To explore how the efficiency of a kettle changes with the volume of water that is used to heat it.
Prediction:
Through a kettle heat is lost through mainly three areas. Firstly, through radiation, this is when heat is thrown out of the kettle via the kettle wall out of the main body of water. The second way is by conduction, this is where heat passes through the main body of water via the kettle wall. And, also by convection, where some of the water particles carrying energy are lost through water vapour through the spout of the kettle. In all these ways of heat loss, everything revolves around the surface area, where all the heat is lost. The bigger the surface area of the volume of water, the more heat energy will be lost.
I have done some preliminary work, which proves that if the volume of the body of water doubles, the surface area is unproportional to the body of water even though you would expect it to double. I have looked at the effects of changing the volume of water and calculated its surface area through the expression 2 ? r(r+h). I am treating the kettle as a cylinder. As the kettle has a volume of 0. 5 litres of water, the surface area is 387 cm2, at 1. 0 litres of water, the surface area is 528 cm2, which demonstrates that doubling the volume does not actually double the surface area.
Preliminary Calculations
Preliminary Calculations:
Volume/Mass of water in a kettle
(Litres/Kilograms)
Depth of water
(cm)
Surface area of cylinder (cm2)
0. 5 l
500 kg
3. 6
387
0. 75 l
750 kg
5. 4
457
1. 0 l
1000 kg
7. 2
528
1. 25 l
1250 kg
9. 0
599
1. 5 l
1500 kg
10. 8
670
Method
Firstly, set up the equipment. In this experiment I am timing to see how long it takes to heat water of given temperatures. I will then fill the kettle up with 500ml of tap water, measuring the water accurately in a measuring cylinder. I will then put the temperature probe in to the kettle at approximately 1cm away from the heating element and ensure that in all my experiments I keep the probe the same distance away from the element because otherwise the temperature of the water will vary throughout the kettle. I will then plug the kettle in and then turn the kettle on and when the temperature reaches 20? C begin the stop clock. Then wait till the water reaches the desired temperature, when it does press the reset button which will give me a lap time, then record the time and again press the rest button which will continue on with the time. When it reaches the next given temperature, press the reset button again and record the lap time and once again press the same button to continue on with timing the heating process, and carrying on repeating this process till I have recorded all the times in my results table.
I will then repeat this experiment 4 more times but to keep the investigation fair, make sure that for every individual experiment………..
* I wash the kettle out thoroughly with cold water to cool the element down, as it will still be warm from the previous experiment.
* I change the water, from hot to cold, so the common starting point remains 20? C for every experiment.
* Use the same common starting point (temperature), because the water’s temperature from the tap can vary slightly.
* I will use the same range of temperatures for each experiment.
In this experiment I am not heating the water to 100? C because it doesn’t always reach this temperature of boiling.
When I have gathered my results, I will need to use some calculations to reach the aim of the experiment, to find out how efficient the kettle actually is.
To work out the efficiency, I can use the simple equation:
Efficiency = Useful energy
Supplied energy – which in my case is:
Efficiency = Heat absorbed by water
Electricity supplied
To calculate the top part of the equation ‘ Heat absorbed by water’, or basically the energy that was successfully used, we use the Specific Heat Capacity of Water, which is 4200J / kg / ? C. This means that for every 4200J of energy taken in by every kilogram of mass, by every temperature in crease of one degree.
An example of working out the ‘ Heat absorbed by the water’ for half a litre ( or half a kilogram ), heated up by 30?, I would do
4200 x 0. 5 x 30 = 6300
To calculate the bottom part of the equation, the ‘ Electricity supplied’, I will use the equation E = P x t, I know what the power (P) rating for a kettle is, 2200, and I will have worked out the time in our experiments.
I just have to divide the ‘ Heat absorbed by water’ by the
‘ Electricity supplied’ and this will give me the efficiency as a decimal, so just times it by 100 to get the percentage.
Results Tables:
For 500ml (0. 5kg) of water.
Temperature Change (? C)
Time (seconds)
Electricity supplied (J)
Energy absorbed
Efficiency (%)
20-35
42
92400
31500
34
20-50
64
140800
63000
45
20-65
77
169400
945000
56
20-80
82
180400
126000
70
20-95
90
195000
157500
80
For 750ml (0. 75kg) of water.
Temperature Change (? C)
Time (seconds)
Electricity supplied (J)
Energy absorbed
Efficiency (%)
20-35
26
57200
49250
82
20-50
49
107800
94500
88
20-65
74
162800
141750
87
20-80
98
215600
189000
88
20-95
125
275000
236250
86
For 1 litre (1kg) of water.
Temperature Change (? C)
Time (seconds)
Electricity supplied (J)
Energy absorbed
Efficiency (%)
20-35
33
72600
63000
87
20-50
64
140800
126000
89
20-65
97
213400
189000
89
20-80
129
283800
252000
89
20-95
162
356400
315000
89
For 1. 25 litres (1. 25kg) of water.
Temperature Change (? C)
Time (seconds)
Electricity supplied (J)
Energy absorbed
Efficiency (%)
20-35
38
83600
78750
94
20-50
77
169400
157500
93
20-65
117
257400
236250
92
20-80
180
352000
315000
89
20-95
199
437800
393750
90
For 1. 5 litres (1. 5kg) of water.
Temperature Change (? C)
Time (seconds)
Electricity supplied (J)
Energy absorbed
Efficiency (%)
20-35
35
121000
94500
78
20-50
101
222200
189000
85
20-65
149
327800
283500
86
20-80
196
431200
378000
88
20-95
244
536800
472500