Determining current through a resistor

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To determine the current required to heat 15 kg of water from 30°C to 90°C in 6 minutes using a 930Ω resistor, first calculate the energy needed using the formula Q=MCΔT, where M is mass, C is specific heat, and ΔT is the temperature change. The power needed can then be found using the relationship power = energy/time. Since power can also be expressed as P = I²R, you can rearrange this to find the current I once you have calculated the required power. Understanding these relationships allows for the calculation of the current necessary to achieve the desired heating.
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Homework Statement



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.
Answer is in Amperes

Homework Equations


Q=MCdeltat I=delta q/delta t


The Attempt at a Solution


I plugged the information into the equations, but don't know how to include R=930Ω into the equations.
 
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rate at which heat is lost by resister = rate at which heat is gained by resistor.

The left side will involve the resistance R.
 
but don't know how to include R=930Ω into the equations

Power = I * V

V= I * R

so

Power = I2 * R
 
CWatters said:
Power = I * V

V= I * R

so

Power = I2 * R

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!
 
Benny851 said:
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!

The details of the water being heated gives you the power, P, required.
 
CWatters said:
Power = I * V

V= I * R

so

Power = I2 * R

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

how so? Can you please elaborate
 
Benny851 said:
how so? Can you please elaborate

How much energy is required to heat 15kg of water from 30C to 90C? How much time is taken?
 
Since, power = energy/time, is the energy the q-value from q=mcΔT?
 
Felicitymaroon said:
Since, power = energy/time, is the energy the q-value from q=mcΔT?

Yes.
 
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Benny851 said:
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!

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.
 
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