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

Click For Summary

Homework Help Overview

The problem involves calculating the final temperature of a gold cube after it has been connected to a capacitor. The cube's dimensions, resistivity, density, and specific heat are provided, along with the capacitor's capacitance and initial voltage. The original poster attempts to determine the temperature change based on energy calculations involving the capacitor's discharge.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster calculates resistance, charge, current, and energy, but questions where their calculations may have gone wrong. Some participants suggest reconsidering the assumptions about energy transfer from the capacitor to the gold cube.

Discussion Status

Participants are engaging in a back-and-forth about the assumptions made in the calculations. Some guidance has been offered regarding the energy stored in the capacitor and its relation to the heating of the gold cube. There is no explicit consensus on the correct approach yet, but the discussion is moving towards clarifying the energy transfer process.

Contextual Notes

The original poster is working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There are also assumptions about energy conservation that are being questioned in the discussion.

yjk91
Messages
73
Reaction score
0

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?
 
Physics news on Phys.org
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
https://www.physicsforums.com/showthread.php?p=3419623#post3419623
 

Similar threads

Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
3K
Replies
2
Views
12K
  • · Replies 41 ·
2
Replies
41
Views
6K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
4
Views
2K