Physics, 6th Edition
Chapter 16. Temperature and Expansion
Chapter 16. Temperature and Expansion 16-1. Body temperature is normal at 98.60. !hat is the correspondin" temperature on the Celsius scale#
= $ 9 %t F − &'0 ( = $ 9 %98.60 − &'0 ( )
tC
t C = &*0C C =
16-'. The +oilin" point o sulur is .$0C. !hat is the correspondin" temperature on the ahrenheit scale# t F
= 9 $ t C + &'0 = 9 $ %.$0 ( + &'0 )
t F = = 8&'0
16-&. steel rail cools rom *0 to &00C in 1 h. !hat is the chan"e o temperature in ahrenheit de"rees or the same time period#
∆t / *0 C &0 C / 0 C ) 0
0
0
9 0 ∆t = 0C 0 ; $ C 0
∆
t / *' 0
16-. t 2hat temperature 2ill 2ill the Celsius and ahrenheit scales ha3e the same numerical readin"# $
9
% x − &'0 ( = 9 $ x + &'0 )
x / -00C or 00
16-$. piece o o charcoal initially at 1800 experiences a decrease in temperature o 1'0 1'0 0. Express this chan"e o temperature in Celsius de"rees. !hat is the the inal temperature on the Celsius scale#
∆t / 1'0 ) 0
0
1'0
$ C0 0 9 0 = 66.* C )
/ 66.* C0 ∆t /
The final temperature is 1800 1'0 0 / 600 which must be converted to 0C: tC
= $ 9 %t F − &'0 ( = $ 9 %600 − &'0 ( ) ''&
t C = 1$.6 0C C =
Physics, 6th Edition
Chapter 16. Temperature and Expansion
16-6. cetone +oils at $6.$0C and li4uid nitro"en +oils at -1960C. Express these speciic temperatures on the 5el3in scale. !hat is the dierence in these temperatures on the Celsius scale# T = t C '*&0m / $6.$0 '*&0)
Acetone: Nitrogen:
T = t C '*&0m / -1960 '*&0)
T / &'9.$ 5 T / **.0 5
∆t / '$'.$ C0
∆t = $6.$0C %-1960C( / '$'.$ C0)
Note: The difference in kelvins is the same as in Celsius degrees.
16-*. The +oilin" point o oxy"en is '9*.&$0. Express this temperature in 7el3ins and in de"rees Celsius. t C = $ 9 %−'9*.&$0 − &'0 () T / tC '*&0 / -18&0C '*&0)
t C / -18&0C T = 90.0 5
16-8. oxy"en cools rom 1'00 to *00, 2hat is the chan"e o temperature in 7el3ins#
∆t / 1'00 *00 / $0 0) 0
$0
$ C0 0 9 0 = '*.8 C )
1 5 / 1 C0)
∆t / '*.8 5
16-9. 2all o ire+ric7 has an inside temperature o &1&0 and an outside temperature o *&0. Express the dierence o temperature in 7el3ins.
∆t / &1&0 *&0 / '0 0) 0
'0
$ C0 0 9 0 = 1&& C )
1 5 / 1 C0)
''
∆t / 1&& 5
Physics, 6th Edition
Chapter 16. Temperature and Expansion
16-10. old melts at 1&&65. !hat is the correspondin" temperature in de"rees Celsius and in de"rees ahrenheit. t C = 1&&6 5 '*&0 / 106&0C) t F
t C = 10600C
= 9 $ t C + &'0 = 9 $ %106&0 ( + &'0 )
t F = 19$00
16-11. sample o "as cools rom 1'0 to -1800C. Express the chan"e o temperature in 7el3ins and in ahrenheit de"rees. : ince 1 5 / 1 C0, the change in kelvins is the same as in C !.;
∆t / -1800C %-1'00C( / -60 C0 ) 9 0 = −108 0 ; 0 $ C
∆t = −60C0
∆T / -60 5 ∆
t / -108 0
Temperature and Expansion 16-1'. sla+ o concrete is '0 m lon". !hat 2ill +e the increase in len"th i the temperature chan"es rom 1'0C to &00C. ssume that α / 9 x 10-6
∆ " = α "0 ∆t = %9 x 10-6 o2 much 2ill it increase in len"th 2hen heated to 800C#
∆ " = α "0 ∆t = %1.* x 10-$ o2 much 2ill it increase in len"th 2hen it is placed into +oilin" 2ater %'1'0(#
∆ " = α "0 ∆t = %1.1 x 10-$ <0 (%1 t(%'1'0 - *00 ( ) ∆= / 0.001$6 t ∆= / 0.001$6 t%1 in.<1' t()
''$
∆= / 0.018* in.
