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Calculate Total Heat Capacity

  1. Mar 11, 2009 #1
    1. The problem statement, all variables and given/known data

    50ml of water at 49.6 C were mixed with 50ml of water at 25.1 C in a calorimeter also at 25.1 C. The final temperature was 30.1 C Assuming that neither the density of water nor its specific heat capacity change with temperature, calculate the total heat capacity of the calorimeter.


    2. Relevant equations

    Density of water = 1.00 g/mL
    Specific heat capacity = 4.18 J / g * K

    3. The attempt at a solution

    q (heat given up by water) = 50ml * (49.6-30.1)
    = 840 cal

    q (heat absorbed by cold water) = 50ml (30.1-25.1)
    = 250 cal

    Heat absorbed by calorimeter = 250 + 840 = 1090 cal

    Ccal = qcal / delta T
    = 590 / (30.1-25.1)
    = 118 K

    The answer is supposed to be 493.24 J/K .... but i am not getting that.
    Please someone please show me how to correct this.

    Thank you.
     
  2. jcsd
  3. Mar 11, 2009 #2

    Astronuc

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    Staff: Mentor

    Woah!

    Heat is lost from the hot water to the calorimeter AND the cold water. One cannot at the heat of the hot and cold water.

    Try ΔQ(hot) = ΔQ(calorimeter) + ΔQ(cold)
     
  4. Mar 11, 2009 #3
    I tried doing that and the answer is 118 (which is still INCORRECT)...i wrote plus there where it should be minus.
     
    Last edited: Mar 11, 2009
  5. Mar 12, 2009 #4

    Borek

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    Staff: Mentor

    Show your detailed work then.
     
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