DEPARTMENT OF CHEMICAL ENGINEERING &
SUSTAINABILITY
KULLIYYAH OF ENGINEERING
BTEN 2182/CHEN 2182
SEMESTER 2, SESSION 2022/2023
CODE & NAME OF EXPERIMENT:
HTL_E1: Heat Transfer by Conduction
DATE OF LAB CLASS: 14.03.2023
GROUP NO: 3(b)
No.
Name
Matric No.
BTEN/CHEN
Task
1.
Ahmad Faruq bin Hisham
2015029
BTEN
Safety Precaution,
Discussion, Conclusion,
2.
Amirul Azizi bin Mohamad
Khairul Amri
2012349
BTEN
Table content, List of
table and figures,
Abstract, Discussion
3.
Nabilah Solehah binti Mohd
Supian
2112762
CHEN
Procedure, Result and
Discussion
4.
Nur Nisha Ismahani Binti Zamzuri
2119962
CHEN
Introduction, Problem
statement, Objective,
Equipment
TABLE OF CONTENT
Abstract
3
1.0 Introduction and Theoretical Background
3
2.0 Problem Statement
4
3.0 Objectives
4
4.0 Equipment
5
5.0 Experimental Procedures/ Methodology
6
6.0 Results and Discussion
8
7.0 Conclusion
23
8.0 Safety and Precaution
23
References
24
1
LIST OF TABLE AND FIGURES
Table 3.1A
8
Table 3.1B
9
Table 3.1C
10
Table 3.1D
11
Table 3.2B
14
Table 3.2C
17
Table 3.2D
20
Figure 1.1
5
Figure 2.1
7
Figure 3.1B
13
Figure 3.2B
15
Figure 3.1C
16
Figure 3.2C
17
Figure 3.1D
19
Figure 3.2D
21
2
ABSTRACT
Heat conduction causes a temperature difference to exist between various body parts. In
this experiment, heat is provided through calorimeters that are filled with both hot and cold water
and flow through a rod made of copper and aluminium for heat conduction. The objective is to
calculate and contrast the heat transfer coefficients of copper and aluminium in a gradient of
constant temperature. The experiment will also look into the heat that is lost to the surroundings
as a result of heat conduction in the rod. Firstly, part A was conducted by taking the temperature
readings of hot water in the calorimeter every 10 seconds for 5 minutes. Based on the results,
heat capacity is calculated. Then, water with temperature 0°C is used in part B to measure the
temperature every minute for 30 minutes. This is to obtain the addition of the heat from the
surroundings. Part C is performed to measure the heat transfer coefficient of copper rod and
aluminium rod for part D. Each rod gives a different value of heat transfer coefficient.
1.0. INTRODUCTION AND THEORETICAL BACKGROUND
Conduction Heat transfer is the movement of heat through matter such as gasses, liquids,
or solids without the matter moving in bulk. Conduction, on the other hand, is the energy transfer
within a substance from more energetic to less energetic particles as a result of particle
interaction. The molecules collide and diffuse as they move randomly, which causes conduction
heat transmission in gasses and liquids. On the other hand, heat transfer in solids occurs as a
result of a combination of free electron energy transport and molecular lattice vibrations.
The mode of heat transfer that takes place in a material due to a temperature gradient is
called thermal conduction. Because of the superfluous convectional heat transfer that both
liquids and gasses exhibit, a strong is used for pure conduction display. Heat conduction happens
in three dimensions in real-world situations, a complexity that frequently necessitates
computational analysis. The fundamental law that links rate of heat flow to temperature gradient
and area must be demonstrated in the lab using a one-dimensional technique.
For this lab report, there are 4 parts that have been conducted which is part A to measure
heat capacity of the calorimeter by taking the temperature readings of hot water in the
calorimeter every 10 seconds for 5 minutes. On the other hand, water with temperature 0°C is
used to measure the heat capacity of calorimeter for part B by taking the temperature every
minute for 30 minutes. Furthermore, part C is performed to measure the heat capacity of copper
rod and aluminium rod for part D.
3
2.0. PROBLEM STATEMENT
The main purposes of this experiment are to determine which material is the best heat
conductor and to study the effects of the environment to the heat transfer in rod by conduction.
As in our daily life application, we use materials with a high thermal conductivity that can
effectively transfer heat such as kettle, cooker, etc. and readily take up heat from their
environment. Poor thermal conductors resist heat flow such as thermos, ice box cooler, etc. and
obtain heat slowly from their surroundings. By performing the experiment and undergoing some
calculations, it is proved that copper rod is the best heat conductor compared to aluminum rod.
