CHAPTER 13: Heat

(4 Hours)

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.

c) Use rate of heat transfer, d) Use temperature-distance graphs to

explain heat conduction through insulated and non-insulated rods, and combination of rods in series.

Learning Outcome:13.1 Thermal Conductivity (1 hour)

x

dTkA

dt

dQ

2

13.1.a) Define heat as energy transfer due to temperature difference.

Heat is energy transferdue to the temperaturedifference.

3

Thermal conduction in a insulator rod(non-metal)

A B

Figure 13.1

B

Molecules at A receiving heat

Molecules vibrate

Collisions cold & hot molecules

Transfer of energy

Until B become hot

4

13.1.b) Explain the physical meaning of thermal conductivity.

A lot of free electrons are available in a metal.

13.1.b) Explain the physical meaning of thermal conductivity.

Thermal conduction in a metal rod

A B

Figure 13.1

It will gain additional thermal energy due to the

heating

they move faster than

before

some of them collide with

colder molecules at B.

cold end (B) of the metal rod become

hot.

B

5

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

13.1.c) Use rate of heat transfer

A

x

1T 2T

21 TT

Figure 13.2

Insulator

x

Situation Thus…

Assuming no heat lost to the surroundings

Heat flow from higher temperature region, T1 to lower temperature, T2.

Rod in steady condition

The rate of heat flows is constant along the rod. 7

The rate of heat flow, dt

dQ

depends on

cross sectional area

temperature gradient through the cross sectional area

the type of material

dQ dTA

dt x

Where:

: cross sectional areaA

: rate of heat flowdQ

dt

: thermal conductivityk

: temperature gradientdT

x

x

dTkA

dt

dQ

• Scalar quantity

• Unit J s1 or watt (W).• -ve: direction decreasing T

8

Thermal Conductivity, k

• Is defined as a rate of heat flows perpendicularly through the unit cross sectional area of a solid, per unit temperature gradient along the direction of heat flow.

• It is a scalar quantity • Its unit is W m1 K1 or

W m1C1.

dQdtkdT

Ax

to determine k

9

Thermal conductivity is a property of

conducting material where good conductors will have higher values of k compared to poor

conductors.

Substancek

( W m1 K1)

Silver 406.0

Copper 385.0

Steel 50.2

Glass 0.8

Wood 0.08

Goo

d co

nduc

tors

10

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 m1 or C m1.

11

Non-insulated rods

1T 2TX Y

21 TT Te,Temperatur

xlength,

1T

Figure 13.4

Non-insulated (X and Y not covered with an insulator)

heat is lost to the surroundings

heat is transferred from X to Y

2T

Te,Temperatur

xlength,0

12

Yat Xat dt

dQ

dt

dQ

where A and k are the same along the rod.

temperature gradient gradually decreases along

the bar

curve graph (temperature gradient at X higher than

that at Y)

1T 2TX Y

21 TT Te,Temperatur

xlength,

1T

Figure 13.4

2T

Te,Temperatur

xlength,0

13

1T 2TX Yinsulator

Temperature gradient in the lagged composite metal bar (insulated)

2T

21 TT Te,Temperatur

xlength,

1T

Te,Temperatur

xlength,0

metal bar XY was completely covered with a good insulator

no heat loss to the surroundings along the bar

the temperature gradient will be constant along the bar

constant dt

dQ along metal bar XY Figure 13.5

insulator

14

Temperature gradient in the lagged composite metal bar

231 TTT and

DC kk

1T 2TMaterial C

Material D

3T

2T

1T3T

Te,Temperatur

xlength,

0Cx DC xx

Figure 13.6

insulator

insulator

15

steady state has

been achieved

the rate of heat flow through

both materials is

same.

From the equation of thermal conductivity,

xdT

k

1Where;

dt

dQ and A

are the same for both C and D.

Because

DC kk therefore

DC

x

dT

x

dT

231 TTT and

DC kk

1T 2TMaterial C

Material D

3T

16

Sticky Notes:

For the temperature change,

a temperature change of 1 K is

exactly equal to a temperature

change of 1 C.

17

Two properly insulated uniform rods D and E make thermal contact at one end as shown below.

insulator

insulator

D EC100

cm 20 cm 40

CT

C80

The cross sectional areas of both rods are the same. The rods are in a steady condition.

