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Thermal Engergy and Water

  1. Nov 13, 2006 #1
    Here's my question:
    Suppose you warm up 550 grams of water (about half a liter, or about a pint) on a stove, and while this is happening, you also stir the water with a beater, doing 4e4 J of work on the water. After the large-scale motion of the water has dissipated away, the temperature of the water is observed to have risen from 24°C to 75°C.
    What was the change in the thermal energy of the water?
    The answer is 1.18e5 Joules

    The equation is Delta E = W + Q = mC*(Delta Temp)
    Solve for Q m=550 C=4.7 change in temp=51 and I'm guessing work is 4e4, but solving for Q using those numbers gives me 2.945 Joules. What am I messing up?
     
  2. jcsd
  3. Nov 13, 2006 #2

    OlderDan

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    It sounds to me like you are misinterpreting the problem. Are you really being asked to find Q?
     
  4. Nov 13, 2006 #3
    yeah, it asks for the change in thermal engergy of the water and the only unkown I have is Q..

    Here's the solution my online hw gave me:
    [​IMG]
     
  5. Nov 13, 2006 #4

    OlderDan

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    What does thermal energy of the water mean?
     
  6. Nov 13, 2006 #5
    well, Delta E thermal for the water is change in the heat engergy of the water, which is Q, is it not?
     
  7. Nov 13, 2006 #6
    wait...i can't believe i didnt get this.

    DeltaE = mC*DeltaT, and I have all of those
    I still get 117810 and the answer is 118000..
     
  8. Nov 13, 2006 #7

    OlderDan

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    In this case ΔE is not Q. The thermal energy of the water is increased by the heat added Q plus the work done by the beater W. That is what your online slolution/explanation is saying. But you are not being asked for Q, you are being asked for ΔE.
     
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