How the Temperature of Water Is Affected By the Time It Is Heated.
How the temperature of water is affected by the time it is heated.
I am going to investigate the how an amount of a substance is affected by a heater made from glass and copper wire.
There are many variables I need to consider.
* The voltage and current - the higher the voltage and current running into the heater means that there will be more energy in. This means that there will be more energy out and so the water will get more energy from the heater in a smaller space of time. This means that the water will heat up quicker. The heating is reversed if there is less voltage.
* The length of wire and number of coils - The longer the copper wire means that the number of coils around the glass will increase. This means that there will be more resistance and so more energy will be lost into the substance whilst it is passing through the wire. This means that there will be more energy put into the substance and it will heat up faster.
* The amount of time - When the heater is in the substance, energy comes from the power pack, through the resister (the heater) and into the substance to be heated. It is this energy that heats up the substance. The longer the heater is left in the water means that more energy can pass through the heater and into the substance and so making it hotter.
* The amount of substance to be heated - The more substance to be heated needs more energy to heat it because there are more particles in the substance. So the more substance I have means that it will take longer to heat up, if I have less substance then it will not take as long.
* The substance I heat - different liquids have different specific heat capacities. Specific heat capacity (as found in my "Key Science - Physics" textbook) is the energy needed to raise the temperature of 1kg of material by 1?C. Specific heat capacity is always a constant for the substance but because different substances have different consistencies and structure then they have different specific heat capacities and so will need more or less energy to heat them up.
I am going to investigate the time that the heater is left in a substance affects the temperature of it. I am going to use 100cm3 of water because it is easy to obtain and I know that the specific heat capacity is 4.2 J/g. I am going to see how the time affects the temperature because I think that it is the simplest experiment to do and easiest to compare results. It can also be easily done in a school classroom, as it does not need any specialist equipment that the school doesn't have.
Method
I am going to make a heater using a glass rod and copper wire. I will coil the copper wire around the glass rod and thus make a heater when I flow electricity through the wire. (As shown below)
I will then set up a circuit, containing a power pack (to power my heater) an ammeter (so I can record the current), a voltmeter (to record the volts) and my heater.
I will place my heater in a 100cm3 beaker of water for different amounts of time and see what the temperature of the water is after each time. I will take the temperature every 2 minutes from 0 to 20 minutes, always leaving the heater in the water. Altogether I will get 11 temperatures. I will then repeat the whole experiment two more times, this will increase the accuracy and mean that I will have more chance of getting reliable results. At the end of my 3 experiments I will have 33 temperatures and I will then work ...
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I will place my heater in a 100cm3 beaker of water for different amounts of time and see what the temperature of the water is after each time. I will take the temperature every 2 minutes from 0 to 20 minutes, always leaving the heater in the water. Altogether I will get 11 temperatures. I will then repeat the whole experiment two more times, this will increase the accuracy and mean that I will have more chance of getting reliable results. At the end of my 3 experiments I will have 33 temperatures and I will then work out an average for each time making my results as accurate as possible. If I get any annolomus results then I will not include these when working out my average because it will affect it too much.
How I will keep the test fair.
I will keep the test fair by not altering any of the variables that I mentioned at the beginning of my investigation that I think will affect the temperature of the water. The only thing that will change each time is the amount of time the heater has been left in the water. I will not change the voltage and so not change the current so no more or less energy will come out of the heater and go into the water, thus affecting the temperature. After taking a preliminary reading I have decided that I will always use 4 V, and so 3.2 A to go into my heater.
I will not change the length of wire or the numbers of coils around it as each time I will be using the same heater. This will mean that each time there will be the same amount of resistance and so the same proportion will be lost each time.
I will not change the substance as each time I will be using tap water, I will use the same water throughout the experiment. I will only change it when I start heating again from the beginning. I will do this so that I am not heating water that has already been heated and so is hot. I will always use tap water so that the temperature that I start heating the water at will not change too much. I will always take the starting temperature of the water before I start. If it is not within 5 degrees of my starting temperature in my first attempt then I will get some more water or heat it so it is closer. On my first attempt it will not matter what my temperature is as long as I record it so I can use it for my results.
I will use a thermometer with the biggest possible scale that I can find so it will easier to get accurate and precise results. When taking the temperature I will not take the thermometer out of the water and will take the temperature exactly at the time and not wait for it to become stable.
I will also always use the same amount of water (100cm3). So that it will always have the same amount of particles to be heated.
I will always use the same beaker, crocodile clips, power pack, voltmeter and ammeter just in case they would affect my results in any way if I did chance them.
Prediction
I think that the longer I leave the heater in the water, the hotter the water will become. I think this because the longer the heater is 'turned on' in the water, the more energy it is receiving from the power pack. This energy is transferred into the water as the copper wire vibrates it causes the molecules in the water to move quicker. These molecules therefore have more energy and so move quicker, hit each other more often and hit each other at a fast speed. This makes the water hotter as the molecules have more energy. Therefore the longer the heater is 'turned on', the more energy it is receiving from the power pack. The more energy the heater has means that when it is active in the water it gives it more energy and so increasing the temperature.
I think that the amount of time and change in heat will be proportional because each time the same amount of energy is going into the heater and so into the water to heat it up.
Formulae I will need to use.
I will use the following formulae to work out the temperature of my water, the energy out of the heater and the energy put into the water.
Energy out of heater
Energy = Mass x Specific heat Capacity x ??(change in temperature)
From my "Key Science - Physics" textbook, I found that Specific heat capacity is the energy needed to raise the temperature of 1kg of material by 1?C. Specific heat capacity is always a constant and the specific heat capacity for water is 4.2 J / g. This means that 4.2 Joules of energy is needed to raise the water temperature by 1oC.