(Given kD = k W m1 C1; kE = ⅔ k W m1 C1)

Example 1 :

18

a. Determine i. The temperature gradient along

the rod D

Solution :

C; C; 80 C; 100 EDED TTTT m 40.0;m 20.0 ED xx

(Given kD = k W m1 C1; kE = ⅔ k W m1 C1)

D

DDE

D x

TT

x

dT

1

D

m C 100

x

dT

20.0

10080

D

x

dT

a. i. the temperature gradient along the rod D

19

a. Determine Solution :

C; C; 80 C; 100 EDED TTTT m 40.0;m 20.0 ED xx

(Given kD = k W m1 C1; kE = ⅔ k W m1 C1)

ii. The temperature T at the free end of the rod E.

a. ii. Since both rods are in the steady condition thus

ED

dt

dQ

dt

dQ

EE

DD

x

dTAk

x

dTAk

20

a. Determine Solution :ii. The temperature T at the free end of the rod E.

a. ii.

C; C; 80 C; 100 EDED TTTT m 40.0;m 20.0 ED xx

(Given kD = k W m1 C1; kE = ⅔ k W m1 C1)

E

DEEE

DD

x

TTk

x

dTk

20 CT

40.0

80100 3

2T

kk

21

C; C; 80 C; 100 EDED TTTT m 40.0;m 20.0 ED xx

(Given kD = k W m1 C1; kE = ⅔ k W m1 C1)

b. Sketch and label a graph to show the variation of the temperature with distance x along D and E.

100

CT

mx0 20.0 60.0

80

20

22

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.

C30 C15

cm 0.1 cm 10

steelcopperinsulator

insulator

The temperature of the exposed surfaces of the copper and steel plates are 30 C and 15 C respectively. Determine

23

(Given k copper = 385 W m1 K1 and k steel = 50.2 W m1 K1)

a. the temperature of theinterface between the copper and steel plates,

m; 10.0m; 01.0 C; 15 C; 30 SCSC xxTT

11S

11C K m W 2.50;K m W 385 kk

SC

dt

dQ

dt

dQ

S

CSSS

C

CCSC x

TTAk

x

TTAk

a. When both rods are inthe steady condition thus

Solution :

24

(Given k copper = 385 W m1 K1 and k steel = 50.2 W m1 K1)

a. the temperature of theinterface between the copper and steel plates,

a.

m; 10.0m; 01.0 C; 15 C; 30 SCSC xxTT

11S

11C K m W 2.50;K m W 385 kk

10.0

152.50

01.0

30385 CSCS TT

C8.29CST

Solution :

25

(Given k copper = 385 W m1 K1 and k steel = 50.2 W m1 K1)

m; 10.0m; 01.0 C; 15 C; 30 SCSC xxTT

11S

11C K m W 2.50;K m W 385 kk

b. the amount of heat flowing from the copper plate to the steel plate in one minute if the cross sectional area for both plates is 50 cm2 and no energy losses to the surroundings.

b. Given

Therefore the amount of heat flowing through the plates is given by

s 60;m 1050 24SC dtAA

C

CCSCC x

TTAk

dt

dQ

Solution :

26

(Given k copper = 385 W m1 K1 and k steel = 50.2 W m1 K1)

m; 10.0m; 01.0 C; 15 C; 30 SCSC xxTT

11S

11C K m W 2.50;K m W 385 kk

b. the amount of heat flowing from the copper plate to the steel plate in one minute if the cross sectional area for both plates is 50 cm2 and no energy losses to the surroundings.

C

CCSCC x

TTAk

dt

dQb.

J 2310dQ

01.0

308.291050385

604dQ

Solution :

27

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 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). ANS : 214 W m1 K1

28

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 150.0C 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, Calculate the temperature of the point where the two metal are joined. (Given k of aluminium = 205 W m1C1 and

k of brass =109 W m1C1)ANS : 90.2 C

29

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,

Learning Outcome:

13.2 Thermal expansion (2 hour)

3;2

30

Thermal expansion• Is defined as a change in dimensions of a body accompanying a change in temperature.

