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Iron and Ice in Aluminum bucket of water

  1. Dec 19, 2014 #1
    1. The problem statement, all variables and given/known data
    Iron ball Mass = 0.8 kg, Cp=484, Ti=80 Celsius edit : Cp = 448
    Aluminum Bucket = 0.5 kg, Cp=900, Ti=20 Celsius
    Water = 0.6 kg, Cp=4186, Ti=20 Celsius
    Ice cube=0.5 Kg, Cp=2090, Lf=3.33e5, Ti=-5 Celsius

    Find the final temperature of the bucket.

    2. Relevant equations
    Delta Q = 0 in equilibrium

    3. The attempt at a solution
    At fist I took into consideration only the water,iron and Aluminum to find the temperature in equilibrium
    Delta Q = 4186*0.6*(Tf-20)+900*0.5*(Tf-20)+448*0.8(Tf-80)=0
    The final temperature I got is 26.47 Celsius

    Next I calculate the amount of energy release from 0.6 kg of water = 600*4.182*26.47=66,482 J
    Then I divided it by the amount of energy needed to bring this amount of water to 0 Celsius - 66,482/333=199.645 Grams of ice.
    Since we have 0.5 kg of ice Tf will be equal to 0 Celsius and 0.3 Kg of the ice will remain in the bucket.

    In order to proof it I wrote the following equation = Delta Q = 4186*0.6*(Tf-26.47)+2090*0.199*(0-(-5)) +0.199*3.33e5+4186*(0.199)*(Tf-0)=0

    I got Tf= -0.65 Celsius which could be a result of roundoff error so it seems like I got the right answer.

    But I think that I did something wrong since I ignored the Iron and the Aluminum in my last equation.

    Thank you for your help!
     
    Last edited: Dec 19, 2014
  2. jcsd
  3. Dec 19, 2014 #2

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    Try it again.
     
  4. Dec 19, 2014 #3
    I think I wrote it by mistake - the given number is 448 for the Cp of the iron
     
  5. Dec 19, 2014 #4

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    Okay. Transcription errors happen to all of us.
    What's the heat capacity of water + bucket + iron? That's the system you're cooling with the ice.
     
  6. Dec 19, 2014 #5
    4.186*600g*26.47+0.450*800*26.47+0.900*500*26.47= 87880 The heat capacity of the system. Do I need to divide it by 333 now? then I get 263.905 grams of ice needed to bring the system to 0 Celsius.
     
  7. Dec 19, 2014 #6

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    These numbers aren't just along for the ride.
    No. 3.33e5, or 3.33 x 105, or 330,000 comes in after one more step; the ice has in initial temperature less than that of the melting point.
     
  8. Dec 19, 2014 #7
    Delta Q =(-87880+3320Tf)+(2090*0.500)*(0-(-5)) +0.500*3.33e5+4186*(0.500)*(Tf-0)=0 (The red part is from the first equation).
    than I get -20 Celsius which is impossible.
    There must be something I'm not aware of. What am I missing?
     
  9. Dec 19, 2014 #8

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    How much heat from bucket, water, and iron to raise the temperature of the ice to 0 C? How much does that lower the temperature of the bucket, water, and iron?
     
  10. Dec 19, 2014 #9
    To raise the temperature of the ice to 0 we need 2.090 x 500 x 5 = 5225 J.
    Bringing the Ice to 0 Celsius will lower the temp of the bucket by 1.6 Celsius ------> 87880-5225=3320Tf => 24.90 C

    Is that correct?
    Than Heat to melt the Ice = 3.33e5x0.5 = 166500 J
     
  11. Dec 19, 2014 #10

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    Yes, and is there that much heat available in the bucket-water-iron?
     
  12. Dec 19, 2014 #11
    Nope, we have only 82655 left in the bucket
     
  13. Dec 19, 2014 #12

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    And the final T is what?
     
  14. Dec 19, 2014 #13
    that means that we can use only 82655 of heat to melt the ice - which leaves us with 166500-82655= 83845
    So it will melt 0.248 kg of the ice and the rest of the ice will remain in the bucket at 0 Celsius.

    Is that correct? or there is a different way to prove it?
     
  15. Dec 19, 2014 #14

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    I haven't checked all your arithmetic, but the only bug I've seen was the transcription error in the original post, so you should be good to go.
     
  16. Dec 19, 2014 #15
    It was a typo. In my notebook I used the right number. Thank you so much!
     
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