Physics, 6th Edition
Chapter 16. Temperature and Expansion
16-1$. The diameter o a hole in a steel plate 9 cm 2hen the temperature is '00C. !hat 2ill +e the diameter o the hole at '000C#
∆ " = α "0 ∆t = %1.' x 10-$
= / 9.0' cm
16-16. +rass rod is '.00 m lon" at 1$0C. To 2hat temperature must the rod +e heated so that its ne2 len"th is '.01 m#
: ∆ " = '.01 m '.00 m / 0.01 m ;
∆ " = α "0 ∆t) ∆t =
∆ " α "o
=
0.010 m -$
0
%1.8 x 10
t = t o # ∆t = 1$0C '*8 C0)
∆t =
)
0
'*8 C
t / '9&0C
16-1*. s4uare copper plate cm on a side at '00C is heated to 1'00C. !hat is the increase in the area o the copper plate# : γ / 'α / '%1.* x 10-$
∆ / γ o∆t / %&. x 10-$
π $ '
=
π %'0 cm('
= &1.16 cm' at '*0C
The change in area must be &1 cm' &1.16 cm'?
∆ A = 'α A0 ∆t)
∆t =
∆ A 'α A0
=
∆ / - 0.16 cm
−0.16 cm -$
0
'%1.' x 10
Thus% the final temperature is '*0C '1.'0C?
''6
= −'1.' C0
t = $.880C
Physics, 6th Edition
Chapter 16. Temperature and Expansion
16-19. !hat is the increase in 3olume o 16.0 liters o ethyl alcohol 2hen the temperature is increased +y &0 C0#
∆& = β &o ∆t = %11 x 10-
∆& / 0.$'8 =
16-'0. Pyrex +ea7er has an inside 3olume o 600 m= at '00C. t 2hat temperature 2ill the inside 3olume +e 60& m=# : ∆& = 60& m= &00 m= / & m= ;
∆& = β &0 ∆t) ∆t =
∆&
β & 0
=
& m= −$
0
%0.& x 10
t = '00 C $$60C)
= $$6 C0
t / $*60C
16-'1. '00 cm& o +en@ene exactly ills an aluminum cup at 00C, and the system cools to 180C, ho2 much +en@ene %at 180C( can +e added to the cup 2ithout o3erlo2in"#
∆& = ∆& ' − ∆&A" = β '&0 ∆t − %&α (&0 ∆t) ∆ t / %18 0( / -'' C0 ∆A / %1'. x 10-
AB / $.1 cm&
16-''. Pyrex "lass +ea7er is illed to the top 2ith '00 cm& o mercury at '00C. >o2 much mercury 2ill o3erlo2 i the temperature o the system is increased to 680C# & o = '00 cm&)
βm / 1.8 x 10-
∆& = ∆&m − ∆&p = β m&0 ∆t − %&α p (&0 ∆t)
∆ t / 680C '00C / 8 C0)
∆A / %1.8 x 10-
''*
AB / 1.6 cm&
Physics, 6th Edition
Chapter 16. Temperature and Expansion
Challenge Problems 16-'&. The diameter o the hole in a copper plate at '00C is &.00 mm. To 2hat temperature must the copper +e cooled i its diameter is to +e '.99 mm. :∆= / %'.99 &.00( / -0.01 mm ;
∆ " = α "0 ∆t) ∆t =
∆ " α "o
=
−0.010 m -$
0
%1.* x 10
t = t o # ∆t = '00C - 196 C0)
)
∆t =
-196 C
0
t / -1*60C
16-'. rectan"ular sheet o aluminum measures 6 +y 9 cm at '80C. !hat is its area at 00C# o / %6 cm(%9 cm( / $ cm')
∆t / 00 '80C / -'80C ) γ / 'α
∆ / 2αo∆t / '%'. x 10-$
/ $&.9 cm'
16-'$. steel tape measures the len"th o an aluminum rod as 60 cm 2hen +oth are at 80C. !hat 2ill the tape read or the len"th o the rod i +oth are at &80# The aluminum rod will expand more than does the steel tape. Thus the tape will give a smaller reading based on the difference in the change of length.