When nearby molecules meet, conduction occurs, where energy is transferred from the more
energetic to the less energetic molecules. Heat moves in the direction of falling temperatures
since molecular energy increases with temperature. Thus, the rod temperature will change by
time whether it decreases or increases because the heat is transferred to or from the environment
by conduction. This experiment is very crucial especially for industries where they need to
choose the best heat conductor materials in order to save the cost for the heat and energy
production.
3.0. OBJECTIVES
The objectives of these experiments are to measure and compare the heat transfer
coefficient of copper and aluminium and to investigate how the heat losses to the environment
during the heat conduction in rod. This experiment also undergoes some calculation by the data
collected during the lab session to determine and prove which material is the best conductor. By
performing this experiment, the students will get a better understanding and visualization of heat
conduction and the effects from the environment.
4
4.0 EQUIPMENT
Figure 1.1
5
5.0 PROCEDURE
6
Figure 2.1
7
6.0 RESULT & DISCUSSIONS
8
9
10
11
DISCUSSION
PART A
1. Using this formula, the heat capacity of the Calorimeter is calculated based on the data of
parameters obtained.
𝐶 = 𝑐𝑤. 𝑚𝑤. 𝜗𝑤 – 𝜗𝑀
𝜗𝑀 – 𝜗R
Where
cw = Specific heat capacity of water
mw = Mass of the water
ϑW = Temperature of the hot water
ϑM = Mixing temperature
ϑR = Room temperature
At T= 300 s, and 𝜗𝑀= 77.8℃,
C = (4184 J⋅kg−1⋅℃−1) (0.26 kg) (100℃ - 77.8℃ ) / ( 77.8℃ - 25.5℃)
= 461.76 J. ℃
12
PART B
1. From equation (6);
∆𝒬 =(𝐶𝑊.𝑚𝑊+𝐶).∆𝑇
where ∆T=T-T0, (T0= Temperature at time t = 0)
∆𝒬 = ( (4184 J⋅kg−1⋅℃−1) (0.26 kg) + 461.76 J. ℃ ) (10℃ )
= 15496 J
2. Graph temperature vs time for the cold water:
Figure 3.1B
13
3. Graph heat from surrounding (Q) vs time
Time (mins)
Temperature
(℃)
Heat (Q) (J)
Time (mins)
Temperature
(℃)
Heat (Q) (J)
0
0
17572.8
16
6.7
5799.0
1
2.4
13355.3
17
7.0
5271.8
2
3.0
12300.96
18
7.2
4920.4
3
3.7
11070.9
19
7.5
4393.2
4
4.1
10368.0
20
7.7
4041.7
5
4.4
9840.8
21
7.9
3690.3
6
4.6
9489.3
22
8.2
3163.1
7
4.7
9313.6
23
8.4
2811.6
8
4.8
9137.9
24
8.6
2460.2
9
4.9
8962.1
25
8.9
1933.0
10
5.1
8610.7
26
9.1
1581.6
11
5.4
8083.5
27
9.3
1230.1
12
5.6
7732.0
28
9.5
878.6
13
5.9
7204.8
29
9.8
351.5
14
6.2
6677.7
30
10.0
0
15
6.4
6326.2
Table 3.2B
By using equation Q = mCΔT we can find the value of heat from surrounding
At t=0
Q= 0.42 kg ( 4184 J⋅kg−1⋅℃−1 ) (10-0)℃
= 17572.8 J
14
We will find all value of heat for the other 30 minutes by using the above formula. The result is tabulated
in the table above.
Figure 3.2B
4. Slope for the graph is calculated:
𝑑𝑄/ 𝑑𝑡 =
4041.7−6326.2
20−15
𝐽
= -456.9 𝑚𝑖𝑛
15
PART C
1. Using equation (6), the heat energy supplied to the lower calorimeter is calculated.
At T= 300 s,
ΔQ = ( (4184 J⋅kg−1⋅℃−1) (0.26 kg) + 461.76 J. ℃) ) (2.96℃)
= 4586.82 J
2. Then, values and the change in the temperature difference metal rod plotted vs time.
Figure 3.1C
3. After that, Q ambient+metal calculated using Equation (6) and plotted vs time.
At T= 0, ΔT= 0.05 ℃,
ΔQ = ( (4184 J⋅kg−1⋅℃−1) (0.42 kg) + 461.76 J. ℃) ) (0.25℃)
= 554.76 J
Then, Q ambient+metal calculated and tabled for the rest of 300 s.
