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Heat Exchange

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  1. Nov 2, 2014 #1

    rlc

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    1. The problem statement, all variables and given/known data
    (c11p34) You pour 160.0 g of hot coffee at 75.0 oC into a 230.0- g glass cup at 24.0 oC. If they come to thermal equilibrium quickly, what is the final temperature (in oC, enter deg in asnwer box)? Assume no heat is lost to the surroundings.

    2. Relevant equations
    (I don't even know, I've seen so many.)

    3. The attempt at a solution
    I've attempted this problem a couple times with different formulas, but none have worked.
    Can someone just tell me what equation to use, please?
     
  2. jcsd
  3. Nov 2, 2014 #2

    BvU

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    Suppose the final temperature is Tf. Then the coffee has cooled off to that temperature and the cup has heated up. They tell you no heat is lost, so apparently it's to do with heat. Where did the heat from the coffee go and where did the heat for the cup come from ?
     
  4. Nov 2, 2014 #3

    rlc

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    Thank you for replying! I actually tried another equation and found the correct answer:
    (mass coffee)(Temp coffee-Tf)(4.18 J/gC)=(mass glass)(Tf-temp glass)(0.840 J/gC)
    (160)(75-Tf)(4.18)=(230)(Tf-24)(0.840)
    Tf=63.57 deg
     
  5. Nov 2, 2014 #4

    BvU

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    It's what we call a heat balance. Heat given off = heat taken up. And you did well to include the cp.
     
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