1. CHAPTER 13:
Heat
(4 Hours)

2. Learning Outcome:
13.1 Thermal Conductivity (1 hour)
At the end of this chapter, students should be
able to:
a) Define heat as energy transfer due to the
temperature difference.
b) Explain the physical meaning of thermal
conductivity.
dQ dT
c) Use rate of heat transfer, kA
dt x
a) Use temperature-distance graphs to
explain heat conduction through insulated
and non-insulated rods, and combination
of rods in series.
2

3. 13.1.a) Define heat as energy transfer due to
temperature difference.
Heat is energy transfer
due to the temperature
difference.
3

4. 13.1.b) Explain the physical meaning of thermal
conductivity.
Thermal conduction in a insulator rod
(non-metal)
B B
A
Molecules at A Figure 13.1 Until B become
receiving heat hot
Molecules Transfer of
vibrate energy
Collisions cold &
hot molecules 4

5. 13.1.b) Explain the physical meaning of thermal
conductivity.
Thermal conduction in a metal rod
B B
A
A lot of free
electrons are Figure 13.1 cold end (B) of the
available in a metal rod become
metal. hot.
It will gain some of them
they move collide with
additional thermal
faster than colder
energy due to the
before 5
molecules at B.
heating

6. Sticky notes:
• The transfer of energy by free
electrons is found to be faster
compared to the transfer of energy
by vibrating molecules in the lattice.
• It is because electrons are lighter
and move faster.
• Conclusion :
Metal much faster conducted the
heat compare to the insulator.
6

7. 13.1.c) Use rate of heat transfer
Situation Thus…
Assuming no Heat flow from
heat lost to the higher Insulator
surroundings temperature
region, T1 to
lower T1 A T2
temperature,
T2. x
Rod in steady The rate of Figure 13.2
condition heat flows is T1 T2
constant along
the rod. 7

8. dQ
The rate of heat flow, dt
depends on dQ dT
kA
cross the type of dt x
sectional area material Where:
temperature gradient through dQ
the cross sectional area : rate of heat flow
dt
A : cross sectional area
• Scalar quantity
• Unit J s 1 or dT
dQ dT : temperature gradient
A watt (W). x
dt x • -ve: direction k : thermal conductivity
decreasing T 8

9. Thermal Conductivity, k
dQ
•Is defined as a rate of dt
k
heat flows perpendicularly dT
A
through the unit cross x
sectional area of a solid,
per unit temperature to determine k
gradient along the
direction of heat flow.
•It is a scalar quantity
•Its unit is W m 1 K 1 or
W m 1 C 1.
9

10. Thermal conductivity is k
Good conductors
Substance
a property of ( W m 1 K 1)
conducting material Silver 406.0
where good
Copper 385.0
conductors will have
higher values of k Steel 50.2
compared to poor Glass 0.8
conductors. Wood 0.08
10

11. 13.1.d) Use temperature-distance graphs to
explain heat conduction through
insulated and non-insulated rods, and
combination of rods in series.
Temperature gradient
• is defined as a temperature difference per unit
length.
• The unit of temperature gradient is K m 1
or C m 1.
11

12. Non-insulated rods
X Y
Non-insulated (X T1 T2
and Y not covered
with an insulator) T1 T2
Tempratu, Temperatur e, T
heat is transferred T1
from X to Y length,x
T2
heat is lost to the
surroundings 0
length, x
Figure 13.4
12

13. temperature gradient X Y
gradually decreases along T1 T2
the bar
T1 T2
Tempratu, Temperatur e, T
curve graph (temperature
gradient at X higher than T1
that at Y)
length,x
where A and
dQ dQ k are the
T2
at X at Y
dt dt same along 0
length, x
the rod. Figure 13.4
13

14. Temperature gradient in the lagged composite
metal bar (insulated)
metal bar XY was completely X insulator Y
covered with a good insulator T1 T2
insulator
no heat loss to the surroundings
along the bar
Tempratu, Temperatur e, T T1 T2
the temperature gradient will be
T1
constant along the bar
dQ along metal
constant 0 Figure 13.5 length, x
dt bar XY 14

15. Temperature gradient in the lagged composite
metal bar
T3
insulator Temperatur e, T
T1 Material C Material D T2 T1
T3
insulator
T1 T3 T2 and kC kD T2
0 xC xC xD
length, x
Figure 13.6
15

16. T3
the rate of
steady heat flow T1 Material C Material D T2
state has through
been both
achieved materials is
T1 T3 T2 and kC k D
same.
From the equation of thermal Because
conductivity, kC kD therefore
Where;
1 dQ and A
k dt dT dT
dT are the same
x x C x D
for both C and
16
D.

