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Calorimeter Constant

  1. Nov 11, 2013 #1
    1. The problem statement, all variables and given/known data
    Calculate the calorimeter constant
    50 ml H2O = 50g
    Starting Temp for hot water = 52.3°C
    Starting Temp for cold water = 21.8°C
    Final Temp = 41.3°C
    Change in Temp = Hot = -11°C / Cold = 19.5°C

    2. Relevant equations
    Q = mcΔT


    3. The attempt at a solution
    Q = (50g)(4.184)(11) = 2301 J
    Q = (50g)(4.184(-19.5) = -4079 J

    2301 - (-4079) = 6380 J

    6380J / 19.5 = 330 J/C

    Now my main questions is since the cold water is gaining heat would the joules be negative or positive.
     
  2. jcsd
  3. Nov 12, 2013 #2
    The transfer of heat was from the warm water to the cold water, so the heat transfer of energy of the cold water is positive.
     
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