# Trend of Electromagnetic Temperature

1. Mar 20, 2012

### Da Apprentice

In order to determine the temperature that an electromagnet reached after a 3 minute period the electromagnet was placed in a constant volume of water. After 3 minutes the change in temperature of the water was measured and from this using the equation Q=mcΔT the energy in joules transferred was calculated. This same process was repeated for the same electromagnet functioning at a different voltage. Graphing the joules exchanged for each experiment against the voltage that the electromagnet was run at produced what appears to be a linear trend.

Why this trend occurred is unknown. It was thought that the trend would be parabolic due to the following;

Joule's First Law: Q=k*I2*R

Ohms Law: V=I*R (hence I=V/R)

Substitution gives: Q=(k*V2)/R

I'm not sure if joules law is the correct law to use in this case and so this is most likely why I'm wrong. Can anyone explain why the result would be linear or confirm that it should in fact have been parabolic.

Thanks
Z.C

2. Mar 20, 2012

### cjl

Did you only run 2 trials? You won't be able to determine whether the trend is linear or not without more data points.

3. Mar 20, 2012

### Da Apprentice

no in total we ran the experiment at 6 different voltages and an additional point was assumed (at 0,0) because with no voltage the magnet doesn't increase in temperature and therefore no energy would be exchanged between the two.

The graph is attached.

Z.C.

#### Attached Files:

• ###### Graph - Energy Exchange.png
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4. Mar 20, 2012

### 256bits

Well, to start off how did you determine that a line of the one you have drawn is the best fit for your date points?
For each voltage setting, one would do several runs to have multiple data points.

You have several sources of error to begin with.
There is the experimental temperature difference between initial and final - and you do not say whether all runs of the experiment started at the same initial water temperature.
Also there is the voltage setting error.
And of course the mass of the water.

Calculating Q, by only using the increase in temperature of the water does not allow for the temperature increase of your electromagnet nor any heat lost from the water to the environment.