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Couldn't find a specific heat for coffee

  1. Jan 9, 2005 #1
    Here's the question:

    I used Q=mc (delta)T.

    So Q + Q = 0. I put the properties of the coffee in one and the properties of the glass in another. Since I couldn't find a specific heat for coffee I just used that of water (I figure they're pretty similar?)...

    Anyway, for my final answer u git 39.42C when it should be something like 73C.

    I don't understand this stuff too well yet, so it's probably a logic mistake on my part, but if someone could explain that to me I'd really appreciate it. Thanks.




    Also... (I figured I'd just add on to this post instead of creating another thread...) How would I go about doing this one:

     
    Last edited: Jan 9, 2005
  2. jcsd
  3. Jan 9, 2005 #2
    [tex]0.15c_{coffee}(86-x) = 0.2c_{glass}(x-22)[/tex]

    Did you do this...ps what did you use for c_glass ???

    regards
    marlon
     
  4. Jan 9, 2005 #3
    The phylosophy is that the total amount of heat is always conserved during the heath-transport. So the coffee will give off heat and the glass will absorbe the heath until equilibrium has been reached...

    Normally, T is in kelvin but once you are talking about temperature differences, it doesn't matter what units you use as long as you use one unit all the time...


    regards
    marlon
     
  5. Jan 9, 2005 #4

    dextercioby

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    For the first part you didn't give us the numbers,meaning the specific heat of glass.The logics seems good,Hot coffee is hot water with a tiny amount of coffee (maybe u like it without (too much)sugar,i hate coffe altogether),so u migh have messed up either the computations,or the units for the physical quantities.

    And for the second problem,show us what u did....

    Daniel.
     
  6. Jan 9, 2005 #5
    For the first problem this is what I did:

    mc(deltaT) + mc(deltaT) = 0
    .15(4187)(deltaT) + .2(840)(deltaT) = 0

    Then I just solved for that.




    For the second problem I didn't do much, I just doing the exact same thing as I did in the first problem for the .25kg of water and the .18kg of water and then I was planning on somehow interpreting that answer into another equation with the .3kg of water and an equilibrium of 30C... But with the first part I ended up with -65719.152C and I knew I was way off...
     
  7. Jan 9, 2005 #6

    dextercioby

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    This notation is misleading.You might think 'delta T' from the LHS is the same thing with the 'delta T' from the RHS.Besides,it should read as an equality between heat absorbed and heat given.

    Okay,the heat balance is
    [tex] 0.15Kg\cdot 4187J Kg^{-1}C^{-1}(86-t)C=0.2Kg\cdot 840J Kg^{-1}C^{-1} (t-22)C [/tex]
    ,which leads to the result
    [tex] t\sim 72.5C [/tex]




    This time u have two equations.The first one is
    [tex] 0.25Kg\cdot 4187J Kg^{-1}C^{-1}(80-t_{1})C=0.180\cdot 4187J Kg^{-1}C^{-1} (t_{1}-10)C [/tex] (1)
    The decond one is
    [tex]0.430Kg \cdot 4187J Kg^{-1}C^{-1}(t_{1}*30)C=0.300Kg \cdot 4187J Kg^{-1}C^{-1} (t_{2}*30) [/tex](2)

    ,where the sign "*" means "mathematically operated" and is a "-",the order of the 2 numbers in the bracket being chosen knowing that the result of the subtraction be >= to 0.

    Daniel.
     
    Last edited: Jan 9, 2005
  8. Jan 9, 2005 #7
    Thanks a lot for the help guys!
     
  9. Jan 9, 2005 #8
    I didn't really understand what you were saying there in your last post dextercioby, about my second problem.

    What I did was take mc(deltaT)+mc(deltaT)=0 for the first 2 portions of water.

    So I have .25(4187)(Tf-80)=-.1(4187)(Tf-10) which solves to 60C.

    The for the second part of the problem where the additional 300g of water is added and the entire portion of water comes to an equilibrium of 30C I did this:

    mc(deltaT)+mc(deltaT)=0
    .43(4187)(30-60)=-.3(4187)(30-Ti)

    ...which solves to -13C.

    The answer is supposed to be 0.33C. Can you see where I went wrong?
     
  10. Jan 9, 2005 #9

    You have the wrong mass for your first equation on the right side of the equation... it's suppose to be .18 kg
     
  11. Jan 9, 2005 #10

    dextercioby

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    U missused the numbers.It's 0.180 Kg in the first case,not 0.100 Kg as u took it.

    Daniel.
     
  12. Jan 9, 2005 #11

    learningphysics

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    How you're doing it is fine. But you can solve with with one equation. Just remember that the total heat change of all 3 groups is 0.

    So
    0.25(4187)(30-80) + 0.180(4187)(30-10) + 0.30(4187)(30-t)=0

    and you get 0.33C.
     
  13. Jan 9, 2005 #12

    dextercioby

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    EDIT:I messed up calculations and wound up with a different result.It's something bad happening with me taoday... :yuck:
    EDIT:This method i proposed leads to the same result as the one proposed by learningphysics.

    Daniel.
     
    Last edited: Jan 9, 2005
  14. Jan 9, 2005 #13

    learningphysics

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    It shouldn't matter when they are mixed. I got the same answer with your equations.
     
  15. Jan 9, 2005 #14
    Thanks guys, stupid mistake on my part!

    I actually had .18 written down but for some reason I erased and wrote .1.
     
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