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The purpose of the investigation is to find the factors affecting the rate cooling. Also to show that there is a relationship between two factors that affect the rate of cooling if there is two factors. Prediction In my experiments I am going to use 6 varied sized beakers, I will keep the volume inside the beakers at a constant 50ml. I would like to prove that when I change the size of the beaker, larger, then the cooling would take less time. This will be because more surface area will be exposed. It is also likely that if I increase the volume but keep the same size beaker the cooling will take longer.
This is because the liquid not exposed to the air will retain its heat for a longer period of time. ? In the first experiment a lot of heat will be lost because the particles have a large amount of energy. ? In the second experiment a fair amount of heat will be lost, at about half the rate of the first experiment. This is because the particles still have energy but most of this energy has been used because of lack of heat. ? In the third experiment not much heat will be lost. This is because there is not much energy left in the liquid.
Heat will be lost when there is lots of energy, when heat is lost the energy decreases causing the rate of cooling to decrease. I predict that when the temperature of the solution is increased, the speed of the reaction will increase. This is because the molecules will have more energy; therefore making them move faster. The molecules will then disperse heat evenly. The graphs I plot of my results will show a relationship between temperature and time. I believe that the graph will begin with a steep gradient, as the heat is high but will then begin to level out as the heat decreases. Method Equipment Used:
50cm3 beaker Stopwatch 100cm3 beaker Bunsen burner 250cm3 beaker Tri-Pod 400cm3 beaker Heat proof-mat 800cm3 beaker Thermometer Water Gauze ? To keep an accurate measurement I will use the same thermometer and end the experiment when the liquid reaches 50oC. ? I will measure the time it takes from when the liquid reaches 95oC. ? To keep accurate and constant results I will ask a friend to measure the time while I measure the temperature of the liquid. I will keep the total volume of water at a constant 50cm3. The stopwatch will be started as soon as the liquid reaches 95oC and when the Bunsen burner has been removed.
For this experiment I am going to change the surface area exposed to the surrounding air. I will use six different sized beakers each with a different radius in order to gather enough information to see a pattern if there is one. Results Experiment Number Area of Beaker Cm2 Capacity of Beaker Time (Minutes) Average Time (Minutes) ____10000____ Average Time (to 1 decimal place) .
Experiment Number Volume Cm3 Temperature oC Average Temperature oC ____1000____ Average Temperature oC (1 decimal place)- Conclusion When looking at the graph below it is quite obvious to see that the speed of the cooling decreases as the surface area is decreases. It is therefore easy to predict that the speed of the reaction is directly proportional to the surface area.
By taking results from the two experiments you cane see that as the surface area increases then the rate of cooling also increases. Therefore it is clear through my results, that rate of cooling and surface area, are directly proportional to each other when using water. For example: when the surface area increases the rate of cooling decreases and when the surface area decreases the rate of cooling increases. Evaluation Through my experiment I attempted to take as much precaution as possible to make the test a fair and accurate test.
However my results and patterns in my graphs show that not all the tests were accurate, as some points do not fit in with the obvious pattern. One of main problem was that I had to estimate how much water to put into the beaker before heating; this was not easy as the water evaporated off and on several occasions we were left with not enough liquid in the beaker. To help with this problem I could have heated more than enough water and the poured the correct amount into another beaker, however, a lot of heat my have been lost in this way.
A waterproof container may have also been helpful as it was have meant that no water to could be lost through evaporation. Another factor affecting my results could have been the humidity within the room on different days. This would have effected the cooling slightly but enough to make a difference. The slight heat would have made the molecules slightly more energetic but not enough to disrupt the experiment I don’t believe. An alternative may have been to use a sealed off laboratory or to do all the experiments within the same climate
Overall I was delighted with how the experiments ran and how my results were gathered and turned out. To perform the experiment again I would need to use many mechanical aids as possible so that I could exclude human error wherever needed. Thus making more accurate and fairer results. Show preview only The above preview is unformatted text This student written piece of work is one of many that can be found in our GCSE Patterns of Behaviour section.