Why time should be kept constant: This is because if there is more time for the heater to convert the electrical energy to heat energy, the water will heat up more. This means if I take a result using the same resistance, voltage and mass of water using the time 1 minute, the temperature will be lower than another result I take using the time 30 minutes.

This explains how voltage, mass of water, time affects the temperature. I must keep them constant and only change one variable (resistance) to keep this experiment a fair test.

Also, room temperature may have an impact on the results obtained but since I am using a computer simulation, this problem is eliminated.

Diagram:

This is how the circuit is supposed to be set up:

Method:

- Open application
- Set all the variables to the required amount (30g water, 6 volts, 15 minutes for time, 3 Ω for resistance)
- Press the play button so that the experiment can start.
- I will make a table and put in the current shown by the ammeter, and the temperature which is read off the thermometer.
- I will repeat each ohm of resistance 3 times to increase reliability unless the 3 times is not sufficient. If I need to do more than 3 times, I will do more until I get 3 results that are similar.
- I will repeat step 3 – 5 again for each different resistance I take (3 Ω, 4 Ω, 5 Ω, 6 Ω, 7 Ω, 8 Ω, 9 Ω, 10 Ω)

Safety:

This experiment is not very dangerous. As long as I don’t let the water contact the power pack directly, nothing dangerous will occur. Also, because I am carrying out this experiment on a computer simulation, there will not be any dangerous situations I will encounter.

Accuracy and Reliability:

I will take 3 measurements of temperature and 3 measurements of current at each ohm. I will then use the arithmetic mean of these measurements in my analysis to calculate the average temperature and average current. This will help me minimize unreliable results. If realize any anomaly results I will redo it again until I have 3 results that are similar.

The measurements themselves will be taken as carefully as possible. I will use a thermometer to measure the temperature (±0.5˚C ) and I will take the reading of the current which is shown on the ammeter (±0.01 amp).

Prediction:

I predict that as I increase the resistance, the raise in temperature will gradually be lower. What I mean by this is that if my resistance is 3 Ω, the temperature low will be at its highest point that I can get with the other variables kept constant, if my resistance is 10 Ω, the temperature of the water will be low.

I have found that when current 'i' passes through resistance R under potential difference V for time t seconds then amount of heat produced Q is equal to: [I got this from the internet: www.pinkmonkey.com]

Q: amount of heat produced (joules)

V: Voltage/potential difference (volts)

i: current (amps)

t: time (seconds)

R: resistance (ohm Ω)

Q = V i t

But V = iR

Therefore Q = i²Rt

= V²t

R

Since R is a denominator in this equation, if R is a small number, Q will be a larger number, whereas if R is a big number, Q will be a smaller number.

Graph should looks like:

Pretest:

The 3 resistance didn’t produce a temperature that is 100˚C and above so it wouldn’t limit my results.

Obtaining:

From the first table, the boxes, numbers that are coloured in red indicates that it is an anomaly; therefore it is not used for calculating the average temperature and average current.

The column for current has ‘/’, this separates the current from the other current measured for the trials. For example:

1.78/3.99/2.54/2.08/2.07/2.17, basically is:

1st trial current/ 2nd / 3rd / 4th / 5th / 6th

The first 3 are in red, so as mentioned above, I will not be using it to calculate my average current because it is an anomaly.

Analysis:

From the graph you can see as the resistance increases, the temperature lowers. I have done a straight line of best fit instead of a curve because I do not think it would be needed. The points of the graph are very near to the line of best fit (showing they are together instead of spread out). This indicates that the correlation of the graph is a strong and negative one.

The formula stated in my prediction clearly proves this point:

Q = V²t

R

So as R increases (since I kept V and t constant), Q decreases.

I would expect the maximum temperature I could get (even when I keep my other variables constant) to be when I use 0Ω of resistance, in other words, no resistance at all. However, I was unable to use 0Ω of resistance in the simulation I used, 3Ω was the least resistance I could use and according to the graph, using 3Ω as the resistance produced the highest temperature compared to the other results.

Since I did not change V and t in the formula above, this means Q and R are inversely proportional. This means that the points should be directly on the line of best fit, but not all of them are. This doesn’t disagree with the fact that Q and R are inversely proportional, it might be because of some uncertainty that occurred in the simulation (could be seen by looking at the table of results for current because they don’t follow the rule of V = iR, take resistance 4 as an example: 1.57 x 4 = 6.28, but the voltage is 6 and not 6.28, therefore proves that the experiment includes uncertainty).

The trend of my graph is quite easy to see, firstly the temperature rises dramatically to 74.2˚C and as I increased the resistance, the increments of temperature steadily decreases compared to the initial temperature (0˚C) – trial 1 increased 74.2˚C, trial 2 increased 65.2˚C, trial 3 increased 60.2, trial 4 increased 57, etc.

However, looking at the graph, it seems that the first 4 points creates a line that is steeper in gradient compared to the one the later 4 points can produce (a relatively flat one). This might also be caused by uncertainty.

The graph does support my original prediction.

Evaluation:

Overall, I think that this investigation went quite well. My range of 8 resistance was adequate to plot a graph which supports my prediction. However, I would like to take more readings between 5Ω - 8Ω because it started to not follow the line of best fit.

As I stated in my analysis, there were uncertainty in my obtaining. Since I used a computer simulation to do this investigation, there are very few errors I am able to identify and no improvements.

I could investigate on other factors:

- Relationship between temperature and mass of water
- Relationship between temperature and voltage

Temperature and mass of water: I might do this next time because I would then be able to gain more understanding on the affects of water molecules and heat.

Temperature and voltage: So then I would be able to use the formula I used to prove my prediction in this investigation and reinforce my prediction by investigating a factor that is a nominator in that formula.