∆== / αl=0∆t / %'. x 10-$
∆=l / 0.0'16 cm
The reading will be less b( the difference in the expansions. )eadin" / 60 cm %0.0&' cm - 0.0'16 cm() )eading / 60.0' cm 16-'6. t '00C, a copper cu+e measures 0 cm on a side. !hat is the 3olume o the cu+e 2hen the temperature reaches 1$00C# : & ! = %0 cm(& / 6,000 cm& )
∆t / 1$0 '0 / 1&0 C0 ;
& = & ! # *α &! ∆t = 6,000 cm& &%1.* x 10-$
''8
& = 6,'0 cm&
Physics, 6th Edition
Chapter 16. Temperature and Expansion
16-'*. Pyrex +ea7er %α / 0.& x 10-$o2 much "lycerine 2ill o3erlo2 the top i the system is heated rom '00C to 1000C#
: & o = '00 m=)
β" / $.1 x 10-
∆& = ∆&m − ∆&p = β m&0 ∆t − %&α p (&0 ∆t)
∆ t / 1000C '00C / 80 C0)
∆A / %$.1 x 10-
AB / 8.0' m=
16-'8. sto3e is at $00. the temperature drops +y $0 7el3ins, 2hat is the ne2 temperature in de"rees Celsius# tC
= $ 9 %t F − &'0 ( = $ 9 %$0 − &'0 ( )
ince 1 5 / 1 C0?
t / '&'.'0 $00
t C / '&'.'0C and
t / 18'0C
*16-'9. 100-t steel tape correctly measures distance 2hen the temperature is '00C. !hat is the
true measurement i this tape indicates distance o 9.6' t on a day 2hen the temperature is &60C# "o∆t / %1.' x 10-$
= / 9.6 t
*16-&0. The diameter o a steel rod is &.000 mm 2hen the temperature is '00C. lso at '00C, the
diameter o a +rass rin" is '.99$ mm. t 2hat common temperature 2ill the +rass rin" slip o3er the steel rod smoothly#
∆= + - ∆=s / &.000 mm '.999 mm / 0.001 mm %1.8 x 10-$
''9
t / *$.'0C
Physics, 6th Edition
Chapter 16. Temperature and Expansion
*16-&1. certain metal cu+e increases its 3olume +y 0.$0 percent 2hen its temperature increases
+y 100 C0. !hat is the linear expansion coeicient or this metal# : 0.$ / 0.00$ ;
∆& = β&0 ∆t = &α&0 ∆t)
∆& 1 = %0.00$0( &∆t & 0 &%100 C0 ( 1
α =
α / 1.6* x 10-$
:
β / &α
and
∆t / 1000C '00C / 80 C0 ;
∆& = β&0 ∆t = &α&0 ∆t) ∆& & 0
∆& & 0
x 10-$
= .&' x 10-& )
+ ∆&,&- / 0.&'
16-&&. round +rass plu" has a diameter o 8.001 cm at '80C. To 2hat temperature must the plu" +e cooled i it is to it snu"ly into an 8.000 cm hole#
∆ " = α "0 ∆t) ∆t =
∆ " α "o
=
−0.001 cm -$
0
%1.8 x 10
t = t o # ∆t = '80C 6.9 C0)
)
∆t =
-6.9 C0
t / '1.10C
6-&. i3e hundred cu+ic centimeters o ethyl alcohol completely ill a Pyrex +ea7er. the temperature o the system increases +y *0 C0, 2hat 3olume o alcohol o3erlo2s# & o = $00 cm&)
βm / 11 x 10-
∆& = ∆&m − ∆&p = β m&0 ∆t − %&α p (&0 ∆t)
∆ t / *0 C0)
∆A / %11 x 10-
'&0
AB / &8.' cm&
Physics, 6th Edition
Chapter 16. Temperature and Expansion