17. Sticky Notes:
For the temperature change,
a temperature change of 1 K is
exactly equal to a temperature
change of 1 C.
17

18. Example 1 :
Two properly insulated uniform rods D and E make
thermal contact at one end as shown below.
(Given kD = k W m 1 C 1; k E = ⅔ k W m 1 C 1)

80 C
insulator
 D E 
100 C TC
insulator
20 cm 40 cm
The cross sectional areas of both rods are the same.
The rods are in a steady condition.
18

19.   
TD 100 C; TDE 80 C; TE T C; xD 0.20 m; xE 0.40 m
(Given kD = k W m 1 C 1; k E = ⅔ k W m 1 C 1)
a. Determine Solution :
i. The temperature a. i. the temperature
gradient along the gradient along the rod D
rod D
dT TDE TD
x D xD
dT 80 100
x D 0.20
dT  1
100 C m
x D
19

20.   
TD 100 C; TDE 80 C; TE T C; xD 0.20 m; xE 0.40 m
(Given kD = k W m 1 C 1; k E = ⅔ k W m 1 C 1)
a. Determine Solution :
ii. The temperature T a. ii. Since both rods are in
at the free end of the steady condition
the rod E. thus
dQ dQ
dt D dt E
dT dT
kD A kE A
x D x E
20

21.   
TD 100 C; TDE 80 C; TE T C; xD 0.20 m; xE 0.40 m
(Given kD = k W m 1 C 1; k E = ⅔ k W m 1 C 1)
a. Determine Solution :
ii. The temperature T a. ii.
at the free end of
the rod E. dT TE TDE
kD kE
x D x E
T 80
k 100 2
3k
0.40

T 20 C
21

22.   
TD 100 C; TDE 80 C; TE T C; xD 0.20 m; xE 0.40 m
(Given kD = k W m 1 C 1; k E = ⅔ k W m 1 C 1)
b. Sketch and label a graph
to show the variation of T C
the temperature with 100
distance x along D and E. 80
20
0 0.20 xm 0.60
22

23. Example 2:
A copper plate of thickness 1.0 cm is sealed to a steel
plate of thickness 10 cm as shown in Figure 13.7.
insulator steel
copper
30 C 15 C
insulator
1.0 cm 10 cm
The temperature of the exposed surfaces of the copper
and steel plates are 30 C and 15 C respectively.
Determine
23

24.  
TC 30 C; TS 15 C; xC 0.01 m; xS 0.10 m;
kC 385 W m 1 K 1; kS 50.2 W m 1 K 1
(Given k copper = 385 W m 1 K 1 and k steel = 50.2 W
m 1 K 1)
Solution :
a. the temperature of the a. When both rods are in
interface between the the steady condition
copper and steel plates, thus
dQ dQ
dt C dt S
TCS TC TS TCS
kC A kS A
xC xS
24

25.  
TC 30 C; TS 15 C; xC 0.01 m; xS 0.10 m;
kC 385 W m 1 K 1; kS 50.2 W m 1 K 1
(Given k copper = 385 W m 1 K 1 and k steel = 50.2 W
m 1 K 1)
Solution :
a. the temperature of the a.
interface between the TCS 30 15 TCS
copper and steel plates, 385 50.2
0.01 0.10

TCS 29.8 C
25

26.  
TC 30 C; TS 15 C; xC 0.01 m; xS 0.10 m;
kC 385 W m 1 K 1; kS 50.2 W m 1 K 1
(Given k copper = 385 W m 1 K 1 and k steel = 50.2 W
m 1 K 1)
b. the amount of heat flowing Solution :
from the copper plate to b. Given
the steel plate in one AC AS 50 10 4 m 2 ; dt 60 s
minute if the cross Therefore the amount of heat
sectional area for both flowing through the plates is
plates is 50 cm2 and no given by
energy losses to the
dQ TCS TC
surroundings. kC AC
dt 26 xC

