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Therm. chem.

  1. Mar 15, 2005 #1
    I don't remember at all how to do these thermochemistry problems. Can someone please guide me through?

    You add 30.0 g of Cu (Specific Heat Capacity = 0.3846 J/K·g) at 92.0°C to 150.0 mL of water at 25.0°C in a very insulating calorimeter. Assuming the calorimeter constant (Cc) is zero, and knowing the Specific Heat Capacity of water is 4.18 J/K·g, what was the final temperature of the combined water/Cu metal solution?
     
  2. jcsd
  3. Mar 16, 2005 #2

    learningphysics

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    Homework Helper

    The heat change in a particular substance (the heat entering the substance) is mc(Tf-Ti), where Tf is the final temperature of that substance, and Ti is in the initial temperature of that substance, and c is the specific heat capacity of that substance.

    The total heat change of any group of isolated substances adds to 0.

    So (m1)(c1)(Tf1-Ti1)+(m2)(c2)(Tf2-Ti2)+.....=0

    Hope this helps, and you can apply it to your particular problem.
     
  4. Mar 16, 2005 #3

    xanthym

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    Let Tf be Final Temperature of combined water/metal solution. Then from conservation of energy:
    {Heat Lost by Metal} = {Heat Gained by Water}
    ::: ⇒ {Metal Mass}*{Metal Heat Capacity}*{(92 + 273) - Tf} =
    = {Water Mass}*{Water Heat Capacity}*{Tf - (25 + 273)}
    ::: ⇒ {30.0 g}*{0.3846 J/(K*g)}*{(92 + 273) - Tf} = {150.0 g}*{4.18 J/(K*g)}*{Tf - (25 + 273)}
    ::: ⇒ {11.54}*{365 - Tf} = {627}*{Tf - 298}
    ::: ⇒ (4212) - (11.54)*Tf = (627)*Tf - (186,800)
    ::: ⇒ (191,000) = (638.5)*Tf
    ::: ⇒ Tf = (299.1 degK) = (26.1 degC)


    ~~
     
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