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Where did i go wrong in fudingin the temperature of the gold cube?

  1. Jul 24, 2011 #1
    1. The problem statement, all variables and given/known data

    A cube of gold (resistivity ?R = 2.44 10-8 O m, density ?D = 1.93 104 kg / m3, and specific heat c = 129 J / kg °C) that is 3.5 mm on a side is connected across the terminals of a 40-µF capacitor that initially has a potential difference of 340.0 V between its plates.

    http://www.webassign.net/bauerphys1/26-p-052.gif

    (b) When the capacitor is fully discharged, what is the temperature of the gold cube? (Use 20°C for room temperature.


    3. The attempt at a solution

    i found R = p L /A

    and got 0.69714 * 10^-5 ohm

    Q = C *V = 40 *10^-6 * 340 V = 136 * 10^-4 C

    I = V/R = 340 / (0.69714 * 10^-5) = 487.7066 * 10^5 A

    time t = Q/I = 136*10^-4 / (487.706*10^5) = 0.27885 * 10^-9s

    E = V^2t/R = 340^2(0.27885*10e-9) / 0.69714e-5
    = 4.6239 J

    m = density * volume = 1.93 * 10e4 * 3.5*e-3^3 =82.74875e-5 kg

    E = mc delta T

    T = E/(mc) = 4.6239 / (82.74875e-5*129) = 43

    Tf = 43+20= 63


    mm where did i go wrong?
     
  2. jcsd
  3. Jul 24, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    You're taking the long way around to calculate the energy deposited in the gold, and there are some assumptions being made that look somewhat dubious.

    Why don't you assume that all the energy initially stored in the capacitor will be dumped into the gold (the capacitor will have zero charge, thus zero energy when its fully discharged, so that energy must have gone into heating the gold).

    Given the capacitance and voltage, what's the energy stored in the capacitor?
     
  4. Jul 24, 2011 #3
    sweet i got it thx
     
  5. Jul 24, 2011 #4


    hey Gneli
    could you help me with this problem? i just need a place to start and some guidance

    thank you
    https://www.physicsforums.com/showthread.php?p=3419623#post3419623
     
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