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UNIT 13 : HEAT
13.1 Thermal Conductivity13.2 Thermal Expansion
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13.1 Thermal ConductivityAt the end of this topic, students should be able
to:
Define heat as energy transfer due to temperature
difference. Explain the physical meaning of thermal
conductivity.
Use rate of heat transfer,
Use temperature-distance graphs to explain heat
conduction through insulated and non-insulated
rods, and combination of rods in series.
2
dx
dTkA
dt
dQ
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Heat always transferred from a hot region (highertemperature) to a cool region (lower temperature) untilthermal equilibrium is achieved.
Heat is transferred by three mechanisms,
1) Conduction
2) Convection
3) Radiation
Thermal Conduction is defined as the process wherebyheat is transferred through a substance from a regionof high temperature to a region of lower temperature.
Heat
is defined as the energythat is t ransferred froma body ata higher temperatureto one at a lower temperature, byconduction, convection or radiation.
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The mechanism of heat conduction through solid
material (for extra knowledge only)
AB
Suppose a rod is heated at one end (A).
Before the rod being heated all the molecules vibrateabout their equilibrium position.
As the rod is heated the molecules at the hot end (A)
vibrate with increasing amplitude, thus the kinetic energy
increases.
While vibrating the hot molecules collide with theneighbouring colder molecules result in transfer of kinetic
energy to the colder molecules.
This transfer of energy will continue until the cold end (B)
of the rod become hot.
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Thermal conductivity, k
Consider a uniform cylinder conductor of length lwith
temperature T1 at one end and T2at the other end as
shown in figure above.
The heat flows to the right because T1 is greater than T2.
21TT
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The rate of heat flow ,
through the conductor is given by:
dt
dQ
dx
dTkA
dt
dQ
l
TT
dx
dT
k
A
dt
dQ
21
:
:
:
points.these
betweendistancethetopointstwobetween
differenceetemperaturtheofratiothe:
gradientetemperatur
tyconductivithermal:
areasectionalcross:
flowheatofrate
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dx
dTkA
dt
dQ
The rate of of heat flow through an object depends on :
1. Thermal conductivity.
2. Cross-sectional area through which the heat flow.
3. Thickness of the material.
4. Temperature difference between the two sides ofthe material.
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dx
TkA
t
Q
dx
dT
A
dt
dQ
k )C(Wm
tyconductivithermal:
1- 1o
k
The negative sign because the temperature T,
become less as the distance, xincreases.
The rate of heat flow is a scalar quantity
and its unit is J s
-1
or Watt (W).
Thermal conductivity , kis defined as the rate of heatf lows perpend icularly through un i t cross sect ional area
of a sol id , per un it temperature gradient along th edirect ion of h eat f low.
Thermal conductivity is a property of conducting material.( the ability of the material to conduct heat) where goodconductors will have higher values ofkcompared to poorconductors.
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Materials with large kare called conductors;
those with small kare
called insulators.
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Heat conduction through insulated rod
A
x
1T 2T 21 TT
Consider heat conduction through an insulated rod which
has cross sectional areaA and lengthxas shown above. If the rod is completely covered with a good insulator, no
heat loss from the sides of the rod.
By assuming no heat is lost to the surroundings, therefore
heat can only flow through the cross sectional area fromhigher temperature region, T1 to lower temperature region,
T2.
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The red lines (arrows) represent the direction of heat flow. When the rod is in steady state (the temperature falls at
a constant rate) thus the rate of heat flows is constant along
the rod.
This causes the temperature gradient will be constant along
the rod as shown in figures above.
1T 2T
insulator
insulator
21 TT 1T
Te,Temperatur
2T
xlength,0
rodthealong
constantdt
dQ
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Heat conduction through non-insulated rod
21 TT
1T
Te,Temperatur
2T
xlength,0
1T 2T
X Y
The metal is not covered with an insulator, thus heat is
lost to the surroundings from the sides of the rod.
The lines of heat flow are divergent and the temperature
fall faster near the hotter end than that near the colder end.
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This causes the temperature gradient gradually
decreases along the rod and result a curve graph where
the temperature gradient at X higher than that at Y as
shown in figure below.
YatXatdtdQ
dtdQ > whereA and kare the
same along the rod.
Less heat is transferred to Y.
