Lab 6
Thermal Conductivity
Aleksandr Shpiler
Section E2
Analysis
In this lab our objective will be to measure the heat conductivity of lexan and
glass. The higher the coefficient k, in the formula Q =
K  A  (T2  T1 )  t
h
will be, the more amount of heat the material will let through….and as well as the lower
the k the less heat the material will let through.
Eventually we will determine which one is a better heat conductor and figure out the
exact indices of thermal conductivity (k).
Data
Part A, Raw Data
Material
Mass of Container,g
Mass of Container with
Water, g
Uncertainty in Mass, g
Time, sec
Uncertainty in Time, sec
Sheetrock
10.7
28.2
0.05
914
0.5
Part A Data converted to SI units
Part A
Material
Mass of Container,kg
Mass of Container with Water,
kg
Uncertainty in Mass, kg
Time, sec
Uncertainty in Time, sec
Sheetrock
0.0107
0.0282
0.00005
914
0.5
Part B Raw Data
Slab Thickness, mm
Uncertainty in Thickness, mm
Initial Diameter of Ice, mm
Uncertainty in the Initial Diameter of Ice,
mm
Final Diameter of Ice, mm
Uncertainty in the Final Diameter of Ice,
mm
Mass of Container with Water, g
Uncertainty in Mass, g
Total Time, sec
Uncertainty in Time, sec
Lexan
5.75
0.11
74.733
Glass
5.672
0.029
70.59
0.674
0.243
70.59
61.258
0.647
0.106
58.7
0.05
518
0.1
125.3
0.05
500
1
Part B Data converted to SI units
Slab Thickness, m
Uncertainty in Thickness, m
Initial Diameter of Ice, m
Uncertainty in the Initial Diameter of Ice,
m
Final Diameter of Ice, m
Uncertainty in the Final Diameter of Ice, m
Mass of Container with Water, kg
Uncertainty in Mass, kg
Total Time, sec
Uncertainty in Time, sec
Lexan
0.00575
0.00011
0.074733
Glass
0.005672
0.000029
0.07059
0.674
0.000243
0.07059
0.000647
0.0587
0.00005
518
0.1
0.061258
0.000106
0.1253
0.00005
500
1
Lexan
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
Glass
Time,
sec
0
25
50
75
100
125
150
175
200
225
250
275
300
325
350
375
400
425
450
475
500
Temperature, deg
cel
98.5
98.5
98.5
98.4
98.6
98.5
98.5
98.5
98.5
98.5
98.6
98.6
98.5
98.6
98.7
98.7
98.5
98.6
98.6
98.5
98.6
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
Time,
sec
0
25
50
75
100
125
150
175
200
225
250
275
300
325
350
375
400
425
450
475
500
Temperature,
deg cel
98.6
98.7
98.6
98.6
98.6
98.6
98.5
98.7
98.6
98.6
98.6
98.5
98.6
98.7
98.7
98.6
98.7
98.6
98.7
98.6
98.7
Analysis
1. To find the amount of heat transferred through the sample, we must first find the
rate RA at which the ice block melts due to heat absorption through the walls of
the ice mold. This rate is given by
RA 
(mcw  mc )
tA
where mcw is the mass of the
container with water found in Part A, mc is the mass of the container itself, and tA is
the time you let the ice melt in Part A. Now we calculate RA.
RA 
(0.0282  0.0107)
kg
= 1.9 * 10 5
914
s
Now using the uncertainty propagation rule applied to R A 
the expressions for the uncertainty components
R A  R A (
RA  1.9 *10 5  (
R A m
cw
,
(mcw  mc )
, I will find
tA
  R A m
c
, and
t
2m
 a)
mcw  mc
ta
0.00005 * 2
0.5
kg

) = 1.20 * 10 7
0.0282  0.0107 914
s
Therefore we obtain that R A  1.9 * 10 5  1.20 *10 7
Kg
s
 R A t
A
Part B (Lexan)
Here we determine the mass of the ice melted due to the thermal conduction through
the slab, m w . This can be found using the equation: mw  mcw  mc  R At B
where m cw is the mass of the container with water found in Part B for either Lexan or
glass, mc is the mass of the container, and tB is time you let the ice melt in Part B.
The term RAtB accounts for the mass of the ice melted due to thermal conduction
through the walls of the ice mold.
m w = 0.0587  0.0107  (518  1.9  10 5 ) = 0.0382 kg
Now we find the uncertainty components,
 m w t
B
m w m
cw
,
 m w m
c
,
 m w  R
A
, and
using the uncertainty propagation rule applied to mw  mcw  mc  R At B
∆ m w = 2m  RAtb  RA tb
∆ m w = 2 * 0.00005  (1.20 *10 7 * 518)  (1.9 *10 5 * 0.5) = 1.72*10-4 kg
Now we find the amount of heat transferred through the slab using equation Q  Lmw
where the coefficient L = 3.335ּ105 J/kg is the latent heat of fusion for ice.
