# Determining current through a resistor

1. Apr 5, 2013

### Felicitymaroon

1. The problem statement, all variables and given/known data

A pot containing 15 kg of water at 30 C is heated until all the water is at 90 C, by placing a resistor (having resistance R=930Ω) in the water and running a current through it. What current must be placed through the resistor if the 15 kg of water is raised from 30 C to 90 C in 6 minutes? The specific heat of water is c=4186 j/kgxC), which is the energy needed to heat one kg of water 1 C.
2. Relevant equations
Q=MCdeltat I=delta q/delta t

3. The attempt at a solution
I plugged the information into the equations, but don't know how to include R=930Ω into the equations.

2. Apr 5, 2013

### rock.freak667

rate at which heat is lost by resister = rate at which heat is gained by resistor.

The left side will involve the resistance R.

3. Apr 6, 2013

### CWatters

Power = I * V

V= I * R

so

Power = I2 * R

4. Apr 6, 2013

### Benny851

How does that help this person solve the problem? You don't know V or P. You only know R. I am working on a similar problem and am also stuck. A little help would be much appreciated!

5. Apr 6, 2013

### Staff: Mentor

The details of the water being heated gives you the power, P, required.

6. Apr 6, 2013

### Benny851

how so? Can you please elaborate

7. Apr 6, 2013

### Staff: Mentor

How much energy is required to heat 15kg of water from 30C to 90C? How much time is taken?

8. Apr 6, 2013

### Felicitymaroon

Since, power = energy/time, is the energy the q-value from q=mcΔT?

9. Apr 6, 2013

### Staff: Mentor

Yes.

10. Apr 8, 2013

### CWatters

As others have said.

First look up Heat Capacity on wiki. That gives you the info to work out how much energy is required to heat that much water by ΔTemperature.

Then power = energy/time.

You know R so you can calculate I.