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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
Lα
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
Vα
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
Lα
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==
=∆