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7/27/2019 Measurement of Spesific Heat
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Measurement of Specific Heat
I. The purpose of the experiment
To determine the specific heat of a metal sample.II. Base theory
Heat is defined as the flow of thermal energy, and as such, has S.I. units of Joules. A
quantity of heat can not be measured directly; a measurement for the amount of thermal energy
in transit (heat) can be made by determining its effects on matter.
When a substance gains heat, its total internal energy is increased, the total internal
energy of a substance being defined as the sum of the potential and kinetic energies of all the
molecules in the substance. Temperature is a measure of the average kinetic energy of themolecules in a substance, and the two are directly proportional.
The greater the average kinetic energy of the molecules (i.e. the faster the molecules
move), the greater the temperature of the substance. Thus, heat transferred to a substance
increases its total internal energy, hence the average kinetic energy of its molecules, hence its
temperature. The temperature change in an object is therefore a measure of the heat flow to or
from that object. The amount of heat, Q, is directly proportional to the temperature change in an
object (T), where units of temperature can be C.
When two different substances are supplied with the same amount of heat, their
temperatures may change by a significantly different amount. This implies that the heat flow and
temperature change in an object are related by some proportionality constant that is unique to the
specific substance. This proportionality constant is the specific heat c, of a substance, and is
measured in units of J/kgC.
Q = m.c. T (1)
The second law of thermodynamics states that when two objects of different temperatures
are brought into contact, heat flows from the warmer object to the cooler object. If these objects
remain in contact for enough time, they will reach equilibrium at the same final temperature, Tf.
The first law of thermodynamics tells us that thermal energy is conserved. In other words, the
same amount of heat flows out of the warmer object as flows into the cooler object, or
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Heat lost = Heat gained (2)
A calorimeter is simply a container that isolates substances from their environment in
order to minimize any heat flow of the surroundings into or out of the system. It allows
measurement of the specific heat of a substance (eg. a metal) by equating the heat lost from themetal with the heat gained by another substance in the calorimeter (usually water) and heat
gained by the calorimeter itself. Equation 2, as applied to that system, is thus expressed as
Q metal = Q water + Q calorimeter
m metal cmetal (T h - T f ) = m water cwater (T f - T c ) + m cal ccal (T f - T c ) (3)
In this experiment, the specific heat of an unknown metal, c metal , will be obtained by
mixing a known mass of a hot metal sample (at temperature T h) with cool water in a calorimeter
(at temperature Tc) and observing the final temperatures of the mixture, T f . Solving for c metal
yields:
( ))(
)(..
chmetal metal
c f cal cal water water metal T T cm
T T cmcmc
+= (4)
III. Apparatus of experiment
- Calorimeter,
- Stirrer,
- Thermometer,( with SMS1 oC)
- Steam generator,
- Balance (SMS of balance 0.01 gr)
- Metal sample (Copper, Lead, Iron)
- Water.
IV. Experimental Procedure
1. Fill the heater with water until cover the metal samples.
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2. While waiting for hot water, then measure the mass of copper and empty cup calorimeter
with Balance.
3. Then put the copper into the heater (after boiling), hanging the copper a few inches from
the bottom of the heater to avoid direct contact with the plate heater. Copper metal must becompletely entered into the water, but do not touch the bottom plate or the wall heater
4. Next step is pouring cold water (temperature below room temperature) into the calorimeter
and filled with enough water (until half of calorimeter).and measure the calorimeter
containing water into the insulation wrap, then measure the water temperature with a
thermometer.
5. Move quickly the Copper metal into the calorimeter from the heating without water wetted
and the stirring the water in the calorimeter, and record the temperature of equilibrium thatoccurs in calorimeter.