Critical Thinking Questions 16-&$. The la+oratory apparatus or measurin" the coeicient o linear expansion is illustrated in i". 16-1*. The temperature o a metal rod is increased +y passin" steam throu"h n enclosed ac7et. The resultin" increase in len"th is measured 2ith the micrometer scre2 at one end. Dince the ori"inal len"th and temperature are 7no2n, the expansion coeicient can +e calculated rom E4. %16-8(. The ollo2in" data 2ere recorded durin" an experiment 2ith a rod o un7no2n metal? "o = 600 mm
t o / '&0C
∆ " = 1.0 mm
t f / 980C
!hat is the coeicient o linear expansion or this metal# Can you identiy the metal#
α =
1.0 mm ∆ " = ) α / '.& x 10-$
luminum
6-&6. ssume that the end points o a rod are ixed ri"idly +et2een t2o 2alls to pre3ent expansion 2ith increasin" temperature. rom the deinitions o oun"Fs modulus %Chapter 1&( and your 7no2led"e o linear expansion, sho2 that the compressi3e orce F exerted +y the 2alls 2ill +e "i3en +y F = α A ∆t 2here A / cross-section o the rod, / oun"Fs modulus, and ∆t / increase in temperature o rod. rom Chap. 1$, youn"Fs modulus is?
=
F" A∆"
)
F
∆" = A ) "
and
∆" = α "0 ∆t
/liminating + ∆ ","-% we have?
'&1
or
∆= =
F
= α ∆t
F = Aα ∆t
Physics, 6th Edition
Chapter 16. Temperature and Expansion
16-&*. Pro3e that the density o a material chan"es 2ith temperature so that the ne2 density ρ is "i3en +y
ρ = 2here
ρ o 1 + β ∆t
ρo / ori"inal density, β / 3olume expansion coeicient, and ∆t / chan"e in
temperature.
& f = & o # β &o ∆t = & o+0 # α∆t-
from which
Now% ρ = m,& or & = m, ρ % so that:
ρo ρ
= 1 + β ∆t )
ρ =
& f & 0
m < ρ m < ρ 0
= 1 − β ∆t )
= 1 + β ∆t
ρ 0 1 + β ∆t
16-&8. The density o mercury at 00C is 1&.6 "
ρ / 1&.$ "
16-&9. steel rin" has an inside diameter o .000 cm at '00C. The rin" is to +e slipped o3er a copper shat that has a diameter o .00& cm at '00C. %a( To 2hat temperature must the rin" +e heated# %+( the rin" and the shat are cooled uniormly, at 2hat temperature 2ill the rin" ust slip o the shat# : ∆=s / .00& cm .00 cm / 0.00& cm ; %a(
∆=s / α=o∆t)
∆t =
0.00& cm -$
0
%1.' x 10
= 6'.$ C0
The steel ring must be heated to: '00C 6'.$0 or
'&'
8'.$0C
Physics, 6th Edition
Chapter 16. Temperature and Expansion 1$-&9. %Cont.(
%+( Find temperature at which ring slips off easil(.
'egin at '00C where " s = .000 cm and =c / .00& cm % next cool both until the diameters are the same. That occurs when the copper rod shrinks more than the steel ring such that: ∆ "copper 2 ∆ " steel / 0.00& cm
α c "c∆t 2 α s " s∆t = 0.00& cm)
∆t =
∆t =
0.00& cm
α c "c − α s "s
0.00& cm %1.* x 10
0
-$
0
= 1$0 C0
Thus the temperature of both copper and steel must decrease b( 1$00C. t f = t ! 3 1$0 C0 / '00C 1$0 C01
'&&
t f = -1&00C
