Calculating Potential Difference for Capacitor Heating of Water

[xswtxoj]

Homework Statement

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?

Homework Equations

Q= mC delta T
W= Q sq/2C
V=W/C

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

 

[LowlyPion]

What's the PE of a charged capacitor?

PE = 1/2*C*V²

You have your 910 J = 1/2*(.00463)*V²

http://en.wikipedia.org/wiki/Capacitor#Energy_storage

 

[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]

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

Check your algebra again.

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]

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]

v= 19826.67 is that the potential difference?

If that's what it calculates out to.