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Lab 3: Thermal Expansion
Phys. 4C
1
Objective
• To find the coefficient of thermal expansion for copper and aluminum from given data, and perform
an error analysis to understand the precision of the experiment.
2
Introduction
Most materials expand when heated through a temperature range that does not produce a phase change.
The added heat increases the average amplitude of vibration of the atoms in the material which increases
the average separation between the atoms.
If an object of length L at some initial temperature Ti is heated to temperature Tf , the increase in length
∆L = Lf − Li is characteristic of the composition of the object. The coefficient of linear expansion α is
given by:
∆L
(1)
L∆T
In this lab we will use data gathered by students in a previous term to calculate α and do an error analysis
to investigate the precision of the lab and see whether the data agrees with the accepted values of α for
copper and aluminum:
α=
αCu = 17.6 × 10−6 (◦ C)−1
αAl = 23.4 × 10−6 (◦ C)−1
Recall the uncertainty (δf ) of a function of x, y, and z is given by
s
2
2
2
∂f
∂f
∂f
2
2
2
δf =
(δx) +
(δy) +
(δz)
∂x
∂y
∂z
(2)
Where δx, δy, and δz (the uncertainties in x, y, and z, respectively) are independent and random.
The uncertainty in a sum or product f = x ± y is determined by adding the uncertainties of the inputs in
quadrature:
q
2
2
δf = (δx) + (δy)
(3)
3
Experimental Procedure
This is an explanation of the experimental procedure followed in previous terms. You will not actually
perform these steps. However, you need to understand what was done so that you can properly analyze the
data and draw conclusions.
The basic idea of the experiment is to heat a metal tube by running steam into it, indirectly measure the
change in its temperature, and also measure the change in its length.
1. The length of the copper tube at room temperature, L, was measured with a meterstick. Care was
taken to measure just the length of the part of the rod whose expansion would be measured. The L
was taken to be the distance from the pin in one end of the rod, which was held fixed in place by the
Steam
generator
Ohmmeter
Metal tube
Dial gauge
Insulation
Bracket
30
0
15
60 50 40
90
70
80
0
10
30
20
Thermistor
Expansion
base
Pin on tube
held fixed
in slot
Figure 1: Schematic of the apparatus for the experiment.
© 1990 PASCO scientific
$2.00
apparatus, to the bracket attached at the far end of the rod. The end of the rod with the bracket could
move horizontally to allow for the expansion, which was measured with a dial gauge.
2. Water was added to the steam generator and heated to a boil.
3. The copper rod was fit into in expansion base as shown in Figure 1. The stainless steel pin on the tube
fit into the slot that fixed it in place, and the bracket at the other end fit against the spring arm of
the dial gauge.
4. A device called a thermistor, which is a resister whose resistance depends on temperature more sensitively than a normal resistor, was attached to the side of the tube, making good contact with it and a
foam insulator is put over it.
5. An ohmmeter was used to measure the resistance of the thermistor at room temperature, Ri .
6. The vent of the steam generator was connected to the inside of the tube using black flexible rubber
tubing. The end of the expansion base that held the tube where it was connected to the steam generator
was propped up so that condensation inside the tube ran down and out the other end, where it was
collected in a small cup.
7. On the dial gauge, each increment is equivalent to 0.01 mm of tube expansion.
8. The value on the ohmmeter, R, and the value on the dial gauge were observed as the tube warmed.
When the thermistor resistance (ohmmeter reading) stabilized, Rhot and the reading on the dial gauge,
∆L, were recorded.
9. This procedure was repeated for aluminum.
4
Data Analysis
1. Derive a symbolic expression for the uncertainty in the expansion coefficient (δα) using the partial
derivative method (Equation 2).
• Your expression should be in terms of the following quantities: L, ∆L, ∆T, δL, δ(∆T ), δ(∆L).
012-04394C
Thermal Expansion Apparatus
➁ Calculate ΔT = Thot – Trm. Record the result in the table.
➂ Using the
equation
ΔL = αLvalues
ΔT, calculate
a for copper, with
steel, and
2. To determine
the
temperature
that correspond
the aluminum.
measured resistance values, assume the
relationship between between R and T is linear (this is a good assumption only over small temperature
αCu = and__________________
ranges),
interpolate between the adjacent data points.
=
α
__________________
steel
• The
uncertainty stated by Pasco for the thermistor values is ±0.2◦ C.
α =
__________________
Al
• Assume
3 sigfigs for given temperature values. For example, assume the temperature value listed
as 25◦ C is actually 25.0◦ C
Questions
➀ Look
up the
accepted
the linear
expansion coefficient
foror
copper,
steel, and
3. Calculate
∆T
± δ(∆T
). values
Recallforthat
the uncertainty
in a sum
difference
is given by Equation 3 in
aluminum. Compare these values with your experimental values. What is the percentage
the introduction.
difference in each case? Is your experimental error consistently high or low?
