18
CH 01 TEMPERATURE AND HEAT 1 CHAPTER 01 TEMPERATURE AND HEAT 1-1 TEMPERATURE SCALES Problem 1-1 Absolute zero is C 0 15 . 273 - . Find absolute zero on the Fahrenheit scale. Solution The Fahrenheit and Celsius scales are related by 9 32 5 - = F C T T Now C T C 0 15 . 273 - = , therefore 9 32 5 15 . 273 - = - F T 32 5 ) 9 )( 15 . 273 ( - = - F T 32 67 . 491 - = - F T F T F 0 67 . 459 32 67 . 491 - = + - = Hence absolute zero on the Fahrenheit scale is F 0 67 . 459 - . Problem 1-2 The melting point of tungsten (W) is C 0 3410 . Express this temperature on the Fahrenheit scale. Solution The Fahrenheit and Celsius scales are related by 5 9 32 C F T T = - 32 5 9 + = C F T T F T F 0 6170 32 ) 3410 ( 5 9 = + =

# Chapter 01 Temperature and Heat (Pp 1-18)

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CH 01 TEMPERATURE AND HEAT 1

CHAPTER 01

TEMPERATURE AND HEAT

1-1 TEMPERATURE SCALES

Problem 1-1

Absolute zero is C015.273− . Find absolute zero on the

Fahrenheit scale.

Solution

The Fahrenheit and Celsius scales are related by

9

32

5

−= FC TT

Now CTC

015.273−= , therefore

9

32

5

15.273 −=

− FT

325

)9)(15.273(−=

−FT

3267.491 −=− FT

FTF

067.4593267.491 −=+−=

Hence absolute zero on the Fahrenheit scale is F067.459− .

Problem 1-2

The melting point of tungsten (W) is C03410 . Express this

temperature on the Fahrenheit scale.

Solution

The Fahrenheit and Celsius scales are related by

59

32 CFTT

=−

325

9+= CF TT

FTF

0617032)3410(5

9=+=

CH 01 TEMPERATURE AND HEAT 2

Problem 1-3

Express the normal body temperature of C037 on

(a) the Fahrenheit scale and

(b) the Kelvin scale.

Solution

(a) The Fahrenheit and Celsius scales are related by

59

32 CFTT

=−

325

9+= CF TT

FTF

06.9832)37(5

9=+=

(b) The Kelvin and Celsius scales are related by

15.273+= CK TT

KTK 15.31015.27337 =+=

Problem 1-4

The boiling point of nitrogen is K35.77 . Express this

temperature in degrees Fahrenheit.

Solution

As 15.273−= KC TT

CTC

08.19515.27335.77 −=−=

Therefore 325

9+= CF TT

FTF

044.32032)8.195(5

9−=+−=

Problem 1-5

At what temperature is the numerical value the same on the

Kelvin and Fahrenheit scale?

Solution

The Fahrenheit and Kelvin scales are related by

5

15.273

9

32 −=

− KF TT

CH 01 TEMPERATURE AND HEAT 3

5

15.273

9

32 −=

− xx

35.245891605 −=− xx

xx 5916035.2458 −=−

35.22984 =x

KorFx 5755754

35.2298 0==

Problem 1-6

At what temperature is the Fahrenheit scale reading is

equal to (a) twice that of the Celsius and (b) half that of the

Celsius?

Solution

The Fahrenheit and Celsius scales are related by

9

32

5

−= FC TT

(a) Now xTF = and xTC 5.0= therefore

9

32

5

5.0 −=

xx

325

)9)(5.0(−= x

x

329.0 −= xx

xx 9.032 −=

x1.032 =

Fx0320

1.0

32==

(b) Now xTF = and xTC 2= therefore

9

32

5

2 −=

xx

160518 −= xx

160518 −=− xx

16013 −=x

CH 01 TEMPERATURE AND HEAT 4

Fx03.12

13

160−=−=

Problem 1-7

At what temperature Celsius is the sum of Celsius

temperature, the Kelvin temperature and the Fahrenheit

temperature equal to 15.495 ?

