# Homework Help: E/m resistance problem

1. Jul 24, 2004

### imationrouter03

A coil is to be used as an immersion heater for boiling water. The coil is to operate at a voltage of 130 V and is to heat an amount of water with a volume of 251 cm^3 by 80.0 degree celsius in a time interval of 6.30 minutes.
Use 4190 J/(kg*K) for the specific heat capacity of water and 1000 kg/m^3 for the density of water.

The question is :
What must the resistance of the coil be (assumed temperature-independent)?

I have tried the following:
i solved for the m=DV=(1000kg/m^3)(2.51*10^-6m^3)=.251 kg
then i solved for Q=mcdeltaT
deltaT=80C+273=353K
Q=.251 kg(4190J/(kg*K))(353K)=371246.57J
then i divided it by 6.30 min
Q=(371246.57J/6.30min)*(1min/60s)=982.13J/s
R=V^2/P=(130^2)/(982.13)=17.207ohms

2. Jul 25, 2004

### maverick280857

First check your calculations and units.

Essentially, the resistive heat energy (the so called "i-squared-r-t heating") heats up the water, so the equation that you need to use is:

$$\frac{V^2}{R}\Delta t = mc\Delta T$$

where t and T denote variables of time and temperature respectively. You can recast this expression in the form you have used in your solution (dividing the right hand side by the time interval).

3. Jul 27, 2004

### imationrouter03

im still not getting the problem right... i've used that formulat but i ended up with my previous response which was wrong. I've double checked my units.. what can i be doing wrong?
they are asking for the resistance of the coil....

4. Jul 27, 2004

### Staff: Mentor

temperature difference is given

It looks like you are treating the 80 degrees as a temperature and then converting it to Kelvin. No! The 80 celsius degrees is the temperature difference.

(Note that 80 C-degrees = 80 K-degrees: the Kelvin and Celsius scale use the same size degree, just a different zero point.)