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Heat Transfer

  1. Apr 16, 2008 #1
    1. The problem statement, all variables and given/known data
    An insulated container is filled with 250g of water at 90degrees Celsius. Which of the following, when added to the hot water, will result in the lowest final temperature of the mixture?
    A. 10g of water at 0 degrees C
    B. 125g of water at 75 degrees C
    C. 15g of water at 0 degrees C
    D. 5g of ice at 0 degrees C
    E. 250g of water at 80 degrees C

    2. Relevant equations
    I used (CM[dT]) + (CM[dT]) = 0
    where the first part is the container's heat and the second is the supposed amount of added substance at a certain temperature.

    3. The attempt at a solution

    I began plugging in numbers, starting from
    A: the final temperature was 86.5
    B. the final temperature was 85
    C. The final temperature was 84.9
    D. The final temperature was 88.23
    E. The final temperature was 85.

    It seems that C would be the answers, but it's just too close to B and E. Also, I think that my equation is wrong because ice has latent heat... so logically, I think that D would be the right one, because it takes heat to first melt the 5g of ice. But could anyone help me with how I should set up the equation due to the latent heat of ice?
     
  2. jcsd
  3. Apr 17, 2008 #2

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    If the final temp is t° C, then,

    heat required to melt ice + heat required raise the temp of melted ice to t = heat given out by the water at 90° C to cool down to t.
     
  4. Apr 17, 2008 #3

    Kurdt

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    You will have to consider latent heat for the part with ice.
     
  5. Apr 17, 2008 #4
    For choice D, this is what I did:
    mL + (ice's)cmdT = (water's)cmdT
    (0.005*3.35*10^5) + (2000*0.005*(T-0)) = 4186*0.250*(90-T)
    and I got T = 87.5 is that right?
     
  6. Apr 17, 2008 #5

    Kurdt

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    What is the 2000 in that calculation?
     
  7. Apr 17, 2008 #6
    I think it's the specific heat of ice.
     
  8. Apr 17, 2008 #7

    Kurdt

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    The ice has turned to water by then though.
     
  9. Apr 17, 2008 #8
    oh heh...
    so then the final T would be 86.6 which is still not the lowest. So the answer would be C then?
     
  10. Apr 17, 2008 #9

    Kurdt

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    That should be correct.
     
  11. Apr 17, 2008 #10

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    I am getting 80.38 °C for D. Recheck your calculation.
     
  12. Apr 17, 2008 #11
    I'm still getting 86.6
    I did what I posted:
    "(0.005*3.35*10^5) + (4186*0.005*(T-0)) = 4186*0.250*(90-T)"
    I believe 80.38 is correct though. But what are you doing exactly then?
     
  13. Apr 17, 2008 #12

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    Sorry...

    You got it right. I'm sorry for my wrong calculations. The ans is 86.67 °C.

    [Just one point though: use CGS for this type of problems. The numbers are so much easier to deal with, provided you don't make careless mistakes like me. E.g., the equation for (D) is:

    5*80 + 5(t-0) = 250(90-t).]
     
    Last edited: Apr 17, 2008
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