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

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Homework Help Overview

The problem involves heating 0.76 kg of water by 15 °C using a 41-ohm resistor connected to a voltage of 230 V. The specific heat of water is provided, and the original poster seeks to determine the time required for this temperature increase.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between power, heat energy, and electrical energy, questioning the validity of equating Q and E in the context of the problem. There is an exploration of the equations for power and heat transfer, with some participants attempting to clarify the definitions and implications of these terms.

Discussion Status

The discussion is ongoing, with some participants affirming the original poster's calculations while others express confusion regarding the relationship between power and heat energy. There is a productive exchange about the assumptions underlying the equations used, particularly in the context of a well-insulated kettle.

Contextual Notes

Participants are navigating the implications of using different forms of energy (electrical vs. thermal) and the assumptions about energy conservation in a closed system, as well as the lack of explicit definitions for certain variables in the equations presented.

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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 ?
 
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You've done it correctly.
 
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 :frown:
 
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.
 
But if we write P=Q/t and P=E/t ..we will get different values for P no?
 
chawki said:
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.
 

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