27.  
TC 30 C; TS 15 C; xC 0.01 m; xS 0.10 m;
kC 385 W m 1 K 1; kS 50.2 W m 1 K 1
(Given k copper = 385 W m 1 K 1 and k steel = 50.2 W
m 1 K 1)
b. the amount of heat flowing Solution :
from the copper plate to b.
dQ TCS TC
the steel plate in one kC AC
minute if the cross dt xC
sectional area for both dQ 4 29.8 30
385 50 10
plates is 50 cm2 and no 60 0.01
energy losses to the
surroundings.
dQ 2310 J
27

28. Exercise 13.1 :
1. A metal plate 5.0 cm thick has a cross sectional
area of 300 cm2. One of its face is maintained at
100 C by placing it in contact with steam and another
face is maintained at 30 C by placing it in contact with
water flow. Determine the thermal conductivity of the
metal plate if the rate of heat flow through the plate is
9 kW.
(Assume the heat flow is steady and no energy is lost
to the surroundings).
ANS : 214 W m 1 K 1
28

29. Exercise 13.1 :
2. A rod 1.300 m long consists of a 0.800 m length of
aluminium joined end to end to a 0.500 m length of
brass. The free end of the aluminium section is
maintained at 150.0 C and the free end of the brass
piece is maintained at 20.0 C. No heat is lost through
the sides of the rod. At steady state, Calculate the
temperature of the point where the two metal are
joined.
(Given k of aluminium = 205 W m 1 C 1 and
k of brass =109 W m 1 C 1)
ANS : 90.2 C
29

30. Learning Outcome:
13.2 Thermal expansion (2 hour)
At the end of this chapter, students should be able to:
a) Define and use the coefficient of linear, area and
volume thermal expansion (Explain expansion of
liquid in a container).
b) Deduce the relationship between the coefficients of
expansion,
2 ; 3
30

31. Thermal expansion
• Is defined as a change in States of Expansion
dimensions of a body
matter
Volume
Linear
accompanying a change
Area
in temperature.
• There are three types of Solid / / /
thermal expansion : Liquid /
i. Linear expansion Gas /
ii. Area expansion
iii. Volume expansion
• At the same temperature,
the gas expands greater
than liquid and solid 31

32. 13.2.a) Define and use the coefficient of linear,
area and volume thermal expansion.
b) Deduce the relationship between the
coefficients of expansion, 2 ; 3
Linear expansion
Consider a thin rod of initial l0
length, l0 at temperature,T0
At T0
is heated to a new uniform l
temperature, T and acquires At T
length, l as shown in Figure l
13.8. Figure 13.8
32

33. l0 l l0 and l T
At T0
l l0 T
l
At T where
l : change in length l l0
l
ΔT: temperature change
Figure 13.8
= T-To
α: coefficient of linear
expansion
Coefficient of linear
expansion, is defined as l
a fractional increase in
length of a solid per unit l0 T
rise in temperature. 33

34. If l = l l0 then If T is…
Value Value length
l l0 1 T ΔT Δl
where - - decreases
l0 : initial length + + increases
l : final length
• l could be the length of a • ΔT is the same in the
rod, the side of a square Kelvin and Celcius scales.
plate or the diameter • the unit of α is °C-1 or K-1.
(radius) of a hole.
34

35. Area expansion
• This type of expansion
involves the expansion
of a surface area of an
object.
• Consider a plate with
initial area, A0 at
temperature T0 is
heated to a new uniform
temperature, T and Figure 13.9
expands by A, as
shown in Figure 13.9.
35

36. A A0 and A T
A A0 T
where
A : change in area A A0
ΔT: temperature change
= T-To
β: coefficient of area
Figure 13.9 expansion
Coefficient of area
expansion, is defined as A
a fractional increase in area
of a solid surface per unit A0 T
rise in temperature. 36

37. If A= A A0 then For isotropic material
(solid) , the area
A A0 1 T expansion is uniform in
all direction, thus the
where relationship between
A0 : initial surface area and is given by:
A : final surface area
2
• the unit of is °C-1 or K-1.
37

38. Proof of = 2
• Consider a square plate with side length, l0 is heated
and expands uniformly as shown in Figure 14.0.
A0 l0
2
eq. 1 l0 l
2
A l eq. 2
l l0 l eq. 3
Subst. eq.3 into eq.2
2
l0 A0
A l0 l
2 2
A l0 2l0 l l l
2
2 l l 2 Figure 14.0
A l0 1 2 and l
l0 l0 0
l0
38