6. Repeat steps 2-5 for Lead, and Iron.
V. An arrangement of data
No Measurement Copper( Cu) Lead (Pb) Iron
1 Mass of cube ..gr ..gr ..gr
2 Mass of calorimeter + stirrer (m 1) ..gr ..gr ..gr
3 Mass of calorimeter + stirrer + water ..gr ..gr ..gr
4 Initial water temperature (t 1) .. oC .. oC .. oC
5 Final temperature observed .. oC .. oC .. oC
6 Final temperature, for cooling (t 2) .. oC .. oC .. oC
7 Temperature of boiling water (t) .. oC .. oC .. oC
VI. Technique of data analysis (Calculation)
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For measurement using balance with smallest scale 0.001 gram, to determine the mass we
using equation:
SMS mmmm21
==
For measurement using Thermometer with smallest scale 1 0C, to determine the mass we
using equation:
SMS t t t t 21
==
To determine the specific heat of metal sample we use equation:
( ))(
)(..
chmetal metal
c f cal cal water water metal T T cm
T T cmcmc
+=
To determine the uncertainly of the specific heat can following equation:
++
++
= 21
2
21
32112
2
21
3211
3213211 )t-t(M
)t-t()cmm(
)t-t(M
)t-t()cmm(mmc
t t mM t t Mt m
++
++ 2
3
2
21
321122
21
3211
2111321 )t-t(M)t-t()cmm(
)t-t(M)t-t()cmm(
t M
t Mt mmt t mm
22
2
21
321121
2
21
3211
311321 )t-t(M
)t-t()cmm(
)t-t(M
)t-t()cmm(t t
t Mt mmt Mt mm
+
++
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++
++= 22
212
32112
1
2
21
3212
2
21
32
)t-t(M
)t-t()cmm(
)t-t(M
)t-t(
)t-t(M
)t-t( M m
cmc
2
2
2
221
32112
1
2
221
32112
3
2
21
11
)t-t(M
)t-t()cmm(
)t-t(M
)t-t()cmm(
)t-t(M
cmm-t t t
++
++
And then to find the relative error we use equation:
RE =
cc
x 100 %
VII. Experiment result
No Measurement Copper Cube Lead Cube Iron Cube
1 Mass of Cube 62.55 gr 67.70 gr 23.50 gr 2 Mass of Calorimeter + Stirrer (m 1) 56.75 gr 81.50 gr 82.82 gr
3 Mass of Calorimeter + Stirrer + Water (m 1 + m ) 178.90 gr 189.41 gr 199.93 gr
4 Initial Water Temperature 26 0C 26 0C 26 0C5 Final Temperature Observer 29 0C 29 0C 29.5 0C
6 Final Temperature, corrected for cooling (t 2) 29 0C 29 0C 28.5 0C7 Temperature of boiling water (t 1) 104 0C 100 0C 100 0C
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VIII. Data Analysis
A. For Lead Cube :
No Measurement Lead Cube
1 Mass of Cube 67.70 gr 2 Mass of Calorimeter + Stirrer (m1) 81.50 gr
3 Mass of Calorimeter + Stirrer + Water (m1 + m ) 189.41 gr
4 Initial Water Temperature 26 0C
5 Final Temperature Observer 29 0C6 Final Temperature, corrected for cooling (t 2) 29 0C7 Temperature of boiling water (t 1) 100 0C
a. Mass of Lead
M M M =
01,021
70.6721
== SMS M M
M = (67.70 0,005) gram
b. Mass of Water
To find mass of water we subtract
mass of calorimeter that contain with
water with mass of calorimeter (189.41-
81.50 = 97.91 gram)
mmm =
0,0121
97.9121 == SMS mm
m= (97.91 0,005) gram
c. Mass of Calorimeter + Stirrer
111 mmm =
0,0121 81.502111== SMS mm
m1 = (81.50 0.005) gram
d. Temperature of initial Lead
