Plan
I am going to collect five different sized beakers – 100cm3, 250cm3, 500cm3, 1000cm3 & 2000cm3 each beaker representing different sized animals.
Next I am going to calculate the surface area and volume for each beaker in order to do the equation surface area: volume ratio.
I am going to get my equipment out. I will fill the kettle with water and boil the water, I will pour the water into the beakers and each person will monitor their beaker. When the temperature drops to 80° then I am going to take the temperature of the water every minute for 20 minutes using a stopwatch and a thermometer.
Equipment list
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5 beakers – 100cm3, 250cm3, 500cm3, 1000cm3 & 2000cm3
- 5 thermometers
- Kettle
- Stopwatch
- Results chart
- Water
Fair testing
I am going to make my experiment a fair test by only having one variable. My only variable will be the size of the beaker, which will be the surface area to volume ratio. I am going to make my experiment a fair test by keeping most things the same in every experiment I do. I will keep these the same:
- All of the beakers will rest on the same surface so that heat loss by conduction will be the same.
- None of the beakers will have lids on them.
- All of the beakers are made of Pyrex.
- The starting temperature in every beaker will be 80°C
- The drop in temperature will be measured every minute for 20minutes.
- None of the beakers will be covered so the heat loss by radiation will be the same for each beaker.
- Each experiment will be done twice with an average result taken from the 2 sets of results for each beaker.
Table showing the average surface area to volume ratio.
I calculated the table using the equations:
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Surface area = 2 ¶ r2 + 2 ¶ rh
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Volume = ¶ r2 h
- Surface area: Volume ratio
- r = radius of beaker
- h = height of glass beaker
- ¶ = 3.141592654
Analysing Evidence
I have shown the average cumulative change in temperature every minute in the table. The starting temperatures of all the beakers are the same because my experiment was a fair test only having 1 variable. 1 minute into the experiment my average cumulative change in temperature table clearly shows that after the first minute of the experiment the smallest beaker has the largest heat and overall temperature loss, but the other beakers had similar loss in temperature with 2 sets of beakers having the same results. At 2 minutes the results are more varied and begin to show the pattern I expected in my prediction. The biggest volume beaker (2000cm3) has lost the least amount of heat energy and the smallest beaker (100cm3) with the biggest surface area to volume ratio has lost the most amount of heat energy.
At 5 minutes the 250cm3 beaker and 500cm3 beakers average cumulative change in temperature is the same. I would identify these two results to be slightly inaccurate I would expect the 500cm3 beakers average cumulative drop in temperature slightly less because the volume of the 500cm3 beaker is double that of 250cm3 beaker therefore making the 500cm3 beaker have a smaller surface area to volume ratio than the 2 smaller beakers that have a larger surface area to volume ratio. At 20 minutes the experiment has finished and the results are clear, they match my prediction, I predicted that the smallest beaker will lose the most amount of heat energy because it has a larger surface area to volume ratio also that the 2000cm3 beaker will have the smallest drop in temperature over 20 minutes because it has the smaller surface area to volume ratio. The experiment has also gone as I predicted as the beakers have lost the most and least heat in the order I predicted.
The largest beaker 2000cm3 beaker lost the least amount of heat over 20 minutes because the largest beaker has the smallest surface area to volume ratio. I found out the surface area to volume ratio by doing several calculations first worked out the surface area by using the equation = 2 ¶ r2 + 2 ¶ rh then I worked out the volume of each beaker using the equation = ¶ r2 h. I then worked out the surface area to volume ratio. I came up with the results in the table at the top of page 4.
In the experiment the beakers lose heat by:
- Radiation – Most heat is lost this way because its radiating heat to the outside area.
- Conduction – to the desk
- Convection – The beaker heats the air around which rises and falls.
Overall my results support my original prediction that I made in page 1.
Evaluation
The quality of the results overall was satisfactory, I didn’t get the amount of results I would have liked. I would have preferred to get 3 sets of results for each beaker and then get a more accurate average result for each beaker. I didn’t record 3 lots of results for each beaker because there wasn’t enough time in the practical lessons. There isn’t any particular result that is anomalous. Although in the average cumulative change in temperature table I recorded a couple of results between the 250cm3 beaker and the 500cm3 beaker that were the same. However the results were averages and not the original, so the original results from the 2 experiments could have been different. The odd result could have occurred because the thermometer could have been read 1 degree above or below the actual temperature, which could have made all the difference to the overall average.
If I did this experiment again I would do things slightly different. I would do 3 experiments then calculate an average from the 3 experiments. 3 experiments for each beaker would be more accurate because you have 3 lots of results and it would be easier to identify if a particular experiment results were incorrect. I would also use a electronic thermometer attached to a data logging piece of software which would automatically record the temperature every 10 seconds, but would highlight the results for each minute. Using software that would automatically record the temperature every 10 seconds would reduce the risk of human error and inaccuracy it would also provide a clearer picture of the temperature during the minutes. I would have a larger amount of data to work with and to compare.