L = 333.5 KJ/Kg
Q = 333.5 * 0.0382 = 12.7397 KJ = 12739.7 J
To find the uncertainty the error propagation rule was used , ∆Q = L  mw
And for lexan it yields the following
∆Q = 333.5 *1.72 *10 - 4 =0.05736 KJ = 57.40 J
d 
the average value for the diameter of the ice block:
d=
(d i  d f )
2
(0.074733  0.07059)
=0.072662 m
2
A
Now we find the cross sectional Area using
A
 (0.072662 )
4
 (0.072662)
2
4
.
2
=0.00415 m2
Now we find the uncertainty ∆A=
∆A=
 (d ) 2
d
2
d
 0.5 = 0.0571 m
The temperature of the sample was calculated using the standard deviation and the
average of the temperatures.
We have T2avg = 98.54 Co
STDEV= 0.07496 Co
∆T =
0.7496 2  0.052 = 0.75 Co
Temperature difference
T21 = (T2 – T1) = T2 – 0 = T2
the melting temperature of ice is 0 degrees Celsius
T21 = 98.54-0 = 98.54 Co
Now we compute the thermal conductivity k for the sample using the equation:
Q
k  A  T2  T1   t
h
kL 
Qh
A  (T2  T1 )  t b
kL =
12739.7 * 0.00575
W
= 0.3458
0.00415 * 98.54 * 518
mK
After using the error propagation rule we obtain the following set of equations…
k 
Qh
Qh
h
Q
h + 2
A +
T
Q +
A  (T2  T1 )  t
A  (T2  T1 )  t
A  (T2  T1 )  t
A  (T2  T1 ) 2  t
+
Qh
t
A  (T2  T1 )  t 2
∆kL= 0.001558 + 0.006615 + 0.758001 + 0.002637 + 0.00005 = 0.76886
W
mK
Part B (Glass)
Here we determine the mass of the ice melted due to the thermal conduction through
the slab, m w . This can be found using the equation: mw  mcw  mc  R At B
where m cw is the mass of the container with water found in Part B for either Lexan or
glass, mc is the mass of the container, and tB is time you let the ice melt in Part B.
The term RAtB accounts for the mass of the ice melted due to thermal conduction
through the walls of the ice mold.
m w = 0.1253  0.0107  (500  1.9  10 5 ) = 0.1051 kg
Now we find the uncertainty components,
 m w t
B
m w m
cw
,
 m w m
c
,
 m w  R
A
, and
using the uncertainty propagation rule applied to mw  mcw  mc  R At B
∆ m w = 2m  RAtb  RA tb
∆ m w = 2 * 0.00005  (1.20 *10 7 * 518)  (1.9 *10 5 * 0.5) = 1.72*10-4 kg
Now we find the amount of heat transferred through the slab using equation Q  Lmw
where the coefficient L = 3.335ּ105 J/kg is the latent heat of fusion for ice.
L = 333.5 KJ/Kg
Q = 333.5 * 0.1051 = 35.0508 KJ = 35050.8 J
To find the uncertainty the error propagation rule was used , ∆Q = L  mw
And for lexan it yields the following
∆Q = 333.5 *1.72 *10 - 4 =0.05736 KJ = 57.40 J
d 
the average value for the diameter of the ice block:
d=
(0.07059  0.061258)
=0.065924m
2
A
Now we find the cross sectional Area using
A
 (0.065924 )
4
2
=0.003413 m2
Now we find the uncertainty ∆A=
d
2
d
 (d ) 2
4
.
(d i  d f )
2
∆A=
 (0.065924)
2
 0.5 = 0.051776 m
The temperature of the sample was calculated using the standard deviation and the
average of the temperatures.
We have T2avg = 98.55 Co
STDEV= 0.075 Co
∆T =
0.075 2  0.052 = 0.0901 Co
Temperature difference
T21 = (T2 – T1) = T2 – 0 = T2
the melting temperature of ice is 0 degrees Celsius
T21 = 98.55-0 = 98.55 Co
Now we compute the thermal conductivity k for the sample using the equation:
Q
k  A  T2  T1   t
h
kg 
Qh
A  (T2  T1 )  t b
kg =
35050.8 * 0.005672
W
= 1.182
0.003413 * 98.55 * 500
mK
After using the error propagation rule we obtain the following set of equations…
k 
Qh
Qh
h
Q
h + 2
A +
T
Q +
A  (T2  T1 )  t
A  (T2  T1 )  t
A  (T2  T1 )  t
A  (T2  T1 ) 2  t
+
Qh
t
A  (T2  T1 )  t 2
∆kL= 0.001934 + 0.006044 + 1.79333 + 0.00108 + 0.001182 = 1.80357
W
mK
Results
Lexan:
Thermal Conductivity:
k = 0.3458  0.76886
W
mK
k/k.= 2.223
Glass:
Thermal Conductivity:
k = 1.182  1.80357
k/k. = 1.5258
W
mK
Conclusion
From the experiment we obtain the result that the lexan sample is a poorer
conductor than the glass. This material is a better insulator and could be used instead of
glass for buildings because it is see-through like glass but conducts heat poorer. In, the
winter Lexan windows would isolate cold better than glass windows.
This can be seen from the obtained k values. Clearly the value for glass is larger
then the value for lexan. Some of the major flaws in the experiment unaccounted for in
the uncertainties could be mostly human error. The student might lose some of the water
mass due to evaporation, drops left on the glass block due to surface tension and also the
melting of the ice while it was not being used.