111 t t t =
121
10021
11 == SMS t t
t1 = (100.0 0,5) 0C
e. Initial temperature of water
t3 = 33 t t
t3 = t 3 21
SMS= (26.0 21
1)
t3 = (26.0 0,5) 0C
f. Equilibrium Temperature of Lead and
Water:
t2 = 22 t t
t2 = t 2 21 SMS=(29.0
21 1)0C
t2 = (29.0 0,5) 0C
c = c =)t-t(M
)t-t(c)mm(
21
3211+
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c = 29.0)-(100.067.7026.0)-(29.0 0,215x81,50)(97.91 +
c = 34.8x10115.72
c = 0.024c gr
kal 0
c = ++
++ 22
21
321121
2
21
3212
2
21
32
)t-t(M
)t-t(c.)mm(
)t-t(M
)t-t(
)t-t(M
)t-t( M m
cm
22
2
2
21
321121
2
2
21
321123
2
21
11
)t-t(M
)t-t().cmm(
)t-t(M
)t-t().cmm(
)t-t(M
)cmm(-ttt
++
++
c = 22
22
)005,0(29.0)-(100.067.70
26.0)-0,215(29.0)005,0(
29.0)-(100.067.7026.0)-(29.0 +
22
22
)5,0(29.0)-(100.067.70
0.215)x81,50(-97.91)005,0(
29.0)-(100.067.7026.0)-(29.0 0.215x81.50)(97.91
++
+
22
2
29.0)-067.70(100.26.0)-(29.00.215x81.50)(97.91
)5,0(29.0)-(100.067.70
26.0)-(29.0 0.215x81.50)(97.91 ++
++
c=25,01002.8105,21040.2105,21034.1105,21024.6
23522524524 +++ x x x x
25,01040.225,01040.22222
++ x x
c = 44581312 1044.11044.11060.11044.11048.41074.9 +++++ x x x x x
c = 41004.3
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c = 0.017c gr
kal 0
c = cc = (0,024 0,017) c gr kal
0
Relative error:
RE =
cc
x 100 %
RE =
024.0017.0
x 100 %
RE = 70%
B. For Copper Cube :
No Measurement Copper Cube
1 Mass of Cube 62.55 gr 2 Mass of Calorimeter + Stirrer (m1) 56.75 gr
3 Mass of Calorimeter + Stirrer + Water (m1 + m ) 178.90 gr
4 Initial Water Temperature 26 0C
5 Final Temperature Observer 29 0C6 Final Temperature, corrected for cooling (t 2) 29 0C7 Temperature of boiling water (t 1) 104 0C
a) Mass of Copper
M = M M
01,021
50.6221 == SMS M M
M = (62.55 0,005) gram
b) Mass of Water
To find mass of water we subtract
mass of calorimeter that contain with
water with mass of calorimeter (178.90-
56.75 = 122.15 gram)
m = mm
m= m 21
SMS = (122.15 21
0,01)
m= (122.15 0,005) gram
c) Mass of Calorimeter + Stirrer
111 mmm =
0,0121
56.7521
11== SMS mm
m1 = (56.75 0,005) gram
d) Temperature of initial Copper
t1 = 11 t t
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t1 = t 1 21
SMS = (104.0 21
1)0Ct1 = (104.0 0,5) 0C
e) Initial temperature of water
t3 = 33 t t
t3 = t 3 21
NST = (26.0 21
1)0C
t3 = (26.0 0,5) 0C
f) Equilibrium Temperature of Copper
and Water:
t2 = 22 t t
t2 = t 2 21
SMS = (29.0 21
1)0C
t2 = (29.0 0,5) 0C
c = c =)t-t(M
)t-t(c)mm(
21
3211+
c = 34.4x10115.4
29.0)-(104.062.5526.0)-(29.0 0.215x56.75)(122.15
=+
c = 0.026c gr
kal 0
c =
+
+
++2
2
21
321121
2
21
3212
2
21
32
)t-t(M)t-t(c.)mm(
)t-t(M)t-t(
)t-t(M)t-t(
M m
c
m
2
2
2
221
32112
1
2
221
32112
3
2
21
11
)t-t(M
)t-t().cmm(
)t-t(M
)t-t().cmm(
)t-t(M
)cmm(-ttt
++
++
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c= 22
2
2
)005,0(29.0)-(104.062.55
26.0)-0,215(29.0)005,0(
29.0)-(104.062.5526.0)-(29.0 +
22
22
)5,0(
29.0)-(104.062.55
0.215)x56.75(-122.15)005,0(
