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Heat/First Law of Thermodynamics/Calorimetry - Have work done

  1. Dec 2, 2008 #1
    Heat/First Law of Thermodynamics/Calorimetry -- Have work done

    A calorimeter of negligible mass contains 920 g of water at 303 K and 48 g of ice at 273 K. Find the final temperature T.
    °C
    Solve the same problem if the mass of ice is 460 g.
    °C

    Formula Ive used
    Mice * Lf + Mice * Cwater * change in T icewater = Mwater * Cwater * change in T water

    (.048kg)(79.7kcal/kg)+(.048kg)(1kcal/kgxK)(Tf - 273K) =
    (.92kg)(1kcal/kgxK)(303K-Tf)

    Ive gotten Tf to be 297.56K which is 24.41 C

    and for part B Ive gotten the answer to be 266.433K which is -6.7167 C

    Is this correct?
     
  2. jcsd
  3. Dec 2, 2008 #2
    Re: Heat/First Law of Thermodynamics/Calorimetry -- Have work done

    I would start with the idea that the -q(water)= q(ice)
    This is because the water is warmer and it is losing heat to the ice. It will continue transferring heat until both are at the same temperature. Since the calorimeter has negligible heat capacity, it does not absorb any of the heat that the water gives off.
    From here you can say that the -ms(t(fin)-t(int)) for water is equal to ms(t(fin)-t(int)) for the ice.
    Hope that helps.
     
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