39. 2
2 l l 2
A l0 1 2 and l
l0 l0 0
l0
2 l l
A l0 1 2 eq. 4 Recap: T
l0 l0
Subst. eq.1 into eq.4
A A0 1 2 T compare with A A0 1 T
Therefore 2
39

40. Volume expansion
Consider a metal cube l0 l
with side length, l0 is
heated and expands
uniformly as shown in
l0
Figure 14.1. l0
l
l
Figure 14.1
40

41. V V0 and V T
l0 l
V V0 T where
l0 ΔV : change in volume
= V-Vo
l0 ΔT : temperature change
l = T-To
l : coefficient of volume
Figure 14.1
expansion
Coefficient of volume
expansion, is defined as V
a fractional increase in V0 T
volume of a solid per unit
rise in temperature. 41

42. If V= V V0 then For isotropic material
(solid) , the volume
V V0 1 T expansion is uniform in
all direction, thus the
where relationship between
V0 : initial volume and is given by:
V : final volume 3
• the unit of is °C-1 or K-1.
42

43. Proof of =3
• Consider a metal cube with side length, l0 is heated
and expands uniformly as shown in Figure 14.1.
3 l
V0 l0 eq. 1 l0
3 eq. 2
V l l0
l l0 l eq. 3
l0
Subst. eq.3 into eq.2 l
3 l
V l0 l Figure 14.1
3 2 2 3
V l0 3l0 l 3l0 l l
2 3 2 3
3 l l l l l
V l0 1 3 3 and 3 0
l0 l0 l0 l0 l0
43

44. 2 3 2 3
3 l l l l l
V l0 1 3 3 and 3 0
l0 l0 l0 l0 l0
3 l
eq. 4 Recap:
l
V l0 1 3 T
l0 l0
V V0 1 3 T compare with V V0 1 T
Therefore 3
44

45. Example 3 : Solution :
A copper rod is 20.0 cm Since their difference in
longer than an aluminum lengths not to change with
rod before heated. How temperature, thus
long should the copper rod
be if the difference in their
l C l A
lengths is to be l
C 0C T l
A 0A T
independent of 5
temperature? 1.70 10 l0C
(Given copper = 5
2.20 10 l0C 0.20
1.70 10 5 C 1 and
aluminum = 2.20 10 5 l0C 0.88 m
C 1)
45

46. Example 4 : Solution :

d
A steel ball has a diameter
1.700 cm; T
0 0 27.0 C;
5 1
of 1.700 cm at 27.0 C. 3.6 10 C
Given that the coefficient of
volume expansion for the a. Given T =77.0 C
steel is 3.3 10 5 C 1,
calculate the diameter of d d0 1 T T0
the steel ball at
T To 50.0
5
a. 77.0 C d 1.700 1 1.1 10 50.0
d 1.701 cm
46

47. Example 4 : Solution :

A steel ball has a diameter d0 1.700 cm; T0 27.0 C;
5 1
of 1.700 cm at 27.0 C. 3.6 10 C
Given that the coefficient of
volume expansion for the b. Given T = 56.0 C
steel is 3.3 10 5 C 1,
calculate the diameter of
d d0 1 T T0
the steel ball at T T 83.0
o
5
b. -56.0 C
d 1.700 1 1.1 10 83
d 1.698 cm
47

48. Example 5 : Solution :
At 20 C a steel ball has a
diameter of 0.9000 cm,
d0S 0.9000 cm;
while the diameter of a hole d 0A 0.8990 cm;
in an aluminum plate is T0S T0A 
20.0 C;
0.8990 cm. Calculate the 
temperature of the steel TS TA T C
ball when its just pass When the ball just pass
through the hole if both ball through the hole, thus
and plate are heated in the dS d A
same time. d0S 1 TS T0S
S
(Given steel = 1.10 10 5
C 1 and aluminum = d0A 1 A TA T0A
48
2.20 10 5 C 1 )

