Electric Heating Coil: Potential Difference in 3.5 mins

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An electric heating coil is immersed in 4.2 kg of water at 22°C. The coil, which has a resistance of 260 Ω, warms the water to 33°C in 3.5 mins. What is the potential difference at which the coil operates?

I was looking through the chapter, and did not see any equations that I can use that involve temperatures, but does this problem involve the equation q(t) = CE[1-e^(-t/T)]?

 

[quark]

What is the quantity of heat absorbed by water? Heat absorbed by water is heat dissipated by heater. Once you know the wattage and resistance of heater, calculate the voltage.

 

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Here's what I did, and I got my answer wrong, but see if you can catch my mistake:

C = Q/mT = 4186 J/(kgK) = Q/(4.2 kg)(306.15 K - 295.15 K) => Q = 193,393.2 C
I = Q/t = 193,393.2 C / 210 s = 920.92 A
V = IR = (920.92 A)(260 ohm) = 239,439.2 V

I thought my overall answer seemed somewhat high, and I was right. Are any of the equations I used incorrect?

 

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Bump! Is anyone able to determine what I am doing wrong?

 

[pseudovector]

The solution for this problem should not involve capacitanc, charges etc.
It's much simpler.
First you calculate the amount of heat the water absorbed.
Heat=specific heat * mass * temperature difference
The heat is the energy that the resistor transferred to the water, in a certain amount of time. Hence the power is
Power = energy / time
now use the formula
(Voltage)^2 / resistance = power
and that's it.

 

[FlipStyle1308]

Awesome, thank you so much!