Specific Heat Capacity Problem

  • Thread starter chops369
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  • #1
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Homework Statement


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.)


Homework Equations


Q = cm∆T


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:
 

Answers and Replies

  • #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.
 
  • #3
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So you're saying I need to use latent heat? If that's the case then I would use Q = cm∆T + mLv, right?
 
  • #4
mgb_phys
Science Advisor
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So you're saying I need to use latent heat? If that's the case then I would use Q = cm∆T + mLv, right?
Yes, and that equals the Q = cm∆T lost by the warm lemonade to get to the final temperature.
 

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