# Check Potential Difference

#### xswtxoj

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

Homework Helper

#### xswtxoj

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

#### LowlyPion

Homework Helper
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

Oh, btw I'd read that as mF as in milliFarads unless your problem said μf and you wrote mf here.

#### xswtxoj

i did it and got 2.10E -3 V and its its in μf

#### LowlyPion

Homework Helper
i did it and got 2.10E -3 V and its its in μf
I still get something different:

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

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

#### xswtxoj

v= 19826.67 is that the potential difference?

#### LowlyPion

Homework Helper
v= 19826.67 is that the potential difference?
If that's what it calculates out to.

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