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What equation am I supposed to use for potential difference?

  1. Sep 11, 2005 #1
    A huge 4.0-F capacitor has enough stored energy to heat 2.5 kg of water from 21 degree Celsius to 95 degree Celsius. What is the potential difference across the plates?

    What equation am I supposed to use for potential difference?
    Is it V = Change in PE / q
    V= - W / q
    V = (PE)(.5)/ Q

    Also, I'm supposed to find how much energy is required to heat the water right? What equation do I use for that?

    I'm just confused with this problem... Can somebody please help. Thanks
     
  2. jcsd
  3. Sep 12, 2005 #2

    quasar987

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    The amount of charge a capacitor can store is directly proportionnal to the potential difference between its plates. This constant of proportionality is the capacitance, C, meaured in Farad (F). You are told here that C = 4 F. Hence, for a capacitor plugged to a battery of potential difference [itex]\Delta V[/itex],

    [tex]Q_{\mbox{final}} = C \Delta V[/tex]

    Where the index "final" refers to the process during which the capacitor charges itself. At any given moment during this process, if the charge (in absolute value) on each plate is q, then the potential difference between the plates is [itex]\Delta V = q/C[/itex].

    Now we contemplate that if we could find Q_final, we'd have won. Fortunately, we are given another clue: at the end of the charging process, the capacitor has enough stored energy to heat 2.5 kg of water from 21 degree Celsius to 95 degree Celsius. Figure out how much energy that it, and compare it to the energy "stored" in the capacitor. The energy stored in the capacitor is the total work done to charge the capacitor from q=0 to q=Q. (Think integral)
     
  4. Sep 12, 2005 #3
    Two things that I think can help you. First you need to find out how much energy is stored in water as it changes it's temperature. This is given by the old thermodynamic equation:

    E = (m)(c)(Tf-Ti)

    Energy - units (joules)
    m - mass (units grams)
    c - specific heat capacity (for water, it's value is 4.186 J/(g)(K)
    Tf - Final temperature
    Ti - initial temperature

    This relation should look familiar too you. Once you get the energy from this, set it equal to one of the energy capacitance equations. The one you are probably looking for is the one that omits Q:

    E = 1/2(Q)(V)

    Substitute for the charge Q with CV in the above equation and you get:

    E = 1/2 (C)(V)^2

    Hope this helped.
     
  5. Sep 12, 2005 #4
    Thanks for the help!
     
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