How long to heat 0.76 kg water by 15 C from a 41-Ohm resistance at 230 V

[chawki]

Homework Statement

There is 0.76 kg of water in a well-insulated electric kettle. The water is being heated by a 41Ohms resistor that is connected to a voltage of 230 V.
The specific heat of water is 4190 J/(kgoC).

Homework Equations

How short is the time needed for the temperature of the water to rise by 15 C ?

The Attempt at a Solution

P=E/t

P=V²/R
P=230²/41
P=1290.24 watt

Q=m*Cp*delta T
Q=0.76*4190*15
Q=47766 J

Is Q here same as E ?
P=Q/t ----> t=Q/P
t=47766/1290.24
t=37.02 s ?

 

[gneill]

You've done it correctly.

 

[chawki]

Dear gneill
I think we are wrong here...we can't write P=Q/t and P=E/t because Q and E are different..
and P which is Power is related to the kettle, means that's it's related to the electrical energy...NOT to the heat content Q..so i think it's wrong to write P=Q/t ?!
I'm confused

 

[gneill]

Well, where do you think the heat is coming from to heat the water? It's not magic.

All energy, in all forms, can be converted into heat energy in one way or another.

 

[chawki]

But if we write P=Q/t and P=E/t ..we will get different values for P no?

 

[gneill]

But if we write P=Q/t and P=E/t ..we will get different values for P no?

Not if the circumstances are such that they must be the same. Here we have a "well-insulated kettle", which means that the electric energy used to heat the kettle's heating element all ends up stating in the water/kettle system. You've defined the rate of electric energy usage to be P = E/t, so P is in watts, or Joules per second, which means that the kettle is providing P watts of heat energy from electric energy. Note that you never provided a number for E; you simply used it to indicate that P was Energy per unit Time.

So heat energy is flowing into the water in the kettle at the rate P joules per second. Now you go to the equation for the temperature rise of the water versus heat added, your

Q=m*Cp*ΔT

Q is in joules. It's the quantity of heat corresponding to the desired temperature rise of ΔT of the water. If heat energy is being provided at rate P joules per second, and you need Q joules to reach the desired temperature, then t = Q/P. If you wish, you can also say that the total electrical energy used during this time is E = P*t = Q.