Temperature gradient , at any point on the rod is
given by the slope of the tangent at that point.
dx
dT
dx
dTkA
dt
dQ= From
1T
Te,Temperatur
2T
xlength,0
dT
dx
X
Y
YatXatdx
dT
dx
dT>
And from the graph
Thus
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Combination 2 metals in series
231 TTT
and
DC kk 1T 2T
insulator
Material C Material D
insulator
3
T
1T
Te,Temperatur
2T
xlength,0
3T
Cx DC xx
xc xD
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When steady state is achieved , the rate of heat
flow through both materials is same (constant).
From the equation of thermal conductivity, we get
dx
dTk 1
DC kk
DC dx
dT
dx
dT
dx
dTA
dtdQ
k
DC dt
dQ
dt
dQ
DC AA
But
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Example 13.1
A metal cube have a side of 8 cm and thermal
conductivity of 250 W m-1 K-1. If two opposite surfaces of the
cube have the temperature of 90 C and 10 C, respectively.Calculate
a) the temperature gradient in the metal cube.
b) the quantity of heat flow through the cube in 10
minutes.(Assume the heat flow is steady and no energy is lost to the
surroundings)
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Solution 13.1
A = l2 = (8x102)2= 64 x104 m2, k = 250 W m-1 K-1,
T1= 90C, T2 = 10C
C90T1
C10T2
cm8x
a) Temperature gradient b) Givent= 10 x 60 = 600 s
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Example 13.2
A 5 mm thick copper plate is sealed to a 10 mm thick
aluminium plate and both have the same cross sectional
area of 1 m2.The outside face of the copper plate is at
100 C, while the outside face of the aluminium plate is at
80 C.
a) Find the temperature at the copper-aluminium
interface.
b) Calculate the rate of heat flow through the cross
sectional area if heat flow is steady and no energy is
lost to the surroundings.
(Use kCu = 400 W m-1C-1 and kAl= 200 W m
-1C-1)
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Solution 13.2
xCu= 5x103m, xAl= 10x10
3m, A= 1 m2, Tcu = 100C, TAl= 80C
mm5
C80TAl
T
C100TCu
mm10 a) The rate of heat flowthrough the copper and
aluminium plate is same,
therefore
Cu Al
dQ dQ
dt dt
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Solution 13.2
xCu= 5x103m, xAl= 10x10
3m, A= 1 m2, Tcu = 100C, TAl= 80C
mm5
C80TAl
T
C100TCu
mm10
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Solution 13.2
xCu= 5x10-3m, xAl= 10x10
-3m, A= 1 m2, Tcu = 100C, TAl= 80C
mm5
C80TAl
T
C100TCu
mm10
CuCu
Cu Cu
T TdQk A
dt x
b) By applying the
equation forrate of heat
flow through the copper
plate, hence
E i
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Exercise
1. A metal plate 5.0 cm thick has a cross sectional area of
300 cm2. One of its face is maintained at 100C by placing it
in contact with steam and another face is maintained at 30
Cby 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).( 214 W m-1K-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
150.0C and the free end of the brass piece is maintained at
20.0C. No heat is lost through the sides of the rod. At steady
state, find the temperature of the point where the two metal
are joined.(Use kof aluminium = 205 W m-1C-1 and kof
brass = 109 W m-1C-1) (90.2C)
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13.2 Thermal expansionAt the end of this topic, students should be able
to:
Define and use the coefficient of linear, area
and volume thermal expansion.Deduce the relationship between the
coefficients of expansion , .
23
3 2
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14.3 Thermal expansion
Thermal expansion is defined as the change in
dimensions of a body accompanying a change in
temperature.
3 types of thermal expansion :
- Linear expansion
- Area expansion- Volume expansion
In solid, all types of thermal expansion are occurred.
In liquid and gas, only volume expansion is occurred.
At the same temperature, the gas expands greater than
liquid and solid.
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Linear expansion
Consider a thin rod of initial length, l0 at temperature,T0is heated to a new uniform temperature, Tand acquires
length, l as shown in figure below.
0l
l
l 0llTl
Tll 0
0: change in lengthl l l 0: change in temperatureT T T
expansionlinearoftcoefficien:
IfTis not too large (< 100o C)
and
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Tll 0
Tl
l
0
The coefficient of linear expansion, is defined asthe change in leng th o f a sol id per uni t length per uni t
r ise change in temperature.
( 1)
o
o
o o
o
l l l
l l l
l l T l
l l T
Unit of is C-1 or K-1.
If the length of the object at a temperature Tis l,
For many materials, every linear dimension changes
according to both equations above. Thus, lcould be the
length of a rod, the side length of a square plate or the
diameter (radius) of a hole.
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For example, as a metal washer is heated, all dimensions
including the radius of the hole increase as shown in
figure below.
