So you should solve for the heat Q directly.

[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

This sadly didn't work...please help me on this level 2 problem

 

[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:

<br /> \frac{V^2}{R}\Delta t = mc\Delta T<br />

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).

 

[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...

 

[Doc Al]

temperature difference is given

then i solved for Q=mcdeltaT
deltaT=80C+273=353K

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.)