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Finding end temperature of mixed substances

  1. Jan 17, 2010 #1
    hi im having a bit of a problem with this anybody care to help me out?

    a calorimeter contains 0.761kg of water. both calorimeter and water have the temperature 4 degrees celsius. 0.054kg of liquid glycerol at 26 degrees celsius is then poured into the calorimeter. find the end temperature of the entire mixture.

    C(calorimeter)=86J/kgK C(water)=4180J/kgK C(glycerol)=2430J/kgK


    so far i've tried the equations Q(calorimeter)+Q(water)=Q(glyserol) and Q(calorimeter)+Q(water)=Q(glycerol)+Q(glycerol turning solid) but i always get some kind of brutal answer, i think the answer lies somewhere between 10 'c and 20 'c but im unsure of how to formulate the equation.
     
  2. jcsd
  3. Jan 17, 2010 #2
    Use the first equation you tried.
    Let the temperature of the final mixture be, t, for example.
    Find the energy gained by the water in rising to that temperature from 4 degrees.
    Find the energy gained by the calorimeter in rising to that temperature from 4 degrees
    Find the energy lost by the glycerol in falling to that temperature from 26 degrees
    Use conservation of energy. Energy gained by Water and calorimeter = energy lost by glycerol. (Assuming no loss to surroundings) Solve for t.
     
    Last edited: Jan 17, 2010
  4. Jan 17, 2010 #3
    Did you remember to convert to kelvin?
     
  5. Jan 17, 2010 #4
    thanks dudes for the help, i just messed up with placing the T in the equation so i got it solved properly now
     
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