Transcript Chapter13-STUDENT

UNIT 13 : HEAT
13.1 Thermal Conductivity
13.2 Thermal Expansion
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13.1 Thermal Conductivity
At 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.
dQ
dT
 Use rate of heat transfer,
 kA
dt
dx
 Use temperature-distance graphs to explain heat
conduction through insulated and non-insulated
rods, and combination of rods in series.
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Heat
• is defined as the energy that is transferred from a body at
a higher temperature to one at a lower temperature , by
conduction, convection or radiation.
• Heat always transferred from a hot region (higher
temperature) to a cool region (lower temperature) until
thermal equilibrium is achieved.
• Heat is transferred by three mechanisms,
1) Conduction
2) Convection
3) Radiation
• Thermal Conduction is defined as the process whereby
heat is transferred through a substance from a region
of high temperature to a region of lower temperature.
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The mechanism of heat conduction through solid
material (for extra knowledge only)
A
B
• Suppose a rod is heated at one end (A).
• Before the rod being heated all the molecules vibrate
about 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 the
neighbouring 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
T1  T2
• Consider a uniform cylinder conductor of length l with
temperature T1 at one end and T2 at the other end as
shown in figure above.
• The heat flows to the right because T1 is greater than T2.
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• The rate of heat flow , dQ
dt
through the conductor is given by:
dQ
dT
dQ
 kA
: rate of heat flow
dt
dx
dt
A : cross sectional area
k : thermal conductivi ty
dT
: temperatur e gradient
dx
: the ratio of the temperatur e difference
between two points to the distance between
these points.
T1  T2
:
l
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dQ
dT
 kA
dt
dx
• 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 of
the material.
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• The negative sign because the temperature T, Q
T
 kA
become less as the distance, x increases.
t
dx
• The rate of heat flow is a scalar quantity
and its unit is J s-1 or Watt (W).
dQ
k   dt
dT
A
dx
k : thermal conductivi ty
(Wm -1 o C1 )
Thermal conductivity , k is defined as the rate of heat
flows perpendicularly through unit cross sectional area
of a solid , per unit temperature gradient along the
direction of heat flow.
Thermal conductivity is a property of conducting material.
( the ability of the material to conduct heat) where good
conductors will have higher values of k compared to poor
conductors.
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Materials with large k
are called conductors;
those with small k are
called insulators.
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Heat conduction through insulated rod
T1
T2 T1  T2
A
x
• Consider heat conduction through an insulated rod which
has cross sectional area A and length x as 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 from
higher temperature region, T1 to lower temperature region,
T2 .
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insulator
T1
Temperatur e,T
T1
T2
insulator
T1  T2
dQ
 constant
dt
along the rod
T2
0
length, x
• 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
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the rod as shown in figures above.
Heat conduction through non-insulated rod
X
T1
Temperatur e,T
T1
T2
0
Y
T2
T1  T2
length, x
• 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
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fall faster near the hotter end than that near the colder end.
• Less heat is transferred to Y.
• 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
Temperatur e,T
shown in figure below.
• From
T1 X
dQ
dT
= kA
dt
dx
dT
• And from the graph
dT
dT
at X >
at Y
dx
dx
• Thus
dQ
dQ
at X >
at Y
dt
dt
Temperature gradient ,
T2
0
dx
Y
length, x
where A and k are the
same along the rod.
dT
at any point on the rod is
dx
given by the slope of the tangent at that point.
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Combination 2 metals in series
xc
T1
T3
Material C
xD
T1  T3  T2
insulator
and
Material D
insulator
T2
kC  k D
Temperatur e,T
T1
T3
T2
0
xC
xC  xD length, x
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• When steady state is achieved , the rate of heat
flow through both materials is same (constant).
dQ dQ

dtC dtD
AC  AD
• From the equation of thermal conductivity, we get
1
k
 dT 


 dx 
But kC  k D
dQ
k   dt
dT
A
dx
 dT   dT 

 

 dx C  dx  D
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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 = (8x102)2 = 64 x104 m2, k = 250 W m-1 K-1,
T1= 90C, T2 = 10C
x  8 cm
T1  90  C
a) Temperature gradient
T2  10  C
b) Given t = 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-1C-1 and kAl = 200 W m-1C-1)
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Solution 13.2
xCu= 5x103m, xAl= 10x103m, A= 1 m2, Tcu = 100C, TAl = 80C
5 mm
TCu  100  C
a) The rate of heat flow
through the copper and
aluminium plate is same,
therefore
TAl  80  C
10 mm
 dQ 
 dQ 

