°C Water = 0

    0.1 kg * 4186 * (final temp-60) + 0.1 kg * 4186 * (final temp - 20)=0

    418.6 (final temp) - 25116 + 418.6 (final temp) - 8372 =0

    837.2 (final temp) = 33488

    final temp = 40

    Therefore, the final temperature of the water is 40 °C
     
  5. Aug 10, 2008 #4

    Ygggdrasil

    User Avatar
    Science Advisor
    2015 Award

    Yes, the final temperature should be 40oC, and your work is correct.
     
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