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Specific Heat and Maybe Latent Heat also

  1. Dec 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Several small piece of aluminum, having total (combined) mass of 2.5 kg are placed in liquid nitrogen and, when removed, are at a temperature of -135C. The aluminum is transferred to an insulating container with 0.50 kg of water. The water is at 25.0C [ Ignore heat lost from the system]

    C Al = 910 J/kg K
    C water = 4190 J/kg K ; Lf = 334000 J/kg ; Freezing pt = 0C
    C ice = 2100 J/ kg K

    2. Relevant equations

    Q = mc(change)T
    Lf = mLf
    Conservation of Energy ( because system is isolated )
    Qcold = -Qhot


    3. The attempt at a solution

    Before I attempted to do this problem, I wasnt sure if there would be a latent phase change for the water to ice because if there was then I would include that eq. Lf = mLf into my solution.

    Gained Energy = Aluminum from water
    Lose Energy = Water from Aluminum

    (2.5 kg)(910 J/kg K)(T - (-135C)) = - (0.5 kg)(4190 J/kg K)( T - 25C)

    2275(T + 135C) + 2095 (T - 25C) = 0 ???

    I am unsure if thats what your suppose to do and is it suppose to be +135C?
     
  2. jcsd
  3. Dec 16, 2008 #2

    rl.bhat

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    You have to include Lf. So heat lost part include, heat lost by water from 25 degree to zero degree + heat lost by ice from zero degree to T degree.
     
  4. Dec 16, 2008 #3
    (2.5 kg)(910 J/kg K)(T - (-135C)) = - (0.5 kg)(4190 J/kg K)( 0 - 25C) +(x)(334000 J/kg) + (2100 J/ kg K)( T - 0C)

    Is this what your talking about? Because then theres 2 variables X because you dont know the amount of water that changes into ice...?
     
  5. Dec 16, 2008 #4

    rl.bhat

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    To check whether the water completely convert in ice or not, calculate amount of heat absorbed by aluminum to reach 0 degree, and amount of heat lost by the water to convert water from 25 degree C to 0 degree C ice. If first one is greater than the second,water is completely converted into ice. Other wise a part of water is converted into ice.
     
  6. Dec 16, 2008 #5
    that is necessary to find the temperature final? And if its partially converted doing what you said will give a ratio of how much ice is in the water?
     
  7. Dec 16, 2008 #6

    rl.bhat

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  8. Dec 16, 2008 #7
    okay so the following i did:

    Qal = (2.5kg)(910 J/kg K) (135C) = 307125 J
    Qw = (0.5kg)(4190 J/kg K) (25C) = 52375 J

    Therefore the amount of heat for water is less than Aluminum so all of the water turns into ice?

    then i can just do :

    (2.5 kg)(910 J/kg K)(T - (-135C)) = - (0.5 kg)(4190 J/kg K)( 0 - 25C) +(0.5 kg)(334000 J/kg) + (2100 J/ kg K)( T - 0C)

    and solve for T?
     
  9. Dec 16, 2008 #8

    rl.bhat

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    Before that check the signs and parenthesis.
     
  10. Dec 16, 2008 #9
    damn i cant get it to work out like i am really bad with the algebra lol anyone help me simplify this please?
     
  11. Dec 16, 2008 #10
    (2.5 kg)(910 J/kg K)(T - (-135C)) = - (0.5 kg)(4190 J/kg K)( T - 25C) +(0.5 kg)(334000 J/kg) + (2100 J/ kg K)( T - 0C)

    i believe all the numbers are correct however im always confused because of the positives and negatives... and the left side is gaining heat and the right side is losing heat. So the left side has a Negative to begin with. Also the Aluminum is at -135 does that cancel out the - from Tf-Ti?
     
  12. Dec 16, 2008 #11

    rl.bhat

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    To avoid confusion in signs, if heat is gained use Tf - Ti. If heat is lost, use Ti - Tf.
    Then use heat gained = heat lost.
     
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