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Transfer of heat question

  1. May 10, 2010 #1
    1. The problem statement, all variables and given/known data

    A large punch bowl holds 3.10 kg of lemonade (which is essentially water) at 23.0^\circ C. A 6.00×10−2-{\rm kg} ice cube at -14.0^\circ C is placed in the lemonade.

    2. Relevant equations

    See below

    3. The attempt at a solution

    I have 3.10 * 4.190 * (23 - tf) = 987.47 - 15.19tf for the water...

    That is set equal to this for the ice...

    .06(14)(2.100) + .06(334) + .06(4.190)(0-tf)

    which overall simplifies to...

    2987.47 - 15.19tf = 21.764 + 20.04 + 2.514tf

    From there i got an answer that doesnt make sense
     
  2. jcsd
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