29.0)-(104.062.55
26.0)-(29.0 0.215x56.75)(122.15 +
++
22
2
0(29.0)-(104.062.55
26.0)-(29.00.215x56.75)(122.15)5,0(
29.0)-(104.062.5526.0)-(29.0 0.215x)75.56(122.15 ++++
c=
25,01020.8105,21046.2105,21038.1105,21040.623522524524
+++ x x x x
25,01046.225,01046.22222 ++ x x
c = 44581311 1052.11052.11068.11052.11076.41002.1 +++++ x x x
c = 4102.3
c = 0.017c gr
kal 0
c = cc
c = (0.0262 0.017)c gr
kal 0
Relative error:
RE =
cc
x 100 %
RE =
0262.0017.0
x 100 %
RE = 64%
C. For Iron Cube :
No Measurement Iron Cube
1 Mass of Cube 23.50 gr 2 Mass of Calorimeter + Stirrer (m1) 82.82 gr
3 Mass of Calorimeter + Stirrer + Water (m1 + m ) 199.93 gr
4 Initial Water Temperature 26 0C
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5 Final Temperature Observer 29.5 0C6 Final Temperature, corrected for cooling (t 2) 28.5 0C7 Temperature of boiling water (t 1) 100 0C
a) Mass of Iron
M = M M
M= M 21
NST = (23.50 21
0,01) gram = (23.50 0,005) gram
b) Mass of Water
To find mass of water we subtract mass of
calorimeter that contain with water with
mass of calorimeter (199.93-82.82 =
117.11 gram)m = mm
m= m 21
NST = (117.11 21
0,01) gr
m= (117.11 0,005) gram
c) Mass of Calorimeter + Stirrer
111 mmm =
0,0121
82.8221
11== SMS mm
m1 = (82.82 0,005) gram
d) Temperature of initial Iron
t1 = 11 t t
t1 = t 1 2
1SMS = (100.0
2
11)0C
t1 = (100.0 0,5) 0C
e) Initial temperature of water
t3 = 33 t t
t3 = t 3 21
SMS = (26.0 21
1)0C
t3 = (26.0 0,5) 0C
f) Equilibrium Temperature of Iron and
Water:
t2 = 22 t t
t2 = t 2 21
SMS = (29.5 21
1)0C
t2 = (29.5 0,5) 0C
c = c =)t-t(M
)t-t(c)mm(
21
3211+
c = 29.5)-(100.023.5026.0)-(29.5 0.215x)82.82(117.11 +
c = 31.65x10150.4
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c = 0.091c gr
kal 0
c = ++
++ 22
21
321121
2
21
32122
21
32
)t-t(M)t-t(c.)mm(
)t-t(M)t-t(
)t-t(M)t-t(
M mc
m
22
2
221
321121
2
221
321123
2
21
11
)t-t(M)t-t().cmm(
)t-t(M)t-t().cmm(
)t-t(M)cmm(- ttt ++++
c = 22
2
2
)005,0(28.5)-(100.023.50
26.0)-0,215(28.5)005,0(
28.5)-(100.023.5026.0)-(28.5 +
22
22
)5,0(28.5)-(100.023.500.215)x82.82(-117.11
)005,0(28.5)-(100.023.5026.0)-(28.5 0.215x82.82)(117.11
+
+
+
22
28.5)-(100.023.5026.0)-(28.50.215x82.82)(117.11
)5,0(28.5)-(100.023.50
26.0)-(28.5 0.215x82.82)(117.11 ++
++
c =
25,01056.2105,21040.6105,21020.3105,21048.122522524523
+++ x x x x
25,01040.625,01040.62222
++ x x
c =
33471211 1002.11002.11064.11002.11056.21048.5 +++++ x x x x x
c = 3102.2
c = 0.046c gr
kal 0
c = cc
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c = (0,091 0,046)c gr
kal 0
Relative error:
RE =
cc
x 100 %
RE =
091.0046.0
x 100 % = 50.5 %
Table Result Calculation
NO Metal Sample Specific Heat (cal /gram oC)
1 Lead Cube 0,024 0,017
2 Copper Cube 0.026 0.017
3 Iron Cube 0,091 0,046
IX. Interpretation
Specific heat from the theory as the table:
NO Material Specific Heat (cal /gram oC)
1 Lead Cube 0,031
2 Copper Cube 0.0923 Iron Cube 0,107
A.
