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Check Potential Difference

  1. Mar 4, 2009 #1
    1. The problem statement, all variables and given/known data

    A 4.63mF capacitor has stored energy to heat 3.00kf of water from 22 degrees C to 94.5 degrees C. What the potential difference?

    2. Relevant equations
    Q= mC delta T
    W= Q sq/2C
    V=W/C


    3. The attempt at a solution
    Q= mC delta T
    Q= 3kg* 4.184* ( 94.5-22) = 910.02 J
    then 910.02 sq/ 2* 4.63X -6= 8.94E 10
    8.94e10 / 910.02= 9.82e7


    is that right my teacher said it wrong
     
  2. jcsd
  3. Mar 4, 2009 #2

    LowlyPion

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  4. Mar 4, 2009 #3
    if PE is 91.02 and the equation is W= 1/2 *C*V sq
    then 910/2 * 4.63x10-6= then sq root then to 9913.33
     
  5. Mar 4, 2009 #4

    LowlyPion

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    Check your algebra again.

    Oh, btw I'd read that as mF as in milliFarads unless your problem said μf and you wrote mf here.
     
  6. Mar 5, 2009 #5
    i did it and got 2.10E -3 V and its its in μf
     
  7. Mar 5, 2009 #6

    LowlyPion

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    I still get something different:

    910 J = 1/2*(4.63*10-6)*V2

    V2 = 2*910/(4.63*10-6)
     
  8. Mar 5, 2009 #7
    v= 19826.67 is that the potential difference?
     
  9. Mar 5, 2009 #8

    LowlyPion

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    If that's what it calculates out to.
     
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