16
Time (s)
ΔT (℃)
ΔQ (J)
0
0.05
110.952
30
0.30
665.712
60
0.88
1952.76
90
1.55
3439.51
120
1.95
4327.13
150
2.32
5148.17
180
2.52
5591.98
210
2.85
6324.26
240
2.77
6146.74
270
2.94
6523.98
300
2.96
6568.36
Table 3.2C
Figure 3.2C
17
4. Slope from the graph calculated to obtain dQ/dt ambient+metal
d𝑄/ 𝑑𝑡 ambient+metal =685 J/s.
5. Heat transfer coefficient, λA calculated using Equation (1) and the values of dQ/dt metal.
𝑑𝑄/ 𝑑𝑡 = −𝜆𝐴. 𝜕𝑇/ 𝜕X
Where
𝑑𝑄/ 𝑑𝑡= The quantity of heat transported with time
𝜆= Heat conductivity of the substance
A= Cross-sectional area
𝜕𝑇/ 𝜕X= Temperature gradient perpendicular to the surface
Using Equation (4),
T(x) = (T2-T1) / l
T(x) = 275.96 K/ 0.315 m
(685 J/s) = -𝜆 (4.91x10−4 m2 ) (876.06 K/m)
𝜆 = -1592.48 W/m.K
6. Lastly, experimental value of 𝜆A compared with theoretical value for heat transfer
coefficient of copper.
Theoretical value of heat conductivity, 𝜆 for copper is -401 W/ m.K. Meanwhile, the
experimental value of 𝜆 for copper is -1592.48 W/ m.K. Therefore,
Percentage error = -1592.48 - (-401) / -401 . 100%
= 297.12%
As shown in the calculation above, the difference between experimental and theoretical
value of heat coefficient of copper, 𝜆, is large. This may be due to a major amount of heat
lost to the surroundings as a result from a mistake in the experiment setup. For example,
some of the lower exposed part of the copper rod may be exposed to air and is not fully
immersed in the cold water. Other than that, there may be mistakes in taking
measurements.
18
PART D
Using equation (6), the heat energy supplied to the lower calorimeter is calculated.
At T= 300 s, ΔT = 4.50 ℃
ΔQ = ( (4184 J⋅kg−1⋅℃−1) (0.26 kg) + 461.76 J. ℃) ) (4.50℃)
= 6973.2 J
Then, values and the change in the temperature difference on the metal rod plotted vs time.
Figure 3.1D
Then, we will calculate the value of Q ambient+metal by using formula 6
At t = 0, ΔT = 0.25
Thus Q ambient+metal;
ΔQ = ( (4184 J⋅kg−1⋅℃−1) (0.42 kg) + 461.76 J. ℃) ) (0.25℃)
= 554.76 J
19
We will find all values of heat for the other 300 seconds by using the above formula. The result is
tabulated in the table below.
Time (s)
ΔT (℃)
ΔQ (J)
0
0.25
554.76
30
0.85
1886.2
60
1.70
3772.4
90
2.34
5192.6
120
2.90
6435.2
150
3.29
7300.6
180
3.64
8077.3
210
3.96
8787.4
240
4.30
9541.9
270
4.50
9985.7
300
4.50
9985.7
Table 3.2D
After that, Q ambient+metal calculated and plotted vs time.
20
Figure 3.2D
From the graph we will get the slope which is
𝑑𝑄/ 𝑑𝑡 =
1886.2−554.76
30−0
𝐽
=44.38 𝑠
Lastly, heat transfer coefficient, λA calculated using Equation (1) and the values of dQ/dt metal.
𝑑𝑄/ 𝑑𝑡 = −𝜆𝐴. 𝜕𝑇/ 𝜕X
Where
𝑑𝑄/ 𝑑𝑡= The quantity of heat transported with time
𝜆= Heat conductivity of the substance
A= Cross-sectional area
𝜕𝑇/ 𝜕X= Temperature gradient perpendicular to the surface
21
From equation (4)
T(x) = 277.65 K/ 0.315 m
From equation (1)
𝐽
(44.38 𝑠 ) = -𝜆 (4.91x10−4 m2 ) ( 881.4K/m)
𝜆 = -102.55 W/m.K
Lastly, experimental value of 𝜆A is compared to the theoretical value for heat transfer coefficient
of aluminium.