1r
2r
0TAt
11rr
22 rr
TTAt 0
A i
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Area expansion
This expansion involving the expansion of a surface area of
an object.
Consider a plate with initial area,A0 at temperature T0 isheated to a new uniform temperature, Tand expands by A,
as shown in figure below.
From this experiment, we get
0AA TA and
TAA 0
0: change in areaA A A 0: change in temperatureT T T
expansionareaoftcoefficien:
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TAA 0
TA
A
0
The coefficient of area expansion, is defined as thechange in area of a sol id surface per un it area per
un it r ise in temperature.
Unit of is C-1 or K-1
The area of the of the surface of object at atemperature Tcan be written as,
T1AA 0
For isotropic material (solid) , the area expansion isuniform in all direction, thus the relationship between and is given by
2
2
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2
l
0l
l0l
0A
Derivation
Consider a square plate with side length, l0 is heated and
expands uniformly as shown in figure below.
2
00 lA 2lA lll 0
20 llA
2
0
2
0 lll2lA
2
where
0l
l2
0
because
2
00
2
0l
l
l
l21lA
0
2
0 Al
0
2
0l
l21lA and Tll
0
T21AA 0 compare with T1AA 0
V l i
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Volume expansion
Consider a metal cube with side length, l0 is heated and
expands uniformly. From the experiment, we get
0VV TV and
0V V T
0: change in volumeV V V
0: change in temperatureT T T expansionvolumeoftcoefficien:
The coefficient of volume expansion, is defined as thechange in volume of a sol id per uni t volum e per uni t
r ise in temperature.
TV
V
0
Unit of is C-1 or K-1.
Th l f bj t t t t T b
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The volume of an object at a temperature Tcan be
written as, T1VV 0
For isotropic material (solid), the volume expansion isuniform in all direction, thus the relationship between and is given by
3
Derivation
Consider a metal cube with side length, l0 is heated and
expands uniformly.
lo
l
l
3
00 lV 3lV lll 0
3
0llV
3202
0
3
0 lll3ll3lV
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3
where
0l
l
l
l3
3
0
2
0
because
3
0
2
00
3
0l
l
l
l3
l
l31lV
0
3
0 Vl
0
3
0l
l31lV
and T
l
l
0
T31VV 0 compare with T1VV 0
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Example 13.3
The length of metal rod is 30.000 cm at 20C and 30.019
cm at 45C, respectively. Calculate the coefficient of
linear expansion for the rod.
l0= 30.000 cm, T0= 20C, l= 30.019 cm, T = 45C
Solution
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Example 13.4
A steel ball is 1.900 cm in diameter at 20.0C. Given that the
coefficient of linear expansion for steel is 1.2 x 105C-1,
calculate the diameter of the steel ball at
a) 57.0C
b) 66.0C
d0= 1.900 cm, T0= 20.0C, = 1.2x105C-1Solution
Tll 10
Tdd 10
a) b)
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Example 13.5
A cylinder of radius 18.0 cm is to be inserted into a brass
ring of radius 17.9 cm at 20.0C. Find the temperature of the
brass ring so that the cylinder could be inserted.(Given the coefficient of area expansion for brass is 4.0 x 105C-1)
rc= 18.0 cm, r0b= 17.9 cm, T0= 20.0C, = 4.0x105C-1
Solution
1-5 C2x10 =
2
=
2When the cylinder pass through
the brassring, thus
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Example 13.6
Determine the change in volume of block of cast iron 5.0
cm x 10 cm x 6.0 cm, when the temperature changes from
15 oC to 47 oC. ( = 0.000010 oC -1 )
Solution
ironcast
E i
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2. 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 is1.7 x 105C1 and the coefficient of linear expansion
for wolfram is 0.43 x 105C1)
Exercise
1. A rod 3.0 m long is found to have expanded 0.091 cm in
length after a temperature rise of 60 o C. What is the
coefficient of linear expansion for the material of the rod ?
78.72C
5.1 x 106o C1
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Thermal Expansion of Liquid in A Container
When a liquid in a solid container is heated, both liquid and
the solid container expand in volume.
Liquid expands more than the solid container.
The change in volume of a liquid is given by
The coefficient of volume expansion of a liquid is defined inthe same way as the coefficient of volume of a solid i.e :
TV
V
0
0V V T
E l 13 7
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Example 13.7
A glass flask with a volume of 200 cm3 is filled to the brim
with mercury at 20 oC. How much mercury overflows
when the temperature of the system is raised to 100 oC ?5 1
5 1
1.2 x 10 K
18 x 10 K
glass
mercury
Solution