To find %error for Lead is following equation:
%100% xValueTheory
ValueTheoryValue Experiment error
=
%100031.0
031.0024.0% xerror
=
%10022.0% xerror =
%22% =error
The magnitude of specific heat of Lead that we get from experiment is 0.024 cal/gr oC.
While, the specific heat of lead from the theory is 0.031 cal/gr oC. Error in this experiment is
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high because the are much mistake when we done this experiment. Error in experiment using
Metal sample lead is 22%. because the error more than 10%, so this experiment cant be
accepted or will be rejected.
In this experiment that we done has relative error 70%, so this result not accurately and
this experiment is failed.
B.
To find %error for Copper is following equation:
%100% xValueTheory
ValueTheoryValue Experiment error
=
%100092.0
092.0026.0% xerror
=
%71%10071.0% == xerror
The magnitude of specific heat of copper that we that from experiment is 0.026 cal/gr oC.
While, the specific heat of lead from the theory is 0.092 cal/gr oC. Error in this experiment is
high because the are much mistake when we done this experiment. Error in experiment using
Metal sample lead is 71%. because the error more than 10%, so this experiment cant be
accepted or will be rejected.
In this experiment that we done has relative error 64%, so this result not accurately and
this experiment is failed.
C.
To find %error for Iron is following equation:
%100% xValueTheory
ValueTheoryValue Experiment error
=
%100107.0
107.0091.0% xerror
=
%14%10014.0% == xerror
The magnitude of specific heat of iron that we that from experiment is 0.091 cal/gr oC.
While, the specific heat of lead from the theory is 0.107 cal/gr oC. Error in this experiment is
high because the are much mistake when we done this experiment. Error in experiment using
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Metal sample lead is 14%. because the error bigger than 10%, so this experiment cant be
accepted or will be rejected.
In this experiment that we done has relative error 50.5%, so this result not accurately and
this experiment is failed.
X. Comment
There are several constraints that occur when we done the experiment, such as when moving
the sample of metal from the heater into calorimeter, the heat from the metal is reduced because
water wetting the metal evaporates into the environment.
Another constraints is about sensitivity of the instrument such as thermometer and balance, in
this experiment the balance that we use not in good condition and the thermometer that we use
has bigger smallest scale(1 0C) so not to accurately.
Data that we analysis is in decimal number so when we calculation that data the risk of error
in the calculation will be greater.
The largest error source is from the among water that we use is too much so the temperature
increase relatively small and error is offset by a large SMS(smallest scale) of the thermometer so
that the relative error and %error of the experiment is very high. Recommended in next
experiments must using a bits of water.
From the experiment that we have done, the result of the experiment is not accurately and the
value doesnt approach with the theory, its caused by some error that we have when this
experiment did. The error is classified to three, there are:
1. Gross Error (error that caused by human):
among them is missreading when read the scale of the balance and thermometer, and fault
when we calculate the data in analysis the data because the data is in decimal.
2. Systematic errors (error that caused by instrument and environment):
a) Instrumental error: the error that occurs because of tools didnt work, in this
experiment the tool (balance) is not exactly in god condition and not in friction less.
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b) Enviromental error: the error that occur because of the disturbance of enviroment such as
the temperature not constant.
3. Random error : due to unknown causes and occur even when all systematic error have been
accounted for
XI. Conclusion
From the experiment that we done we get that the value of specific heat Lead is 0.026 cal/gr oC, with accuracy 22%, and relative error 70%. The value of specific heat Copper is 0.026 cal/gr oC, with accuracy 71%, and relative error 64%.The value of specific heat Iron is 0.091 cal/gr oC,
with accuracy 14%, and relative error 50.5%.