Solution

Let x0C be the desired temperature, then

KxCx )15.273(0 +=

and 59

32 CFTT

=−

59

320xFx

=−

328.1325

90 +=+= xxFx

Now 15.49500 =++ FxKinxCx

15.495)328.1()15.273( =++++ xxx

3213.27315.4958.1 −−=++ xxx

1908.3 =x

Cx050

8.3

190==

Problem 1-8

The melting and boiling points of gold are C01064 and

C02660 respectively. (a) Express theses temperature in

Kelvins. (b) Compute the difference between these

temperatures in Celsius degrees and Kelvins.

Solution

(a) KKCTMELTING 1337)2731064(10640 =+==

KKKTBOILING 2933)2732660(26600 =+==

(b) CTT MELTINGBOILING

0159610642660 =−=−

KTT MELTINGBOILING 159613372933 =−=−

CH 01 TEMPERATURE AND HEAT 5

1-2 LINEAR EXPANSION

Problem 1-9

By how much would a m10 long aluminum bar change its

length in the process of going from C030 to C

050 ? The

temperature coefficient of linear expansion for aluminum is 105105.2 −−× C .

Solution

The change in length is given by

TLL ∆=∆ 0α

)3050)(10)(105.2( 5 −×=∆ −L

mmmL 5105 3 =×=∆ −

Problem 1-10

A m100 steel measuring tape was marked and calibrated

at C020 . At a temperature of C

030 , what is percentage error

in a m100 distance when using this tape? The value of

coefficient of linear expansion for steel is 105102.1 −−×= Cα .

Solution

Now TLL ∆=∆ 0α

)2030)(100)(102.1( 5 −×=∆ −L

mL 012.0=∆

Percentage error %1000

×∆

=L

L

%012.0100100

012.0=×=

Problem 1-11

An aluminum flagpole is m33 high. By how much does its

length increases as the temperature increases by C015 ? The

coefficient of linear expansion for aluminum is 105105.2 −−×= Cα .

CH 01 TEMPERATURE AND HEAT 6

Solution

The change in length is given by

TLL ∆=∆ 0α

)15)(33)(105.2( 5−×=∆L

mmmL 12102375.1 2 ≅×=∆ −

Problem 1-12

Steel railroad tracks are laid when the temperature is

C05− . A standard section of rail is then m0.12 long. What

gap should be left between rail sections so that there is no

compression when the temperature gets as high as C00.42 ?

The coefficient of linear expansion for steel is 105102.1 −−×= Cα .

Solution

The linear expansion of a section of steel for given temperature

change is calculated as

TLL ∆=∆ 0α

)}0.5()0.42){(0.12)(102.1( 5 −−×=∆ −L

mmmL 8.610768.6 3 ≅×=∆ −

Hence the desired gap between two rail sections is mm8.6 .

Problem 1-13

A circular hole in an aluminum plate is cm725.2 in

diameter at C020 . What is its diameter when the

temperature of plate is raised to C0140 ?

Solution

Now TLL ∆=∆ 0α

)20140)(725.2)(105.2( 5 −×=∆ −L

cmL 008.0=∆

The new diameter will be

LLL ∆+= 01

cmL 733.2008.0725.21 =+=

CH 01 TEMPERATURE AND HEAT 7

Problem 1-14

A copper telephone wire has essentially no sag between

poles m0.35 apart on a winter day when the temperature is

C00.20− . How much longer is the wire on a summer day

when CTC

00.35= ? The coefficient of linear expansion for

copper is 105107.1 −−×= Cα .

Solution

The increase in length is given by

TLL ∆=∆ 0α

)}0.20(0.35){0.35)(107.1( 5 −−×=∆ −L

cmmL 27.31027.3 2 =×=∆ −

Problem 1-15

A brass rod m5.1 long expands mm89.1 when heated from

C020 to C

090 . Find the coefficient of linear expansion of

brass.

Solution

The change in length is given by

TLL ∆=∆ 0α

1053

0

108.1)2090)(5.1(

1089.1 −−−

×=−

×=

∆= C

TL

CH 01 TEMPERATURE AND HEAT 8

1-3 SUPERFICIAL OR SURFACE EXPANSION

Problem 1-16

A glass window is m200 by m300 at C010 . By how much

has its area increased when its temperature is C040 ?

Assume that the glass is free to expand. The coefficient of

linear expansion for glass is 106109 −−× C .

