Specific Heat Capacity of Tap Water

advertisement
Feyza Selçuk
9-C
İHSAN DOĞRAMACI FOUNDATION
BΔ°LKENT ERZURUM LABORATORY SCHOOL
2nd Semester
2011-2012 Academic Year
9th Grade Physics
LAB REPORT
Name: Feyza Selçuk
Topic: Specific Heat Capacity of Normal Tap Water
1
Feyza Selçuk
9-C
Research Question: Specific heat capacity of pure water is known as 4200 J/(kgoC). But
what is the specific heat capacity of normal tap water?
Hypothesis: In my opinion, normal tap water will have a different specific heat capacity
than water because it is not pure.
Introduction:
The specific heat capacity of a solid or liquid is defined as the heat required raising unit mass
of substance by one degree of temperature. It is Formula is:
π‘’π‘›π‘’π‘Ÿπ‘”π‘¦ π‘‘π‘Ÿπ‘Žπ‘›π‘ π‘“π‘’π‘Ÿπ‘Ÿπ‘’π‘‘ = π‘šπ‘βˆ†π‘‡
Where m is mass of substance, c is specific heat capacity, and βˆ†π‘‡ is change in temperature.
The aim of this experiment is to find the specific heat capacity of normal tap water.
To find the energy transferred, this formula can be used:
π‘π‘œπ‘€π‘’π‘Ÿ:
π‘’π‘›π‘’π‘Ÿπ‘”π‘¦
π‘‘π‘–π‘šπ‘’
where power is known as 1000 W. (the kettle which is used to heat the water in experiment
has 1000 W rating)
2
Feyza Selçuk
9-C
Experimental Setup:
Before start to the experiment, area of experiment, equipments and safety should be
controlled.
List of Equipment:
οƒΌ Kettle (1⁄2 liter capacity, 1000 W rating)
οƒΌ Thermometer
οƒΌ Stopwatch
οƒΌ Tap water (1⁄2 kg)
To get more reliable results, the experiment should be repeated at least 4 times (or more.)
For each experiment, make sure that all items are ready.
Procedure to Follow:
ο‚· Firstly, prepare the apparatus needed during the experiment.
ο‚· Fill the kettle with water. Make sure that it is ½ kg.
ο‚· Then, measure the temperature of water before heating it, and record.
ο‚·
Monitor time by using stopwatch. Make sure that the water does not boil to prevent
evaporating.
ο‚·
After some time, measure the temperature of hot water and record it again.
ο‚·
Repeat these procedures for 4 times and for the each repetition, determine a different
time.
ο‚·
Decide a temperature and heat water until it reaches this target. Measure the times
passed. (to get more reliable result.)
ο‚·
Last, use your recorded data to calculate the specific heat capacity.
3
Feyza Selçuk
9-C
Data:
First Experiment:
Time:
0 second
45th second
Temp.:
26oC
44 oC
βˆ†π‘‘π‘–π‘šπ‘’
45s
18oC
βˆ†π‘‡
The specific heat capacity of tap water found in this experiment is:
π‘’π‘›π‘’π‘Ÿπ‘”π‘¦ π‘‘π‘Ÿπ‘Žπ‘›π‘ π‘“π‘’π‘Ÿπ‘Ÿπ‘’π‘‘ = π‘šπ‘βˆ†π‘‡
π‘π‘œπ‘€π‘’π‘Ÿ =
π‘’π‘›π‘’π‘Ÿπ‘”π‘¦
π‘‘π‘–π‘šπ‘’
1000 W= energy/45s
Energy = 45 000 J
45 000 J = ½ kg x c x 18oC
c = 5000 J/kg oC
Second Experiment:
Time:
Temp.:
0 second
60th second
17 oC
41 oC
βˆ†π‘‘π‘–π‘šπ‘’
60s
24oC
βˆ†π‘‡
4
Feyza Selçuk
9-C
The specific heat capacity of tap water found in this experiment is:
π‘’π‘›π‘’π‘Ÿπ‘”π‘¦ π‘‘π‘Ÿπ‘Žπ‘›π‘ π‘“π‘’π‘Ÿπ‘Ÿπ‘’π‘‘ = π‘šπ‘βˆ†π‘‡
