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Thermodynamics, energy, work, heat problem

  1. Sep 10, 2009 #1
    a system consisting of 73.2g of liquid water at 298k is heated using an immersion heater at a constant pressure of 1.00bar. if a current of 2.25A passes through the 10 ohm resistor for 125s, what is the final temp of the water?

    im having trouble finding temp?
    i found power and work using P = IR^2 and work = P/t
    but im stuck, i initially thought i should use PV = nRT but i have too many unknowns
    please help asap



    Moderator's Note: Multiple Threads Merged
     
    Last edited by a moderator: Sep 11, 2009
  2. jcsd
  3. Sep 10, 2009 #2

    Redbelly98

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    You should re-check that equation for work. It's not quite right.

    The key is that the work is entirely converted into heat energy. So you'll need an equation that relates heat energy to temperature change.

    Nope. That equation is for an ideal gas. The water is a liquid here.
     
  4. Sep 10, 2009 #3
    i know Heat = mass * Cp * change in temperature but how does the power relationship tie into this?
     
  5. Sep 10, 2009 #4

    mukundpa

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    What is specific heat of a substance? Do you know the definition?
     
  6. Sep 10, 2009 #5
    Q=nC(Tf-Ti) i believe C is the specific heat
     
  7. Sep 10, 2009 #6

    mukundpa

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    Rather for water (liquid) we can write Q = mC (Tf - Ti)

    You have calculated Q, See for specific heat of water, it is a constant and calculate !!
     
  8. Sep 10, 2009 #7

    Redbelly98

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    Good, that's right. Keep that one in mind.

    There's another useful equation relating energy, power and time. See my previous comment about re-checking an equation you wrote in post #1.
     
  9. Sep 11, 2009 #8

    Borek

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    You start with liquid water at 298 K - what will happen to it when it is heated?
     
  10. Sep 11, 2009 #9
    The formula you have Q = mC (Tf - Ti) only works if the heat capacity is constant throughout the range of temperatures. It might be close enough approximation in this case, if thats what your professor told you then use that equation. But in a general snse, you have to integrate from Tf to Ti...
     
  11. Sep 11, 2009 #10

    Redbelly98

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    We usually consider C to be a constant in introductory physics problems. :smile:

    (And it is constant to within about 0.1% in this case.)
     
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