measurement of spesific heat

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




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





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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


=

%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.



Metal sample lead is 71%. because the error more than 10%, so this experiment cant be


In this experiment that we done has relative error 64%, so this result not accurately and


C.

To find %error for Iron is following equation:

%100% xValueTheory


=

%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.




15/20

Metal sample lead is 14%. because the error bigger than 10%, so this experiment cant be


In this experiment that we done has relative error 50.5%, so this result not accurately and


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.


17/20

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


=

%100031.0

031.0024.0% xerror

=


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%22%10022.0% == xerror

To find %error for Copper is following equation:

%100% xValueTheory


=

%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


19/20

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.html


20/20

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

measurement of spesific heat

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