Solution

The increase in surface area is given by

TATAA ∆=∆=∆ αβ 2 αβ 2=Θ

)1040)}(10300)(10200){(109(2 226 −×××=∆ −−−A

231024.3 mA−×=∆

Problem 1-17

A square hole cm0.8 along each side is cut on a sheet of

copper. (a) Calculate the change in the area of this hole if

the temperature of the sheet is increased by K00.50 . (b)

Does this change represent an increase or a decrease in the

area enclosed by the hole?

Solution

(a) The change in the area of the hole is given by

TATAA ∆=∆=∆ αβ 2 αβ 2=Θ

)0.50()100.8)(107.1(2 225 −− ××=∆A

2510088.1 mA−×=∆

224 109.0101088.0 cmmA ≅×=∆ −

(b) As the length of each side of the square hole increases with

increase in temperature, therefore the above A∆ represents an

increase in the area of the hole.

CH 01 TEMPERATURE AND HEAT 9

1-4 CUBICAL OR VOLUMETRIC EXPANSION

Problem 1-18

A quartz sphere is cm75.8 in diameter. What will be its

change in volume if it is heated from C030 to C

0200 ? The

coefficient of volume expansion for quartz is106101 −−× C .

Solution

The change in volume is given by

TVV ∆=∆ 0γ

)30200(2

75.8)(

3

4)101(

3

43

63 −

×=∆

=∆ − ππγ TrV

3832 1096.51096.5 mcmV−− ×=×=∆

363 1011 mcm−×=Θ

Problem 1-19

Find the change in volume of an aluminum sphere of cm10

radius when it is heated from C00 to C

0100 . The coefficient

of linear expansion for aluminum is 105105.2 −−×= Cα .

Solution

The change in volume is given by

TVTVV ∆=∆=∆ 00 3 αγ

The volume of a sphere is given by

3

03

4rV π=

Hence

TrTrV ∆=∆

=∆ 33 43

43 αππα

)0100()1010)(105.2)((4 325 −××=∆ −−πV

335 4.311014.3 cmmV =×=∆ −

363 1011 cmm ×=Θ

CH 01 TEMPERATURE AND HEAT 10

Problem 1-20

A bowl made of Pyrex glass is filled to the very brim with 3100 cm of water at C

010 . How much will overflow when

the temperature of the filled bowl is raised to C050 ? The

coefficient of volume expansion for water is 1041007.2 −−× C .

Solution

The change in volume is given by

TVV ∆=∆ 0γ

)1050)(10100)(1007.2( 64 −××=∆ −−V

337 828.01028.8 cmmV =×=∆ −

363 1011 cmm ×=Θ

Problem 1-21

The mercury in a thermometer at C00 has a volume of

350.0 cm . What will be its volume at C0100 ? The

coefficient of volume expansion of mercury is 1041082.1 −−× C .

Solution

The new volume ‘V’ is given by

TVVVVV ∆+=∆+= 000 γ

)1(0 TVV ∆+= γ

)}0100)(1082.1(1){1050.0( 46 −×+×= −−V

337 5091.010091.5 cmmV =×= − 363 1011 cmm ×=Θ

Problem 1-22

What is the volume of a lead ball at C012− if its final

volume at C0160 is 3530 cm ? The coefficient of linear

expansion for lead is105109.2 −−×= Cα .

Solution

Let 1V and 2V be the volume of given lead ball at C012− and

C0160 respectively, then

)31(3 111112 TVTVVVVV ∆+=∆+=∆+= αα

CH 01 TEMPERATURE AND HEAT 11

)}12(160){109.2(31

530

)31( 5

21

−−×+=

∆+=

−T

VV

α

3

2 522 cmV =

Problem 1-23

The volume of a metal block increases by %15.0 when it is

heated through C050 . Calculate the coefficient of linear

expansion for this metal.

Solution

The change in volume is given by

TVTVV ∆=∆=∆ 00 3 αγ

×

=

∆=

503

1

100

15.0

3

1

0 TV

105101 −−×= Cα

Problem 1-24

A brass cube has an edge length of cm2.33 at C020 . Find

(a) the increase in surface area and

(b) the increase in volume when it is heated to C075 .

The coefficient of linear expansion for brass is 105109.1 −−× C .