4. Use ➁
Equation
1 tooffind
On the basis
yourα.
answers in question 1, speculate on the possible sources of error in
your experiment. How might you improve the accuracy of the experiment?
5. Use the expression derived in step 1 of the data analysis instructions to find δα.
➂ From your result, can you calculate the coefficients of volume expansion for copper,
aluminum,
andtheoretical
steel? (i.e. ΔV
= αvolV ΔT)
6. Compare
with the
α value.
THERMISTOR CONVERSION TABLE:
Temperature versus Resistance
Res.
(Ω)
351,020
332,640
315,320
298,990
283,600
269,080
255,380
242,460
230,260
218,730
207,850
197,560
187,840
178,650
169,950
161,730
153,950
146,580
139,610
133,000
126,740
120,810
115,190
109,850
104,800
100,000
Temp.
(°C)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
Res.
(Ω)
95,447
91,126
87,022
83,124
79,422
75,903
72,560
69,380
66,356
63,480
60,743
58,138
55,658
53,297
51,048
48,905
46,863
44,917
43,062
41,292
39,605
37,995
36,458
34,991
33,591
32,253
Temp.
(°C)
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
5
Figure 2
Res.
(Ω)
Temp.
(°C)
30,976
29,756
28,590
27,475
26,409
25,390
24,415
23,483
22,590
21,736
20,919
20,136
19,386
18,668
17,980
17,321
16,689
16,083
15,502
14,945
14,410
13,897
13,405
12,932
12,479
12,043
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
Res.
(Ω)
11,625
11,223
10,837
10,467
10,110
9,767.2
9,437.7
9,120.8
8,816.0
8,522.7
8,240.6
7,969.1
7,707.7
7,456.2
7,214.0
6,980.6
6,755.9
6,539.4
6,330.8
6,129.8
5,936.1
5,749.3
5,569.3
Temp.
(°C)
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
5
Tables
Copy and complete the following tables. Write the symbolic δα expression on the same page as the tables,
they will be uploaded together in question 1 on the Canvas quiz.
5.1
5.2
5.3
Data
L (mm)
∆L (mm)
Ri (kΩ)
Rhot (kΩ)
Copper
705.0
0.830
96.3
7.04
Aluminum
701.0
1.165
107.0
7.44
Ti (◦ C)
Thot (◦ C)
∆T (◦ C)
Uncertainties
δL (mm)
δ∆L (mm)
0.5
0.005
δRi (kΩ)
δRhot (kΩ)
δTi (◦ C)
δThot (◦ C)
0.2
0.2
Results
Accepted α, 10−6 (◦ C)−1
Copper
17.6
Aluminum
23.4
α ± δα, 10−6 (◦ C)−1
δ∆T (◦ C)
Thermal Expansion Lab Experimental Setup
April 25, 2020
1/5
Thermistor Connection
April 25, 2020
2/5
Dial Gauge
April 25, 2020
3/5
➂ From your result, can you calculate the coefficients of volume expansion for copper,
aluminum, and steel? (i.e. ΔV = αvolV ΔT)
THERMISTOR CONVERSION TABLE:
Temperature versus Resistance
Res.
(Ω)
351,020
332,640
315,320
298,990
283,600
269,080
255,380
242,460
230,260
218,730
207,850
197,560
187,840
178,650
169,950
161,730
153,950
146,580
139,610
133,000
126,740
120,810
115,190
109,850
104,800
100,000
Temp.
(°C)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
Res.
(Ω)
95,447
91,126
87,022
83,124
79,422
75,903
72,560
69,380
66,356
63,480
60,743
58,138
55,658
53,297
51,048
48,905
46,863
44,917
43,062
41,292
39,605
37,995
36,458
34,991
33,591
32,253
Temp.
(°C)
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
Res.
(Ω)
Temp.
(°C)
30,976
29,756
28,590
27,475
26,409
25,390
24,415
23,483
22,590
21,736
20,919
20,136
19,386
18,668
17,980
17,321
16,689
16,083
15,502
14,945
14,410
13,897
13,405
12,932
12,479
12,043
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
Res.
(Ω)
11,625
11,223
10,837
10,467
10,110
9,767.2
9,437.7
9,120.8
8,816.0
8,522.7
8,240.6
7,969.1
7,707.7
7,456.2
7,214.0
6,980.6
6,755.9
6,539.4
6,330.8
6,129.8
5,936.1
5,749.3
5,569.3
Temp.
(°C)
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
5
April 25, 2020
4/5
Temperature vs. Resistance Thermistor Data
April 25, 2020
5/5