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Homework Help: Heat Transfer Question

  1. Apr 26, 2009 #1
    1. The problem statement, all variables and given/known data

    A piece of metal with a mass of 1.50 kg, specific heat of 200 J/kg*oC, and initial temperature of 100oC is dropped into an insulated jar that contains liquid with a mass of 3.00 kg, specific heat of 1,000 J/kg*oC, and initial temperature of 0oC. The piece of metal is removed after 5 seconds, at which time its temperature is 20oC. Neglect any effects of heat transfer to the air or insulated jar.

    The temperature of the liquid after the metal is removed is ... ?

    2. Relevant equations
    [delta]U = Q + W

    3. The attempt at a solution

    I'm not sure how to apply an equation to solve this. Any guidance?
    Last edited: Apr 26, 2009
  2. jcsd
  3. Apr 26, 2009 #2
    The options for the temperature of the liquid after the metal is removed is:
  4. Apr 26, 2009 #3


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    Staff Emeritus
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    Homework Helper

    Hello science.girl,

    Hmmm, that equation you wrote doesn't seem to apply here.

    Do you have your physics textbook or class notes handy? You're looking for an equation that contains the specific heat, mass, and temperature change in it -- often denoted by c, m, and ΔT, respectively.
  5. Apr 26, 2009 #4
    The only equation I can think of is

    c = Q/(m[Tf - Ti])

    But, I'm not given a value Q. I would be looking for Tf, right?
  6. Apr 26, 2009 #5
    Any help? Is there an equation that I'm missing?
  7. Apr 26, 2009 #6

    Q=mc*(change in temperature)

    Where: Q absorbed = Q released

    So you can set a side equal to each part of the equation then plug in...

    m*c*(Tf-Ti) for the liquid = m*c(Tf-Ti) for the metal
  8. Apr 26, 2009 #7
    Now I see how the equation is applied. Thank you so much for your help. :smile:
  9. Apr 26, 2009 #8
    no prob :biggrin:
  10. Apr 27, 2009 #9
    Also before you do any problem in thermo like this use the 1st law and reduce it down, its good practise to start doing this as it helps to visualise the problem. These are relatively simple but if you ever do anything with the steady flow energy equation you'll see why doing this is useful.

    so in this case you take

    dU = dQ + qW

    we know there is no work, but there is heat transfer.

    dU = dQ

    This means that the heat flow from the bar into the liquid is the only thing to affect the internal energy. so this means we can apply the Q=mcdt as above.
  11. Apr 27, 2009 #10
    Interesting. This is very useful, Chris. Thank you for this explanation... I will try to apply it to my further studies of physics! :smile:
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