# Electric heating coil

1. Sep 13, 2006

### FlipStyle1308

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)]?

2. Sep 13, 2006

### 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.

3. Sep 13, 2006

### FlipStyle1308

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?

Last edited: Sep 14, 2006
4. Sep 15, 2006

### FlipStyle1308

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

5. Sep 16, 2006

### 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 transfered 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.

6. Sep 16, 2006

### FlipStyle1308

Awesome, thank you so much!