Chapter13 Student

7/27/2019 Chapter13 Student

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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

Chapter13 Student

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