49. Example 5 : Solution :
At 20 C a steel ball has a
diameter of 0.9000 cm,
d0S 0.9000 cm;
while the diameter of a hole d 0A 0.8990 cm;
in an aluminum plate is 
T0S T0A 20.0 C;
0.8990 cm. Calculate the 
temperature of the steel TS TA T C
ball when its just pass 5
0.9000 1 1.10 10 T 20
through the hole if both ball 5
and plate are heated in the 0.8990 1 2.20 10 T 20
same time. 
T 121 C
(Given steel = 1.10 10 5
C 1 and aluminum =
49
2.20 10 5 C 1 )

50. Thermal expansion of a liquid
• When the liquid in a vessel is heated both liquid
and vessel expand in volume.
• This expansion is called apparent expansion and
always less than the true expansion of the liquid.
• The coefficient of volume expansion of a liquid is
defined in the same way as the coefficient of
volume of a solid where
V
V0 T
• The volume expansion of a liquid whether true or
apparent depend on the change in density of the
liquid. 50

51. Variation of Liquid Density The densities of the liquid at
the two temperature are
m m
Consider V0 and V are the 0 and
volumes of the liquid at T0 V0 V
and T then
The mass of the liquid always
constant when it is expand so
V V0 1 T T0 that m m and
1 T T0
0
T T0 ΔT
0 where
1 T : final density
0 : initial density
51

52. Example 6 : Solution :
3
A glass flask whose volume V0g V0m 1000.0 cm ;
is 1000.0 cm3 at 0.0 C is Voverflow 15.5 cm3 ;
completely filled with 
mercury at this
T0g T0m 0.0 C;
temperature. When the Tg Tm 100.0 C;
flask and mercury are 18.0 10 5 K 1
m
warmed to 100.0 C, 15.5
cm3 of mercury overflow. If
the coefficient of volume V0m
expansion of mercury is
18.0 10 5 K 1, determine Voverflow
the coefficient of volume
 
expansion of the glass. 0.0 C 100.0 C
52

53. V0g V0m 1000.0 cm3 ; The change in volume of
3
Voverflow 15.5 cm ; the mercury after it is
T0g T0m 0.0 C; heated equals to
Tg Tm 100.0 C;
5 1
Vm V
m 0m T
m 18.0 10 K
5
Vm 18.0 10 1000
V0m 100.0 0.0
3
Vm 18.0 cm
Voverflow
0.0 C 100.0 C 53

54. Thus the change in volume
The change in volume of of the glass is
the mercury after it is Vg Vm Voverflow
heated equals to Vg 18.0 15.5
Vg 2.5 cm3
Vm V
m 0m T
The coefficient of volume
5
Vm 18.0 10 1000 expansion for glass flask is
Vg V T
g 0g
100.0 0.0
3 2.5 g 1000 100.0 0.0
Vm 18.0 cm 5 1
g 2.5 10 K
54

55. Exercise 13.2 :
1. The length of a copper rod is 2.001 m and the length
of a wolfram rod is 2.003 m at the same temperature.
Calculate the change in temperature so that the two
rods have the same length where the final
temperature for both rods is equal.
(Given the coefficient of linear expansion for copper
is 1.7 10 5 C 1 and the coefficient of linear
expansion for wolfram is 0.43 10 5 C 1)
ANS. : 78.72 C 55

56. 2. A metal sphere with radius of 9.0 cm at 30.0 C is
heated until the temperature of 100.0 C . Determine
the percentage of change in density for that sphere.
(Given the coefficient of volume expansion for metal
sphere is 5.1 10 5 C 1)
ANS. : 0.36 %
56

57. 3. a. An aluminum measuring rod which is correct at
5 C measures a certain distance as 88.42 cm
at 35 C. Determine the error in measuring the
distance due to the expansion of the rod.
b. If this aluminum rod measures a length of steel
as 88.42 cm at 35 C, calculate the correct length
of the steel.
(Given the coefficient of linear expansion for
aluminum is 22 10 6 C 1)
ANS. : 0.06 cm; 88.48 cm 57

58. 4. A glass flask is filled “to the mark” with 50 cm3 of
mercury at 18 C. If the flask and its content are
heated to 38 C, how much mercury will be above the
mark?
(Given 6 C 1 and
glass= 9.0 10
6 C 1)
mercury = 182 10
ANS. : 0.15 cm3
58

59. THE END…
Next Chapter…
CHAPTER 14 :
Kinetic theory of gases
59