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UNIT 13 : HEAT

13.1 Thermal Conductivity

13.2 Thermal Expansion

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.

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dx

dTkA

dt

dQ

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• 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.

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.

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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

• 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.

21 TT

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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

Adt

dQ

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:

:

:

points. these

between distance the to points two between

difference etemperatur the of ratio the :

gradient etemperatur

tyconductivi thermal :

area sectional cross :

flow heat of rate

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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 of

the material.

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dx

TkA

t

Q

dxdT

A

dtdQ

k )C(Wm

tyconductivi thermal : 1- 1o

k

• The negative sign because the temperature T, become less as the distance, x increases.• The rate of heat flow is a scalar quantity and its unit is J s-1 or Watt (W).

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

A

x

1T 2T 21 TT

• 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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• 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

rod the along

constantdt

dQ

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Heat conduction through non-insulated rod

21 TT

1TTe,Temperatur

2T

xlength,0

1T 2TX 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.

YatX atdt

dQ

dt

dQ> where A and k are 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

1TTe,Temperatur

2T

xlength,0

dTdx

X

Y

YatX atdx

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

3T

1TTe,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

dxdT

k

1

DC kk

DC dx

dT

dx

dT

dxdT

A

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 13.1A = l2 = (8x102)2 = 64 x104 m2, k = 250 W m-1 K-1,

T1= 90C, T2 = 10C

C90T1

C10T2

cm8x

a) Temperature gradient 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-1C-1 and kAl = 200 W m-1C-1)

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Solution 13.2

xCu= 5x103m, xAl= 10x103m, A= 1 m2, Tcu = 100C, TAl = 80C

mm 5

C80TAl

T

C100TCu

mm 10 a) The rate of heat flow through the copper and

aluminium plate is same, therefore

Cu Al

dQ dQ

dt dt

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Solution 13.2

xCu= 5x103m, xAl= 10x103m, A= 1 m2, Tcu = 100C, TAl = 80C

mm 5

C80TAl

T

C100TCu

mm 10

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Solution 13.2

xCu= 5x10-3m, xAl= 10x10-3m, A= 1 m2, Tcu = 100C, TAl = 80C

mm 5

C80TAl

T

C100TCu

mm 10

CuCu

Cu Cu

T TdQk A

dt x

b) By applying the equation for rate of heat flow through the copper plate, hence

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Exercise1. 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 30C 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 150.0C 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 k of aluminium = 205 W m-1C-1 and k of brass = 109 W m-1C-1) (90.2C)

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 , .

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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,T0 is heated to a new uniform temperature, T and acquires length, l as shown in figure below.

0l

l

l 0ll Tl

Tll 0

0: change in length l l l

0: change in temperature T T T

expansion linear of tcoefficien :

• If ΔT is not too large (< 100o C)

and

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Tll 0

Tl

l

0

• The coefficient of linear expansion, is defined as the change in length of a solid per unit length per unit rise 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 T is l,

• 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 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

11 rr

22 rr

ΔTTAt 0

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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

0AA TA and

TAA 0

0: change in area A A A

0: change in temperature T T T expansion area of tcoefficien :

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TAA 0

TA

A

0

• The coefficient of area expansion, is defined as the change in area of a solid surface per unit area per unit rise in temperature.

• Unit of is C-1 or K-1

• The area of the of the surface of object at a temperature T can be written as,

T1AA 0

• 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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2

l

0l

l0l

0A

Derivation

Consider a square plate with side length, l0 is heated and expands uniformly as shown in figure below.

200 lA 2lA lll 0

20 llA

20

20 lll2lA

2

where

0l

l2

0

because

2

00

20 l

l

l

l21lA

020 Al

0

20 l

l21lA

and Tl

l

0

T21AA 0 compare with T1AA 0

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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 volume V V V 0: change in temperature T T T

expansion volume of tcoefficien :

The coefficient of volume expansion, is defined as the change in volume of a solid per unit volume per unit rise in temperature.

TV

V

0 Unit of is C-1 or K-1.

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• The volume of an object at a temperature T can be written as, T1VV 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.

lo

l

Δl

300 lV 3lV lll 0

30 llV

320

20

30 lll3ll3lV

33

3

where

0l

l

l

l3

3

0

2

0

because

3

0

2

00

30 l

l

l

l3

l

l31lV

030 Vl

0

30 l

l31lV

and Tl

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 = 45CSolution

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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 105 C-1, calculate the diameter of the steel ball at

a) 57.0Cb) 66.0C

d0= 1.900 cm, T0= 20.0C , = 1.2x105C-1 Solution

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 105 C-1)

rc= 18.0 cm, r0b= 17.9 cm, T0= 20.0C , = 4.0x105C-1

Solution

1-5 C 2x10 =2

=

2When the cylinder pass through the brass ring, 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

iron cast

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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 is 1.7 x 105 C1 and the coefficient of linear expansion for wolfram is 0.43 x 105 C1)

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 106 o 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 in the same way as the coefficient of volume of a solid i.e :

TV

V

0

0V V 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 ?

5 1

5 1

1.2 x 10 K

18 x 10 K

glass

mercury

Solution