π‘π‘œπ‘€π‘’π‘Ÿ =
π‘’π‘›π‘’π‘Ÿπ‘”π‘¦
π‘‘π‘–π‘šπ‘’
1000 W= energy/60s
Energy = 60 000 J
60 000 J = ½ kg x c x 24oC
c = 5000 J/kg oC
Third Experiment:
Time:
Temp:
0 second
75th second
13 oC
47 oC
βˆ†π‘‘π‘–π‘šπ‘’
75s
34oC
βˆ†π‘‡
The specific heat capacity of tap water found in this experiment is:
π‘’π‘›π‘’π‘Ÿπ‘”π‘¦ π‘‘π‘Ÿπ‘Žπ‘›π‘ π‘“π‘’π‘Ÿπ‘Ÿπ‘’π‘‘ = π‘šπ‘βˆ†π‘‡
π‘π‘œπ‘€π‘’π‘Ÿ =
1000 W= energy/75s
5
π‘’π‘›π‘’π‘Ÿπ‘”π‘¦
π‘‘π‘–π‘šπ‘’
Feyza Selçuk
9-C
Energy = 75 000 J
75 000 J = ½ kg x c x 34oC
C = 4411 J/kg oC (approximately)
Fourth Experiment:
Time:
Temp:
0 second
90th second
19 oC
56oC
βˆ†π‘‘π‘–π‘šπ‘’
90s
37oC
βˆ†π‘‡
The specific heat capacity of tap water found in this experiment is:
π‘’π‘›π‘’π‘Ÿπ‘”π‘¦ π‘‘π‘Ÿπ‘Žπ‘›π‘ π‘“π‘’π‘Ÿπ‘Ÿπ‘’π‘‘ = π‘šπ‘βˆ†π‘‡
π‘π‘œπ‘€π‘’π‘Ÿ =
1000 W= energy/90s
Energy = 90 000 J
90 000 J = ½ kg x c x 37oC
c = 4864 J/kg oC (approximately)
6
π‘’π‘›π‘’π‘Ÿπ‘”π‘¦
π‘‘π‘–π‘šπ‘’
Feyza Selçuk
9-C
Fifth Experiment:
Time:
Temp:
0 second
166th second
12oC
80 oC
βˆ†π‘‘π‘–π‘šπ‘’
166s
βˆ†π‘‡
68oC
The specific heat capacity of tap water found in this experiment is:
π‘’π‘›π‘’π‘Ÿπ‘”π‘¦ π‘‘π‘Ÿπ‘Žπ‘›π‘ π‘“π‘’π‘Ÿπ‘Ÿπ‘’π‘‘ = π‘šπ‘βˆ†π‘‡
π‘π‘œπ‘€π‘’π‘Ÿ =
π‘’π‘›π‘’π‘Ÿπ‘”π‘¦
π‘‘π‘–π‘šπ‘’
1000 W= energy/166s
Energy = 166 000 J
166 000 J = ½ kg x c x 68oC
c = 4882 J/kg oC
Data Result:
Experiment:
Specific Heat Capacity Found: (J/kg oC)
1st (for βˆ†T of 18oC)
5000 J/kg oC
2nd (for βˆ†T of 24oC)
5000 J/kg oC
3rd (for βˆ†T of 34oC)
4411 J/kg oC
4th (for βˆ†T of 37oC)
4864 J/kg oC
5th (for βˆ†T of 68oC)
4882 J/kg oC
7
Feyza Selçuk
9-C
According to these experiments, average specific heat capacity of tap water found is:
4831 J/kg oC.
Conclusion and Evaluation:
As a conclusion, we see the result is different from the specific heat capacity of pure
water. Because tap water is not pure and it contains some other additives, its specific heat
capacity is different. During the experiment, volume of water should be kept constant to get
more reliable result because in this experiment time is our changeable factor. So change in
volume of water would cause unreliable results. The other important thing is using kettle
instead of Bunsen burner. If we used Bunsen burner, then the experiment would not be done
fairly because we cannot provide the same conditionals for each repetition. Also, we make
sure that thermometer just touches to water molecules because to get just water temperature,
thermometer should be used very carefully. In other ways, thermometer will give us different
results. To get more reliable result, this experiment can be re-done but by changing constant
and independent values such as volume of water and time.
In this experiment, average specific heat capacity of tap water is 4831 J/kg oC but by
eliminating human errors and using more precise measuring techniques, we will be able to get
more accurate and more consistent readings.
8
Download