Heat produce by electric current

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Homework Help Overview

The discussion revolves around the heating effect of electric current as observed in a lab experiment involving a resistor and water heating. The original poster reports varying percentages of error in heat production between two runs at different current levels, raising questions about energy loss and measurement accuracy.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculations used for determining heat and work done, questioning the reliability of the resistor's value under different current conditions. There are inquiries about the experimental setup, including temperature stabilization and the mass of water used.

Discussion Status

The conversation is ongoing, with various participants offering insights into potential sources of error and suggesting alternative measurement approaches. There is no explicit consensus, but several productive lines of questioning have emerged regarding the assumptions made during the experiment.

Contextual Notes

Participants note potential constraints such as the accuracy of measurements, the stability of connections in the circuit, and the conditions under which the experiments were conducted, including time allowed for temperature stabilization and the use of consistent water mass.

bnk0430
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I recently finished a lab experiment on heating effect of an electric current(heating water by running current through a resistor). I did two runs of the test first at 2 amp and second at 3 amp. The first run shows a small % error however the second run gave me 26% error. What would explain the loss of heat energy per work done? I'm guessing it's not from a faulty resistor and the current was stable throughout the experiment.
 
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How did you calculate the heat? Did you use the resistor's value? It could change with more current because the resistor is heating up. One should rather use the current and voltage values.
 
I used the W = I2Rt to calculate the work done and Q = (mwcw+mccc+mhch)(Tf-Ti) to calculate the heat produce. In my understanding the 2nd run with higher current should produce more heat than the first correct? However my result is showing lower heat produced than it should for the amount of work done. The only conclusion I have now is the bad insulator in the calorimeter causing the loss of heat.
 
The discrepancy (W<Q) can be explained if R was larger, which usually happens with resistors - their value increases as they heat up (with larger current). Did you use a wire element to heat the water up?
 
Basic_Physics is suggesting that you shouldn't really rely on R being constant or the same at all temperatures. Would have been better to measure V as well as I and use energy = VIt rather than I2Rt.

There are lots of ways to make mistakes when doing calorimetry. Sorry if any of my suggestions are too obvious an error for you to have made.

The equation..

Q = (mwcw+mccc+mhch)(Tf-Ti)

assumes the water, calorimiter and heater all undergo the same temperature rise Tf-Ti. Were any precautions taken/necessary to ensure that was the case? Was enough time allowed both at the start and the end for temperatures to stabilize ? Was second run rushed because lab session ending?

Was Tf anywhere near boiling point? Any steam emerging?

Was new cold water used for the second run? Are you sure the mass of water used was same for both runs or if different was that included in the calculation?

How big was Tf-Ti? More than a few degrees? Accuracy of thermometer? Thermometer battery Ok?
 
In our own lab I came to the conclusion that a bad connection in the electric circuit led to incorrect results, that is that the current was intermittent which means in effect that t should be smaller. I agree fully with CWatters - these type of experiments can be tricky - there is a lot to consider.
 

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