 

 dt Cu  dt  Al
T
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Solution 13.2
xCu= 5x103m, xAl= 10x103m, A= 1 m2, Tcu = 100C, TAl = 80C
5 mm
10 mm
TAl  80  C
TCu  100  C
T
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Solution 13.2
xCu= 5x10-3m, xAl= 10x10-3m, A= 1 m2, Tcu = 100C, TAl = 80C
5 mm 10 mm
TAl  80  C
TCu  100  C
b) By applying the
equation for rate of heat
flow through the copper
plate, hence
 T  TCu 
 dQ 


   kCu A 
 dt Cu
 xCu 
T
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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 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).( 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
150.0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, find the temperature of the point where the two metal
are joined.(Use k of aluminium = 205 W m-1C-1 and k of
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brass = 109 W m-1C-1) (90.2C)
13.2 Thermal expansion
At 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   2 ,   3 .
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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,T0
is heated to a new uniform temperature, T and acquires
length, l as shown in figure below.
• If ΔT is not too large (< 100o C)
l0
l
l
l  T and l  l0
l  l0 T
l : change in length  l  l0
T : change in temperature  T  T0
 : coefficient of linear expansion
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l  l0 T
l

l0 T
•Unit of  is C-1 or K-1.
• The coefficient of linear expansion,  is defined as
the change in length of a solid per unit length per unit
rise change in temperature.
• If the length of the object at a temperature T is l,
l  l  lo
l  l  lo
l   lo T  lo
l  lo (T  1)
• For many materials, every linear dimension changes
according to both equations above. Thus, l could be the
length of a rod, the side length of a square plate or the 26
diameter (radius) of a hole.
• For example, as a metal washer is heated, all dimensions
including the radius of the hole increase as shown in
figure below.
r1  r1
At T0  ΔT 
r1
r2
r2  r2
At T0
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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 is
heated to a new uniform temperature, T and expands by A,
as shown in figure below.
•
From this experiment, we get
A  A0 and A  T
A  A0 T
 A : change in area  A  A0
T : change in temperature  T  T0
 : coefficient of area expansion
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A  A0 T
A

A0 T
• Unit of  is C-1 or K-1
• The coefficient of area expansion,  is defined as the
change in area of a solid surface per unit area per
unit rise in temperature.
•
The area of the of the surface of object at a
temperature T can be written as,
A  A0 1  T 
• For isotropic material (solid) , the area expansion is
uniform in all direction, thus the relationship between  and
 is given by
  2
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Derivation
  2
Consider a square plate with side length, l0 is heated and
expands uniformly as shown in figure below.
l0
l0
l
A0
l
A0  l02
A  l2
l  l0  l
A  l0  l 
2
Al
2
0
2


 2l0 l  l
2





l

l
A  l02 1  2     because
l0  l0  


l 
2
A  l0  1  2  where l02  A0
l0 

A  A0 1  2T  compare with
  2
2
 l 
   0
 l0 
and
l
 T
l0
A  A0 1  T 
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Volume expansion
Consider a metal cube with side length, l0 is heated and
expands uniformly. From the experiment, we get
V  V0
and
V  T
V   V0 T
V : change in volume  V  V0
T : change in temperature  T  T0
 : coefficient of volume expansion
V

Unit of  is C-1 or K-1.
V0 T
The coefficient of volume expansion,  is defined as the
change in volume of a solid per unit volume per unit
rise in temperature.
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•
The volume of an object at a temperature T can be
written as,
V  V 1  T 
0
• For isotropic material (solid), the volume expansion is
uniform 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.
l  l0  l
V0  l03 V  l 3
V  l0  l 
3
V  l03  3l02 l  3l0 l 2  l 
3
Δl
lo
l
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2
3
2
3


 l   l 
l  l   l 
3
V  l0 1  3  3      because 3      0
l0

 l0   l0 
 l0   l0  
l
l 
3
3
 T
V  l0  1  3  where l0  V0 and
l0 
l0

V  V0 1  3T  compare with V  V0 1  T 
  3
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Example 13.3
The length of metal rod is 30.000 cm at 20C and 30.019
cm at 45C, respectively. Calculate the coefficient of
linear expansion for the rod.
Solution
l0= 30.000 cm, T0= 20C , l= 30.019 cm, T = 45C
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Example 13.4
A steel ball is 1.900 cm in diameter at 20.0C. Given that the
coefficient of linear expansion for steel is 1.2 x 105 C-1,
calculate the diameter of the steel ball at
a) 57.0C
b) 66.0C
Solution
d0= 1.900 cm, T0= 20.0C ,  = 1.2x105C-1
l  l0 1  T 
a)
d  d 0 1  T 
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 20.0C. 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 105 C-1)
Solution
rc= 18.0 cm, r0b= 17.9 cm, T0= 20.0C ,  = 4.0x105C-1
When the cylinder pass through
the brass ring, thus
  2

 = = 2x10
2
5 
C-1
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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. ( cast iron = 0.000010 oC -1 )
Solution
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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 ?
5.1 x 106 o C1
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 is
1.7 x 105 C1 and the coefficient of linear expansion
for wolfram is 0.43 x 105 C1)
78.72C
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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
V   V0 T
•The coefficient of volume expansion of a liquid is defined in
the same way as the coefficient of volume of a solid i.e :
V

V0 T
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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 ?
 glass  1.2 x 105 K 1
 mercury  18 x 105 K 1
Solution
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