Solution

The initial surface area and volume of the glass cube are given

by 21222

0 10613.6)102.33(66 mxA−− ×=×==

32323

0 10659.3)102.33( mxV−− ×=×==

(a) TATAA ∆=∆=∆ 00 2 αβ

)2075)(10613.6)(109.1(2 15 −××=∆ −−A

2310382.1 mA−×=∆

(b) TVTVV ∆=∆=∆ 00 3 αγ

)2075)(10659.3)(109.1(3 25 −××=∆ −−V

3410147.1 mV−×=∆

CH 01 TEMPERATURE AND HEAT 12

Problem 1-25

The coefficient of volume expansion of glycerin is 1041005.5 −−× C . What will be the fractional change in its

density for a C050 rise in temperature?

Solution

The change in density is given by

V

m

V

m−=−=∆

'' ρρρ

∆+

∆−=

−∆+

=∆)(

11

0000 VVV

Vm

VVVmρ

0

0

00

2

0V

V

V

V

V

m

V

Vm ∆−=

−=

∆−≅∆ ρρ

TV

TV

V

V∆−=

∆−=

∆−=

∆γ

γ

ρ

ρ

0

0

00

24

0

10525.2)50)(1005.5( −− ×−=×−=∆

ρ

ρ

Problem 1-26

When the temperature of a metal cylinder is raised from

C060 to C

0100 , its length increases by %092.0 .

(a) Find the percentage change in density.

(b) Identify the metal.

Solution

(a) The fractional change in density is given by

%276.0%)092.0(33 −=−=∆

−=∆

−=∆

L

L

V

V

ρ

ρ

(b) Now TLL ∆=∆ α

105103.260100

%)092.0(1 −−×=−

=

∆= C

TL

The given metal is aluminum.

CH 01 TEMPERATURE AND HEAT 13

1-5 SPECIFIC HEAT

Problem 1-27

What is the specific heat of a metal if kJ135 of heat is

needed to raise kg1.5 of the metal from C020 to C

030 ?

Solution

Now TCmQ ∆=

1013

2647)2030)(1.5(

10135 −−=−

×=

∆= CkgJ

Tm

QC

Problem 1-28

The temperature of a silver bar rises by C010 when it

absorbs kJ23.1 of energy by heat. The mass of the bar is

g525 . Determine the specific heat of silver.

Solution

The heat absorbed is given by

TCmQ ∆=

101

3

3

234)10)(10525(

1023.1 −−

−=

×

×=

∆= CkgJ

Tm

QC

Problem 1-29

The brake linings of the wheels of a car have total mass

kg8.4 and specific heat capacity 111200 −− KkgJ . Calculate

the maximum possible temperature rise of the brake linings

when the car (of mass kg800 ) traveling at 115 −sm is

brought to rest by applying the brakes.

Solution

Assuming no lasses we have

Heat generated at brake linings = Kinetic energy of the car

2

2

1vMTCm =∆

KCm

vMT 6.15

)1200)(8.4(2

)15)(800(

2

22

===∆

CH 01 TEMPERATURE AND HEAT 14

Problem 1-30

(a) Compute the possible increase in temperature for

water going over Niagara fall, m4.49 high.

(b) What factors would tend to prevent this possible

rise?

(Given that 114190 −−= KkgJC for water)

Solution

(a) If the change in potential energy of water appears as rise

in its internal energy, then

hgmTCm =∆

KC

hgT 116.0

4190

)4.49)(8.9(===∆

(b) The water will evaporate during its fall accompanied by

cooling, therefore the above rise in temperature i.e.

0.116 K is not observed.

Problem 1-31

A kW2 electric heater is used in a well-insulated hot water

tank of capacity litres200 to raise the temperature of water

from C05 to C

060 . How long will it take to change the

temperature of water in the tank?

Solution

Now

TCmtPQ ∆==

P

TCV

P

TCmt

∆=

∆=

ρ

3

33

102

)560)(4190)(10200)(101(

×

−××=

t

shst 5min24623045 ==

CH 01 TEMPERATURE AND HEAT 15

Problem 1-32

Calculate the minimum amount of heat required to

completely melt g130 of silver (melting point K1235 )

initially at C0

0.16 . The specific heat and latent heat of

fusion of silver are 11236 −− KkgJ and 151005.1 −× KJ

respectively. Assume that specific heat does not change with

temperature.

