Electric Heating Coil: Potential Difference in 3.5 mins

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Homework Help Overview

The problem involves an electric heating coil used to heat water, focusing on calculating the potential difference at which the coil operates. The context includes specific parameters such as the mass of water, initial and final temperatures, and the resistance of the coil.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between heat absorbed by water and the energy dissipated by the heating coil. Questions arise regarding the appropriate equations to use, particularly concerning temperature and heat transfer. Some participants express confusion over the calculations and seek to identify errors in their reasoning.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to the problem. Some guidance has been offered regarding the simplification of the calculations, emphasizing the relationship between heat, power, and voltage. However, there is no explicit consensus on the correct method or resolution of the problem.

Contextual Notes

Participants note that the solution should not involve concepts like capacitance or charge, indicating a focus on thermal energy and electrical power instead. There is also a mention of the time constraint of 3.5 minutes for the heating process.

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An electric heating coil is immersed in 4.2 kg of water at 22°C. The coil, which has a resistance of 260 Ω, warms the water to 33°C in 3.5 mins. What is the potential difference at which the coil operates?

I was looking through the chapter, and did not see any equations that I can use that involve temperatures, but does this problem involve the equation q(t) = CE[1-e^(-t/T)]?
 
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What is the quantity of heat absorbed by water? Heat absorbed by water is heat dissipated by heater. Once you know the wattage and resistance of heater, calculate the voltage.
 
Here's what I did, and I got my answer wrong, but see if you can catch my mistake:

C = Q/mT = 4186 J/(kgK) = Q/(4.2 kg)(306.15 K - 295.15 K) => Q = 193,393.2 C
I = Q/t = 193,393.2 C / 210 s = 920.92 A
V = IR = (920.92 A)(260 ohm) = 239,439.2 V

I thought my overall answer seemed somewhat high, and I was right. Are any of the equations I used incorrect?
 
Last edited:
Bump! Is anyone able to determine what I am doing wrong?
 
The solution for this problem should not involve capacitanc, charges etc.
It's much simpler.
First you calculate the amount of heat the water absorbed.
Heat=specific heat * mass * temperature difference
The heat is the energy that the resistor transferred to the water, in a certain amount of time. Hence the power is
Power = energy / time
now use the formula
(Voltage)^2 / resistance = power
and that's it.
 
Awesome, thank you so much!
 

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