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Thermodynamics Problem - the lake :P

  1. Jun 15, 2011 #1
    Hi.. looks like simple problem but the result i got is unreasonable.

    1. The problem statement, all variables and given/known data

    There is lump of ice, that falling down into a lake that his temperature is 0 celsius degrees.

    K of the lump getting melt ( = M that melt/M )

    What is the height that from him, the lump fell ?


    2. Relevant equations

    Ep = mgy

    Q = mc(Delta t)

    Q = mL

    3. The attempt at a solution

    I advocate that

    mgy = m*k*(L + t of melting)
    (c of the water that melt = 1)
    t of melting of water = 0

    So

    y = KL/g

    But when I place the numerical data, the result is completily different from what the writers wrote, as I sad above.

    Thanks in advance !
     
  2. jcsd
  3. Jun 15, 2011 #2

    tiny-tim

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    Homework Helper

    Hi Planck const! :smile:

    A lump of ice falls into a lake whose temperature is 0 degrees Celsius.

    A proportion K of the mass of the lump melts ( = M that melt/M )

    What is the height from which the lump fell ?

    Your method looks ok :confused:

    perhaps the units needed converting?

    Can you show us your own calculations, and also the figure given for L, and the result given in the book?
     
  4. Jun 15, 2011 #3
    Thanks for the reply!

    The initial temperature of the lake water = 0 C deg'
    L of water = 79.7 cal/gr
    K in the book = 5*10^-3 (no units)
    g = 980 cm/sec^2

    The answer in the book is the the height, h = 171 m = 17,100 cm (!!!)

    Now all the potential energy become kinetic and its become thermal energy.
    The thermal energy that goes from the body to the lake caused from the melting and the change in the temerature...
    On those things my calculations were based .
     
  5. Jun 15, 2011 #4

    tiny-tim

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    Hi Planck const! :smile:

    (have a degree: ° and try using the X2 icon just above the Reply box :wink:)

    How did you convert each of those units (cal/gr and cm/s2) to SI units (or cgs units)?
     
  6. Jun 15, 2011 #5
    Let's see...
    cal = 4.2 J
    so I need to multiply L in 4.2
    and I get:
    h (cm) = K*L*4.2 / g = 4.2 * 5 * 10^-3 * 79.7 / 980
    Using pocket calculator, h = 1.83 * 10^-3 (cm) !!!!!!!!!!!!
    In the book - 1.7 * 10^+3

    About the math singles in this forum... Im used to the way that I write. Thanks anyway :)
     
  7. Jun 15, 2011 #6

    gneill

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    Staff: Mentor

    [tex] \frac{79.7 cal}{g} \times \frac{4.187 J}{cal} \times \frac{1000 g}{kg} = ? [/tex]
     
  8. Jun 16, 2011 #7
    Ok

    Need erg and not Jaul ...

    1 cal = 4.2 J = 4.2*10^7 erg

    K*L / g = 5*10^-3 * 79.7 * 4.2*10^7 / 980 = 17,025 cm = 170.25 m

    Thank you very much!
     
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