Solution

The desired amount of heat is given by

fLmTCmQ +∆=

)( fLTCmQ +∆=

Now

kggm 130.0130 ==

11236 −−= KkgJC

151005.1 −×= KJL f

KKCT 289)27316(0.16 0

1 =+==

KT 12352 =

KTTT 946289123512 =−=−=∆

Hence

)}1005.1()946)(236){(130.0( 5×+=Q

kJJQ 67.4210267.4 4 =×=

Problem 1-33

What mass of steam at C0100 must be mixed with g150 of

ice at C00 , in a thermally insulated container, to produce

liquid at C050 ? The specific heat, latent heat of fusion and

latent heat of vapourization of water are 114190 −− KkgJ , 151034.3 −× KJ and 1610256.2 −× KJ respectively.

CH 01 TEMPERATURE AND HEAT 16

Solution

Let ‘m’ be the desired mass of steam, then heat librated by

steam to produce water at 50 0C will be

)( vv LTCmLmTCmQ +∆=+∆=

)}10256.2()50100)(4190{( 6×−−= mQ

JmQ )104655.2( 6×=

The above heat will be absorbed by kggm 150.01501 == of

ice to produce water at 50 0C. Hence

)(111 ff LTCmLmTCmQ +∆=+∆=

)}1034.3()050)(4190){(150.0()104655.2( 56 ×+−=×m

)101525.8()104655.2( 46 ×=× m

gkgm 33033.0104655.2

101525.86

4

==×

×=

Problem 1-34

How much energy is required to change a g40 ice cube

from ice at C010− to steam at C

0110 ?

Solution

The net heat to complete the above change consists of the

following processes.

Q = Heat needed to convert C010− ice into C

00 ice +

Heat needed to convert C00 ice into C

00 water +

Heat needed to convert C00 water into C

0100 water +

Heat needed to convert C0100 water into C

0100 steam +

Heat needed to convert C0100 steam into C

0110 steam.

321 TCmLmTCmLmTCmQ STEAMvWATERfICE ∆++∆++∆=

)( 321 TCLmTCLTCmQ STEAMvWATERfICE ∆++∆++∆=

)100)(4190()1034.3()10)(2090){(040.0( 5 +×+=Q

)}10)(92010()10256.2( 6 +×+

JQ 51022.1 ×=

CH 01 TEMPERATURE AND HEAT 17

1-6 THERMAL CONDUCTION

Problem 1-35

Calculate the rate at which heat would be lost on a very cold

winter day through a mm 8.32.6 × brick wall cm32 thick.

The inside temperature is C026 and the outside temperature

is C018− ; assume that the thermal conductivity of the brick

is 1174.0 −−

KmW .

Solution

The quantity of heat ‘Q’ transferred from one face to the other

is given by

tx

TAkQ ∆

∆−=

x

TAk

t

QH

∆−=

∆=

Now 1174.0 −−= KmWk

22 56.238.32.6 mmA =×=

CTTT 0

21 442618 −=−−=−=∆ (Note that the difference in

temperature will be same on Celsius and Kelvin scales)

mcmx 32.032 ==∆

Hence

WH310397.2

32.0

44)56.23)(74.0( ×=

−=

Problem 1-36

A steel plate measures cm20 by cm20 and is mm0.5

thick. One face is maintained at C0140 . How much power

would be needed to maintain the other face at C0159 ? The

value of thermal conductivity for steel is 1146 −−KmW .

Solution

The desired power is given by

x

TAkH

∆−=

Now 1146 −−= KmWk

CH 01 TEMPERATURE AND HEAT 18

222422 104104004002020 mmcmcmA−− ×=×==×=

KCTTT 1010150140 0

21 ==−=−=∆

mmmx31050.5 −×==∆

Hence

WH 3680105

10)104)(46(

3

2 =

×

−×−=

Problem 1-37

How thick a concrete wall would be needed to give the same

insulating value as cm10 of fiberglass? The values of

thermal conductivity of concrete and fiberglass are 111 −−

KmW and 11042.0 −−

KmW respectively.

Solution

Now

CONCRETEFIBERGLASS HH =

C

C

F

Fx

TAk

x

TAk

∆−=

∆−

F

F

C

C xk

kx ∆

=∆

mcmxC 38.2238)10(042.0

1==

=∆