Theoretical value of heat conductivity, 𝜆 for copper is -247 W/ m.K. Meanwhile, the
experimental value of 𝜆 for aluminium is -102.55 W/ m.K. Therefore,
Percentage error =-102.55 - (-247) /-247 . 100%
= -58%
As shown in the calculation above, the difference between experimental and theoretical value of
heat coefficient of aluminium, 𝜆, is quite large. This may be due to a major amount of heat lost to
the surroundings as a result from a mistake in the experiment setup. For example, some of the
lower exposed part of the aluminium rod may be exposed to air and is not fully immersed in the
cold water. Other than that, there may be mistakes in taking measurements.
Despite the fact that both copper and aluminium are excellent heat conductors, copper has a
higher thermal conductivity than aluminium. Copper has a thermal conductivity of about 400
W/mK, compared to about 200 W/mK for aluminium.
Accordingly, copper is a better material choice than aluminium for applications requiring rapid
heat dissipation, such as heat sinks, heat exchangers, and electrical wiring. Copper also transfers
heat more effectively than aluminium.
Copper has a higher thermal conductivity than aluminium, despite the fact that both metals are
excellent heat conductors. In comparison to aluminium, which has a thermal conductivity of
about 200 W/mK, copper has about 400 W/mK.
22
As a result, copper is a better material option than aluminium for uses like heat sinks, heat
exchangers, and electrical wiring that require quick heat dissipation. Additionally, copper
conducts heat more efficiently than aluminium.
In conclusion, copper conducts heat more effectively than aluminium, making it a better choice
for applications where effective heat dissipation is crucial. But aluminium is a lighter and more
affordable substitute, making it a better option for applications where weight and price are more
important factors.
7.0 CONCLUSION
To conclude, the results and discussion stated above shows that the objective of the
experiments, which is to calculate and compare the heat transfer coefficients of copper and
aluminium in a constant temperature gradient, as well as investigating the heat losses to the
atmosphere during the heat conduction in rods, has been achieved.
The experiment shows that heat conduction happens when temperatures in separate parts
of a body differ from one another. The heat conduction will flow in one direction only, from a
location with higher temperature to a lower one. Some amount of heat is lost to the environment
during heat conduction due to the differences in temperature of the conductor and air
surrounding it. The amount of heat transferred in both copper rod and aluminium rod is shown to
follow a temperature gradient. As shown in the calculation above, the value of 𝜆 of copper rod is
higher than aluminium rod, indicating that copper is a better heat conductor than aluminium.
Thus, Copper is often the greatest metal option for thermal conductivity. However in some cases
where some factors, such as price and weight, are crucial when selecting the right metal,
aluminium is a better option than copper.
23
8.0 SAFETY PRECAUTION
In this experiment, the mediums which provide and receive heat - hot water and cold
water - during heat transfer are being controlled in order to accurately determine the amount of
heat transfer of the copper rod and aluminium rod. Thus, a number of safety precautions must be
followed so that there are no external variables that will affect the heat transfer process. The first
one is to ensure that the thermometer doesn't touch the inner surface of the calorimeter when
recording the temperature of water in order to avoid measurement errors. Next is to remove all
ice before starting the timer for Part B and Part C of the experiment so that the ice will not affect
(lowers) the temperature of cold water during the heat transfer process. Before beginning the Part
C of the experiment, stir the ice water until the temperature is as close to 0°C as possible.
Always remember to put the calorimeter under running tap water after finishing each part of the
experiment to reset the calorimeter’s temperature back to room temperature. Lastly, in Part C of
the experiment, make sure that the exposed lower part of the metal rods are fully immersed with
cold water to minimize heat loss to the surrounding during the heat transfer process.
References
Incropera, F. P., DeWitt, D. P. (2002), Fundamentals of Heat and Mass Transfer, John Wiley &
Sons, Inc. (Chapter 2-5)
Mokhena, T. C., Mochane, M. J., Sefadi, J. S., Motloung, S. V., & Andala, D. M. (2018).
Thermal Conductivity of Graphite-Based Polymer Composites. Impact of Thermal Conductivity
on Energy Technologies. doi:10.5772/intechopen.75676
Uswah Solehah Sh. (2020, July 12). Experiment 4A and 4B - Heat Capacity of Calorimeter &
Ambient Heat [Video]. Youtube. https://www.youtube.com/watch?v=AQJG6XwDsSQ
Uswah Solehah Sh. (2020, July 12). Experiment 4C and 4D - Thermal & Electrical Conductivity
of Metal Rods [Video]. Youtube. https://www.youtube.com/watch?v=RAAzFHRU0fI
24
Wilson Power Solutions. (n.d.). Retrieved March 21, 2023, from
https://www.wilsonpowersolutions.co.uk/app/uploads/WPS_Aluminium-v-Copper_white-paper_
-2021.pdf
25