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Joule machine

  1. Mar 24, 2008 #1


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    [SOLVED] joule machine..

    1. The problem statement, all variables and given/known data
    In joule's apparatus below, the mass of each block is 1.50kg, and the insulated tank is filled with 200g of water. What is the increase in the emperature of he water after the bocks fall through a distance of 3.00m?


    2. Relevant equations
    [tex]\Delta U_{total}= W_{total} [/tex]

    3. The attempt at a solution

    Hm...not quite sure. I would think that I need to find the calories but I'm not sure how to find that either.

    I know that the work done would = potential energy from when the blocks drop the 3m but to find the change in temperature would I use this equation ([tex]Q= C\Delta T[/tex] and equate the the Q to the [tex] \Delta U[/tex] )?

    I know:
    h= 3.00m
    m= 1.50kg
    g= 9.8m/s

    mass of water= 2.00g=> 0.002kg
    help please..
    Last edited: Mar 24, 2008
  2. jcsd
  3. Mar 24, 2008 #2
    Try conservation of energy

    The change in gravitational potential energy equals the heat gained by the water.
  4. Mar 24, 2008 #3


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    so would it be:

    [tex]mgh= mc\Delta T[/tex] ?
  5. Mar 24, 2008 #4
    Yes. The first m should be the mass of both blocks since they both fall through that distance.
  6. Mar 24, 2008 #5


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    so it's 2mgh
  7. Mar 24, 2008 #6
    if you want m to be 1.5 kg, then yes
  8. Mar 24, 2008 #7


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    okay I found the temp change to be 10.54 deg C
    so that sounds alright to me.

    Thanks for your help mike :smile:
  9. Mar 24, 2008 #8
    Hmm, are you sure that your units are consistent on both sides?

    The specific heat of water is approximately 4.18 J/(g*C) which is equal to 4,180 J/(kg*C).

    The temperature change that I get is about 0.105 degrees C.
  10. Mar 24, 2008 #9


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    uh oh..I thought that it was 2g of water. That is what was wrong with it.

    I get the same thing after I finish correcting.

    Thank you for catching that :redface:
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