The specific heat of iron is bigger than copper and the specific heat of copper bigger than
Lead.
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Questions and Solution:
1. What can you conclude about measuring specific heat of metal?
Solution:
From the experiment that we done we get that the value of specific heat Lead is 0.026 cal/gr oC, with accuracy 22%, and relative error 70%. The value of specific heat Copper is 0.026 cal/gr oC, with accuracy 71%, and relative error 64%.The value of specific heat Iron is 0.091 cal/gr oC,
with accuracy 14%, and relative error 50.5%.The specific heat of iron is bigger than copper and
the specific heat of copper bigger than Lead.
2. Explain the errors of this experiment!
Solution:
1. Gross Error (error that caused by human): among them is missreading when read the scale
of the balance and thermometer, and fault when we calculate the data in analysis the data
because the data is in decimal.
2. Systematic errors (error that caused by instrument and environment):
a) Instrumental error: the error that occurs because of tools didnt work, in this
experiment the tool (balance) is not exactly in god condition and not in friction less.
b) Enviromental error: the error that occur because of the disturbance of enviroment such
as the temperature not constant.
3. Random error : due to unknown causes and occur even when all systematic error have been
accounted for
3. Calculate the percentage errors (the order of accuracy) of this experiment!
To find %error for Lead is following equation:
%100% xValueTheory
ValueTheoryValue Experiment error
=
%100031.0
031.0024.0% xerror
=
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%22%10022.0% == xerror
To find %error for Copper is following equation:
%100% xValueTheory
ValueTheoryValue Experiment error
=
%100092.0
092.0026.0% xerror
=
%71%10071.0% == xerror
To find %error for Iron is following equation:
%100% xValueTheory
ValueTheoryValue Experiment error =
%100107.0
107.0091.0% xerror
=
%14%10014.0% == xerror
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Reference
Djonoputro, B.D. 1977. Teori Ketidakpastian. Bandung: Universitas ITB.
Halliday, D., Resnick, R., and Walker, J. (1993), Fundamentals of Physics , 4th edn (extended),
John Wiley & Sons, New York.
Unname. Measure of specific heat. (online) http://www.utc.edu/Faculty/Harold-Climer/
Sheatlab.pdf be access on April 14 th2011.
Unname. Lab report .(online). http://www.physics.uc.edu/~bortner/labs/sample_lab_report/Lab
%20 Report.htm be access on April 14 th2011
Unname. Specific of Heat. (online) http://phoenix.phys.clemson.edu/labs/223/spheat/index.html
be access on April 14 th2011.
http://www.utc.edu/Faculty/Harold-Climer/%20Sheatlab.pdfhttp://www.utc.edu/Faculty/Harold-Climer/%20Sheatlab.pdfhttp://www.physics.uc.edu/~bortner/labs/sample_lab_report/Lab%20%20http://www.physics.uc.edu/~bortner/labs/sample_lab_report/Lab%20%20http://phoenix.phys.clemson.edu/labs/223/spheat/index.htmlhttp://www.utc.edu/Faculty/Harold-Climer/%20Sheatlab.pdfhttp://www.utc.edu/Faculty/Harold-Climer/%20Sheatlab.pdfhttp://www.physics.uc.edu/~bortner/labs/sample_lab_report/Lab%20%20http://www.physics.uc.edu/~bortner/labs/sample_lab_report/Lab%20%20http://phoenix.phys.clemson.edu/labs/223/spheat/index.html7/27/2019 Measurement of Spesific Heat
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Measurement of Specific Heat
(Physics Laboratory II)
Lab ReportWRITTEN BY,
KOMANG GEDE YUDI ARSANA (NIM. 1013021018)
PHYSICS DEPARTMENT OF EDUCATION
FACULTY OF MATHEMATIC AND SCIENCE
GANESHA UNIVERSITY OF EDUCATION
April 2011
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