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Specific Heat Capacity Problem

  1. Mar 21, 2009 #1
    1. The problem statement, all variables and given/known data
    How much ice was added to a 0.75 kg jug of lemonade at 15 degrees celcius if the final temperature of the jug is 8 degrees celcius? (Assume the lemonade has the specific heat capacity of water.)


    2. Relevant equations
    Q = cm∆T


    3. The attempt at a solution
    Well I think I might have to rearrange Q = cm∆T to solve for mass, so m = Q/c*∆T. But I don't have the heat, and then how would I account for the mass of the jug? Help!! :bugeye:
     
  2. jcsd
  3. Mar 21, 2009 #2

    mgb_phys

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    You don't need the mass of the jug - it means the final temperautre of the mixture in the jug.
    Remember the energy needed to melt ice.
     
  4. Mar 21, 2009 #3
    So you're saying I need to use latent heat? If that's the case then I would use Q = cm∆T + mLv, right?
     
  5. Mar 21, 2009 #4

    mgb_phys

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    Yes, and that equals the Q = cm∆T lost by the warm lemonade to get to the final temperature.
     
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