Energy into heater
Energy = Voltage x Current (I) x Time in seconds)
Using the energy into water formula and the energy from heater formula I can work out how much energy is lost each time and the efficiency of my heater.
Temperature of the water
By using the above formulae and rearranging them I can predict what the temperature rise will be between each time. This means that I will be able to predict what the final temperature of the water will be after 20 minutes of heating. This formula will be:
Temperature rise = Energy from heater
M x C
Temperature rise = V x I x T
M x C
Temperature after a certain amount of time (T) = starting temperature + VIT
MC
I will now work out how much my water should rise by each time and thus work out the temperature at each time. I took the temperature of 100cm3 of water that I got from the tap that I will use as my starting temperature. The water was 20?C.
Time(s)
Temperature (oC) (3dp)
Temperature change (?C)
0
20.000
20
23.657
3.7
240
27.314
3.7
360
30.971
3.7
480
34.629
3.7
600
38.286
3.7
720
41.943
3.7
840
45.600
3.7
960
49.257
3.7
080
52.914
3.7
200
56.571
3.7
TOTAL TEMPERATURE CHANGE
37 ?
I can see that after heating my water for 1200 seconds (20minutes), my temperature should be around 57?C. This shows a temperature rise of 37?C. I will draw a graph of these results to see if they are proportional in any way.
This shows that the temperature rise must be proportional with the time that the heater is left in the water because the graph is a straight line.
Results
Here are my results; I have called my 1st 2nd and 3rd attempts, one, two and three respectively.
Temperature (oC)
Time (seconds)
One
Two
Three
Average
Temperature change
0
20.25
21.5
24
21.91667
20
23.5
24.5
29
25.66667
3.8
240
28.5
29
33.5
30.33333
4.7
360
32.5
35.5
37.5
35.16667
4.8
480
38.5
41.5
40
40
4.8
600
41.5
49
43
44.5
4.5
720
45
53
46
48
3.5
840
48.5
58
49
51.83333
3.8
960
51
65
51.5
55.83333
4.0
080
54
67.5
54
58.5
2.7
200
55
69
55.5
59.83333
.3
TOTAL TEMPERATURE CHANGE
37.9
I can also work out how efficient my heater was and the energy lost each time using the two formulae for energy in and out that I explained at the beginning.
Efficiency = Energy out of heater
Energy into heater
Energy in (J)
Energy out (J)
Efficiency (%)
536
2000
30.2
3072
800
58.6
4608
900
41.2
6144
700
27.7
7680
600
20.8
9216
440
5.6
0752
560
4.5
2288
160
9.4
3824
680
4.9
I will draw a graph for each of the results tables, as it will be easier to examine the results from there.
Conclusion
After looking at my results and graph I can see that I was right. As I said in my prediction, the longer the heater was kept in the water, the hotter it got. This is because the power pack gives the heater energy, this energy is transferred into the water as the copper wire vibrates it causes the molecules in the water to move quicker. These molecules therefore have more energy and so move quicker, hit each other more often and hit each other at a fast speed. This creates more pressure in the water but it also makes it hotter as the molecules have more energy. Therefore the longer the heater is 'turned on', the more energy it is receiving from the power pack. The more energy the heater has means that when it is active in the water it gives it more energy and so increasing the temperature.
The reason I have a curve and not a straight line as I predicted is because as the water gets hotter there is more energy. The molecules are moving faster and so it is easier for the heat to escape. The temperature change gets lower and lower as more and more heat is lost. This is also why my heater gets less and less efficient.
The temperature and time were not exactly proportional in my experiment. I got a very slight curve that curved more as the time went on and the water got hotter. It ended up quite close to what I predicted as my final temperature was 60?C. It would have been closer I think but in my experiment the water was slightly hotter to start with than that that I used in my prediction. The total temperature change though is very similar;it was 36.571 ?C in my prediction and 37.917 ?C in my experiment.
As more and more energy went into the heater and it heated up the water more, it got less and less efficient. The efficiencies dropped quite rapidly until at 20 minutes when the heater was only 4.9% efficient. The energy in my heater that was lost could have gone to a number of different places, most of it would've gone into the air, this is because I did not insulate my beaker in any way. Some would go into the glass and some into the copper, not much of the energy actually goes into heating the water.
Analysis.
After doing my experiment I was quite pleased as my temperatures were close to what I predicted. The reason why they are not exactly and why my temperature change gets less and less towards the end of my experiment is because the formulae I used do not take into account energy loss. As my water is heated not all the energy goes into the water. Some goes through the heater and back into the power pack. Some goes into the air, as my water got hotter more and more went into the air because there is more energy moving around. Heat is also lost into the container and so not all the energy out of the heater goes into heating the water up.
To make my results more reliable I could use a thermometer with a larger scale so it would be easier to read or use a digital thermometer, which would be even more accurate. If I had more money or resources I could use a data log where it would automatically record the temperature at each required time.
I would always start with the same temperature of water to make it easier when working out my temperature changes. Also it will be easier to compare my results. I would also insulate my container and so hopefully reduce heat loss. This would hopefully mean if I drew a graph my results would be closer to a straight line. I would also increase accuracy but doing my experiment more times. This would mean that I have a larger range of results to take an average of and so would mean they would be more accurate and easier to spot anomalous results.
I could also investigate more temperatures (e.g. up to 40 minutes) or take the temperature more often. This would mean that I would get a better range of results and so have a better idea of how the water is heated.
I could extend the investigation by changing some of the variables I thought of at the beginning of my investigation. For example, I could see how using a longer piece of wire and so putting more coils on my heater affects the overall temperature of the water. I could also try and see what happens if I used more or less water and if then my heater would become more efficient. I could also try using a different substance with a different specific heat capacity and see if that changes the results. I could change energy into the heater and see if it becomes more or less efficient.