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Help review for exam (last question)

  1. Mar 4, 2009 #1
    Question is
    Problem 7: Joule’s paddle wheel experiment consists of a falling weight which spins a
    paddle-wheel in an insulated barrel of water. As the weight falls, the water warms up.
    Assume idealized conditions with no friction in the pulley, string, and bearings. How
    much does the temperature of the water change? Assume there is 0.4 kg of water, the
    weight has m=100 kg and it falls 1 m. The specific heat capacity of water is 4.2 J/g/K.

    here is a website to a pic of what a Joule's paddle wheel experment looks like.
    has something to do with the energy from the weight moving the 1 m. this turns paddle making thermal energy but dont know how to find this or how it is related

    http://upload.wikimedia.org/wikipedia/commons/c/c3/Joule's_Apparatus_(Harper's_Scan).png

    A. 0.19 degrees
    B. 0.29 degrees
    C. 1.16 degrees
    D. 0.58 degrees (correct answer)
    E. – 3.6 degrees
     
  2. jcsd
  3. Mar 4, 2009 #2

    Andrew Mason

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    Can you determine how much energy (in Joules) the weight delivers to the water in its fall? How much heat energy (in Joules) does it take to raise the 400 g of water 1 degree C.?

    AM
     
  4. Mar 4, 2009 #3
    i know how to find the energy of the weight, but not the energy it takes to heat the water 1 degree C.

    only formula i know is Cm(Tf - Ti) + Cm(Tf - Ti) = 0 but this is when you have 2 items at different temps and want to find equilibrum point in C.
     
  5. Mar 4, 2009 #4

    lanedance

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    the specific heat capacity is the energy it takes to heat one g of water by 1K by definition...
     
  6. Mar 4, 2009 #5
    so...
    4.2 is energy it takes to heat 1 g by 1k 1k = -272.15 C so rewrtie it as
    4.2 is energy it takes to heat 1 g by -272.15 C ??
    then divide by -272.15 to get energy need to heat 1g by 1 C ?? i get -.0154327
    i have 400 g of water so X previous by 400 to get -6.173066 J ??
    why is it negative? did i mess up somewhere?
     
  7. Mar 4, 2009 #6

    lanedance

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    K & degC differ in absolute magnitude by

    T(K) = T(degC) + 273.15

    but for a change in temp
    dT = T1-T0
    then

    dT(K) = dT(degC)

    so the difference in a temps is the same in Kelvin or Celsius
     
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