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Specific Heat Capacity? Calorimeter

  1. Mar 3, 2009 #1
    im doing a calorimeter experiment and need to find the specific heat capacity of the calorimeter, but always gets a negative answer.

    Initial temp = 23 C
    Final temp = 60 C
    Change in temp = 37 C

    Mass of calorimeter = 0.027645kg
    Mass of water = 0.1kg

    Specific heat capacity water = 4200
    Specific heat capacity calorimeter = ???

    Heat supplied = 6349.551J (using q= VxIxT =7.93x1.57x510)

    i assume you find by q(total)=q(cal)+q(water)
  2. jcsd
  3. Mar 3, 2009 #2
    the temp for water or cal????

    q(total)=0 so q(cal)=q(water) so mc(t2-t1){for cal}=mc(t2-t1){for water}
  4. Mar 4, 2009 #3
    is it possible for the specific heat capacity of the calorimeter be more than that of water
  5. Mar 6, 2009 #4
    i think more and say you later
  6. Mar 6, 2009 #5
    ElectricalEnergySupplied = HeatIncreaseInWater + HeatIncreaseInCalorimeter

    HeatIncreaseInWater = (WaterMass*WaterSpecificHeat*TemperatureRise)

  7. Mar 6, 2009 #6


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    Science Advisor

    Be careful with units and temperature differences, and also carefully state the problem.

    The specific heat of water is 1 cal/gm °C or 4186 J/kg K (ΔT in °C = ΔT in K), but in one's problem the value has been rounded to 4200. The specific heat of metal is lower, e.g. 0.092cal/gm °C or 386 J/kg K for copper.

    In the problem, does the heat transfer from the water to the calorimeter, i.e. does the water heat the calorimeter, or is the heat supplied (6349.551J) enter the water and the calorimeter, i.e. both water and calorimeter start fromt the same initial temperature and achieve the same final temperature? In the former case, Δq(water)+Δq(calorimeter) = 0, whereas in the latter case q(heat supplied) = q(water)+q(calorimeter).

    See also - http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html
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