• There are three types of thermal expansion :i. Linear expansionii. Area expansioniii. Volume expansion

• At the same temperature,

the gas expands greater than liquid and solid

States of matter

Expansion

Linear

Area

Volume

Solid / / /

Liquid /

Gas /

31

13.2.a) Define and use the coefficient of linear, area and volume thermal expansion.b) Deduce the relationship between the coefficients of expansion, 3;2

Linear expansion

Consider a thin rod of initial

length, l0 at temperature,T0

is heated to a new uniformtemperature, T and acquireslength, l as shown in Figure13.8.

l

l

Figure 13.8

0l

At T0

At Tl

0l

32

l

Figure 13.8

At T0

At T

l

0l

0 length in change : lll where

Tl 0ll

Tll 0and

ΔT: temperature change = T-To

α: coefficient of linear expansion

Coefficient of linear expansion, is defined as a fractional increase in length of a solid per unit rise in temperature.

Tl

l

0

33

If l = ll0 then

Tll 10

length final:l

where

length initial:0l

If T is…

Value ΔT

Value Δl

length

- - decreases

+ + increases

• ΔT is the same in the Kelvin and Celcius scales.• the unit of α is °C-1 or K-1.

• l could be the length of a rod, the side of a square plate or the diameter (radius) of a hole.

34

• 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 expands by A, as shown in Figure 13.9.

Area expansion

Figure 13.9

35

0 areain change : AAA where

TAA 0TA 0AA and

ΔT: temperature change = T-Toβ: coefficient of area expansion

Coefficient of area expansion, is defined as a fractional increase in area of a solid surface per unit rise in temperature.

TA

A

0

Figure 13.9

36

If A= AA0 then

where

• the unit of is °C-1 or K-1.

TAA 10

area surface final:A

area surface initial:0A

For isotropic material (solid) , the area expansion is uniform in all direction, thus the relationship between and is given by:

2

37

Proof of = 2• Consider a square plate with side length, l0 is heated

and expands uniformly as shown in Figure 14.0.

0l

0l

0A

l

l

Figure 14.0

lll 0

2lA

200 lA eq. 1

eq. 2eq. 3

Subst. eq.3 into eq.2

2

00

20 21

l

l

l

llA

202

0 2 llllA 20 llA

02

0

l

land

38

2

00

20 21

l

l

l

llA 0

2

0

l

land

0

20 21

l

llA Recap: T

l

l

0

eq. 4

2

TAA 210 compare with TAA 10

Therefore

Subst. eq.1 into eq.4

39

Volume expansion

Consider a metal cube

with side length, l0 is

heated and expands

uniformly as shown in

Figure 14.1.

Figure 14.1

0l

l

0l

0l

l

l

40

whereTVV 0TV 0VV and

0l

l

0l

0l

l

lFigure 14.1

ΔV : change in volume = V-Vo

ΔT : temperature change = T-To

: coefficient of volume expansion

Coefficient of volume expansion, is defined as a fractional increase in volume of a solid per unit rise in temperature.

TV

V

0

41

If V= VV0 then

where

• the unit of is °C-1 or K-1.

: final volumeV0 : initial volumeV

For isotropic material (solid) , the volume expansion is uniform in all direction, thus the relationship between and is given by:

TVV 10

3

42

Proof of =3• Consider a metal cube with side length, l0 is heated

and expands uniformly as shown in Figure 14.1.

0l

l

0l

0l

l

lFigure 14.1

lll 0

3lV

300 lV eq. 1

eq. 2

eq. 3

Subst. eq.3 into eq.2

3

0

2

00

30 331

l

l

l

l

l

llV

320

20

30 33 llllllV 30 llV

033

0

2

0

l

l

l

land

43

3

0

2

00

30 331

l

l

l

l

l

llV 03

3

0

2

0

l

l

l

land

3

0

30 31

l

llV

TVV 310 compare with TVV 10

Therefore

Recap:eq. 4 Tl

l

0

44

Solution :Example 3 :

A copper rod is 20.0 cm longer than an aluminum rod before heated. How long should the copper rod be if the difference in their lengths is to be independent of temperature?

