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

AI Thread Summary
The discussion revolves around calculating the temperature increase of a gold cube after being connected to a capacitor. The initial calculations involve determining resistance, charge, current, and energy, but the approach is criticized for being overly complex. A simpler method suggested is to directly use the energy stored in the capacitor to find the temperature change in the gold. The energy stored can be calculated using the capacitance and voltage, leading to a more straightforward solution. Ultimately, the focus is on simplifying the calculations to accurately determine the final temperature of the gold cube.
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Homework Statement



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.

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= 63mm where did i go wrong?
 
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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?
 
sweet i got it thx
 
gneill said:
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?



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

thank you
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