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Help! Can't solve this problem

  1. Dec 5, 2008 #1
    1. The problem statement, all variables and given/known data
    A 175 g aluminum cylinder is removed from a liquid nitrogen bath, where it has been cooled to -196°C. The cylinder is immediately placed in an insulated cup containing 26.5 g of water at 15.0°C, and the system is allowed to come to equilibrium. Determine if all of the water freezes. If so, find the equilibrium temperature. If not, find the amount of water that has frozen. The average specific heat of aluminum over this temperature range is 653 J/(kg·K).


    2. Relevant equations[/
    I need some serious help

    3. The attempt at a solution
     
  2. jcsd
  3. Dec 5, 2008 #2

    rl.bhat

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    First of all calculate the amount of heat to be removed from the water to convert it into ice. Then find the amount of heat needed to reduce the temperature of the aluminum cylinder to zero. Compare these two and decide what happens.
     
  4. Dec 5, 2008 #3
    Here are my calculations, any help as to what I’m doing wrong would be appreciated.

    Q (15 --> 0) = (.0265)(4186)(15)
    = 1663.9 J
    1663.9 J = (.175)(653) ∆T
    ∆T (15 --> 0) = 14.56

    Q (freeze) = (.0265)(33.5e4)
    Q (freeze) = 8877.5 J

    8877.5 J = (.175)(653) ∆T
    ∆T (freeze) = 77.685

    *is -79.4 degrees the correct final answer? I went a little further with my calculations, but i'm not sure if what I did makes sense;

    8877.5 J = (.175)(653)(final temperature - (-196))
    final temperature = -79.4 degrees

    *any help where to go from here?
     
    Last edited: Dec 5, 2008
  5. Dec 5, 2008 #4

    rl.bhat

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    Find the total amount of heat needed to freeze the water. This amount of heat is less than the amount of heat needed to increase the temperature of aluminum from -196 degree C to 0 degree C.
    Now wright the equation for heat lost by aluminum = heat gained by ice. Assume the average specific heat of ice is 2100 J/kg.k
     
  6. Dec 5, 2008 #5
    How do you find the total amount of heat needed to freeze the water?
    Q (water to freeze) = (.0265)(33.5e4)
    = 8877.5 J
     
  7. Dec 5, 2008 #6

    rl.bhat

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    Total heat = Q ( water to freeze ) + Q(water to cool --15--->0)
     
  8. Dec 5, 2008 #7
    Total Heat = (8877.5) + (1663.9) = 10541.4 J
     
  9. Dec 5, 2008 #8
    is this right..
    Q (al) = (.175)(653)(196)
    = 22,397.9 J
     
  10. Dec 6, 2008 #9

    rl.bhat

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    Q(al) is greater than total heat lost by water.
    So the water will loss heat further to increase the temperature of Aluminum.

    Heat gained by aluminum = Heat lost by ice. From this find the equilibrium temperature.
     
  11. Dec 6, 2008 #10
    m(a) * c(a) * (T(f) - T(i)) = m(i) * c(i) * (T(f) - T(i))
    (.175 kg) (653) (T(f) + 196) = (.0265) (2100) (T(f) - 0))

    *I'm sure I'm just not getting the algebra right or something...
     
  12. Dec 6, 2008 #11
    Conservation of energy problem! HELP PLEASE!

    1. The problem statement, all variables and given/known data
    A 175 g aluminum cylinder is removed from a liquid nitrogen bath, where it has been cooled to -196°C. The cylinder is immediately placed in an insulated cup containing 26.5 g of water at 15.0°C, and the system is allowed to come to equilibrium. Determine if all of the water freezes. If so, find the equilibrium temperature. If not, find the amount of water that has frozen. The average specific heat of aluminum over this temperature range is 653 J/(kg·K).

    *I know that all of the water will freeze, so i'm not looking for the amount of water that froze. I'm looking for the equilibrium temperature. I know it is a negative number, but not as negative as -196 degrees celsius. I just need somebody who knows how to solve this to walk me through it, numbers and all, because I AM LOST! Any help is appreciated! Thanks.
     
  13. Dec 6, 2008 #12

    LowlyPion

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  14. Dec 6, 2008 #13

    turin

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    Re: Conservation of energy problem! HELP PLEASE!

    How do you know this? Hint: you can't know this without calculating a few things, or you are extremely intuitive about these sorts of problems, much more so than me. Even if you are extremely intuitive, can you humor me with a couple of equations that you might think are relevant. Hint: "(specific) heat capacity", "latent heat".
     
  15. Dec 6, 2008 #14
    Re: Conservation of energy problem! HELP PLEASE!

    Q (15-->0) = (.0265)(4186)(15)
    = 1663.9 J
    1663.9 J = (.175)(653) ∆T *the 653 is the specific heat of aluminum over this temp. range
    ∆T (15-->0) = 14.56

    Q (freeze) = (.0265)(33.5e4) *the 33.5e4 is the latent heat of fusion
    Q (freeze) = 8877.5 J

    8877.5 J = (.175)(653)∆T
    ∆T (freeze) = 77.685

    *any help where to go from here? I'm just looking for some help.
     
    Last edited: Dec 6, 2008
  16. Dec 6, 2008 #15

    turin

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    What is the temperature of the rod when all the water has frozen into ice? What is the temperature of the ice at this point? Is the rod/ice system in thermal equilibrium at this point? If not, what should happen? Hint: what is the specific heat of ice?
     
  17. Dec 6, 2008 #16
    Temp. of the rod when all ice has frozen: -196 + 14.56 + 77.685 = -103.755 degrees

    The temp. of the ice at this point would be 0 degrees celsius right?

    The system is not in thermal equilibrium at this point, so the rod should continue to cool the ice past zero degrees. I just dont know how much, or how to find that. The heat gained by the aluminum at this point is the heat lost by the ice. This is what i tried, but i think its wrong;

    m(a) * c(a) * (T(f) - T(i)) = m(i) * c(i) * (T(f) - T(i))
    (.175 kg) (653) (T(f) + 103.755) = (.0265) (2100) (T(f) - 0))

    *the specific heat of ice I used was 2100.
    * this would give me a final answer of -69.77 degrees. I'm not sure if that right though!
     
    Last edited: Dec 6, 2008
  18. Dec 6, 2008 #17

    turin

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    You need to be careful with signs. When material gains heat it gets one sign, and when a material loses heat it gets the opposite sign. It doesn't matter which sign you choose, but you need to make the same choice for all materials.
     
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