(Given copper = 1.70105 C1 and aluminum = 2.20105 C1)

Since their difference in lengths not to change with temperature, thus

AC ll

50C1.70 10 l

TlTl 0AA0CC

0C 0.88 ml 5

0C2.20 10 0.20l

45

Solution :Example 4 :

A steel ball has a diameter of 1.700 cm at 27.0 C. Given that the coefficient of volume expansion for the steel is 3.3 105 C1, calculate the diameter of the steel ball at

a. 77.0 C

0 01.700 cm; 27.0 C;d T

5 13.6 10 C

a. Given T =77.0 C

00 1 TTdd

51.700 1 1.1 10 50.0d

cm 701.1d

50.0oT T

46

Solution :Example 4 :

0 01.700 cm; 27.0 C;d T

5 13.6 10 C A steel ball has a diameter of 1.700 cm at 27.0 C. Given that the coefficient of volume expansion for the steel is 3.3 105 C1, calculate the diameter of the steel ball at

b. -56.0 C

b. Given T = 56.0 C

51.700 1 1.1 10 83d

cm 698.1d

00 1 TTdd

83.0oT T

47

Solution :Example 5 :

At 20 C a steel ball has a diameter of 0.9000 cm, while the diameter of a hole in an aluminum plate is 0.8990 cm. Calculate the temperature of the steel ball when its just pass through the hole if both ball and plate are heated in the same time.

(Given steel = 1.10105

C1 and aluminum = 2.20105 C1 )

0S 0.9000 cm; d

0S 0A 20.0 C;T T

S A CT T T

0A 0.8990 cm;d

When the ball just pass through the hole, thus

AS dd 0S S S 0S1d T T

0A A A 0A1d T T 48

Solution :Example 5 :

At 20 C a steel ball has a diameter of 0.9000 cm, while the diameter of a hole in an aluminum plate is 0.8990 cm. Calculate the temperature of the steel ball when its just pass through the hole if both ball and plate are heated in the same time.

(Given steel = 1.10105

C1 and aluminum = 2.20105 C1 )

0S 0.9000 cm; d

0S 0A 20.0 C;T T

S A CT T T

0A 0.8990 cm;d

50.9000 1 1.10 10 20T

C 121 T

50.8990 1 2.20 10 20T

49

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

• The volume expansion of a liquid whether true or apparent depend on the change in density of the liquid.

TV

V

0

50

51

Variation of Liquid Density

Consider V0 and V are the volumes of the liquid at T0 and T then

00 1 TTVV

The densities of the liquid at the two temperature are

The mass of the liquid always constant when it is expand so that

00 V

m and V

m

00

1 TTmm

and

ΔTTT 0

T

10 where

density final :density initial :0

Solution :Example 6 :

A glass flask whose volume is 1000.0 cm3 at 0.0C is completely filled with mercury at this temperature. When the flask and mercury are warmed to 100.0C, 15.5 cm3 of mercury overflow. If the coefficient of volume expansion of mercury is 18.0 105 K1, determine the coefficient of volume expansion of the glass.

30g 0m 1000.0 cm ; V V

0g 0m 0.0 C;T T

15m K 100.81

3overflow 15.5 cm ;V

g m 100.0 C;T T

0mV

C 0.0

overflowV

C 0.100 52

30g 0m 1000.0 cm ; V V

0g 0m 0.0 C;T T

15m K 100.81

3overflow 15.5 cm ;V

g m 100.0 C;T T

0mV

C 0.0

overflowV

C 0.100

The change in volume of the mercury after it is heated equals to

TVV m0mm

5m 18.0 10 1000

100.0 0.0

V

3

m cm 0.18V

53

The change in volume of the mercury after it is heated equals to

TVV m0mm

5m 18.0 10 1000

100.0 0.0

V

3

m cm 0.18V

Thus the change in volume of the glass is

overflowmg VVV 5.150.18g V

3g cm 5.2V

The coefficient of volume expansion for glass flask is

15g K 105.2

TVV g 0gg 0.00.10010005.2 g

54

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

ANS. : 78.72 C 55

2.A metal sphere with radius of 9.0 cm at 30.0C is heated until the temperature of 100.0C . Determine the percentage of change in density for that sphere.

(Given the coefficient of volume expansion for metal sphere is 5.1 105 C1)

ANS. : 0.36 %

56

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 22 106 C1)

ANS. : 0.06 cm; 88.48 cm57

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 glass= 9.0 106 C1 and

mercury = 182 106 C1 )

ANS. : 0.15 cm3

58

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

59