12 Chapter Solutions
Physics 102 Dr. M. Rulison Monday, January 29
11.)
L = L0T L0 = LT =
12
0.53 m 6 10-/ C o 30 C o
= 1472. 2m
15.)
L = L0T L0 =
m Co L T
=
( 8.47 10 m )
-4 o 75 .0 C
= 1.1293 -5 C o 10 m
L = L0T = 1.1293 10-5
(
)
-25.0C o = -2. 8233 10-4 m
17.)
Ltot = LBr + LAl = L0, Br Br T + L0, Al Al T = 0, Br Br + L0, Al Al T L 1.3 10-3 m = T = 21.3C o
19 ( .2 0m )
10-6 /C o + ( 1.0m ) 10-6 /C o T 23
T f = Ti + T = 28o C + 21.3C o = 49.3o C
27.)
V = V0 T = 4 r03 T = 4 ( 2 10-2 m ) 10-6 /C o 129C o 57 3 3 3
= 2. 464 10-7 m3
31.)
Vspill = Vgas - Vsteel = V0, gas gas T - V0,steel steel T = V0 T gas - steel
-6 o Vspill =( 20 gal ) C o 18 - 36 10-6 /C o = 0.329 gal 950 10 / C
35.)
Applied pressure must counteract the potential expansion due to heating. V V = V0 T 207 -6 /C o 10C o = 0.00207 10 V0 = T = P = - B - = ( ) B
V V0 V V0
= ( 2.2 109 Pa ) ( 0.00207 ) = 4.554 106 Pa = 45.1atm
37.)
Vnet ,alc Vnet , Hg V T ( - ) = V0T ( alc - pyrex ) =
0 Hg pyrex
alc - pyrex Hg - pyrex
10 C C 10 = 120010-/6 /C- -.99.99 10-6 //C o = 1200--99..99 = 6. 915 o 182 182
o o
-6
-6
41.)
Q = mcT - 2000 J = ( 0.6kg ) 4186 J /kg - C o T T = -0.8C o T f = Ti + T = 37.0o C - 0.8C o = 36.2o C
43.)
Q = mcT
(3
105 J /h ) ( 0.5h ) = ( 1200kg ) 4186 J /kg - C o T
T = 0.03C o T f = Ti + T = 21.00o C + 0.03C o = 21.03o C
45.)
Qnet = 0 Q forg + Qoil = 0 m forg c forg T forg + moil coil Toil = 0 (75kg )(430 J /kg - C o )(47o C - Ti , forg ) + (710kg )(2700 J /kg - C o )(47 o C - 32o C ) = 0 Ti , forg = 939o C
63.)
Qnet = 0
QAl + Qgly + Qunk = 0
- Al c Al TAl + mgly cgly Tgly = munk L f ,unk + munk cunk Tunk m - 0.150kg ) J /kg - C o + ( 0.100kg )...