Practice Problems
17.
The formula assumes that there is no thermal
expansion. This is a good approximation for solids
and liquids but it is not for gases.
1.
D
Q = mcT; the smaller mass will have the greater
change in temperature.
2.
A
Q = mcT; mass and specific heat affect how great T
given the same Q.
3.
A
Q = mcT; the soil has a smaller c  when losing heat
in the evening, it will have a greater T.
5.
A
Q = mcT
Q = (0.100 kg)(130 J/kg•K)(5 K) = 65 J
19.
Q = mcT; given the same Q and m, the one with the
lower c will have the greater T.
4.
A
18.
The beach cools faster than the ocean  the warmer
ocean air has a greater pressure.
20.
L = LoT
L = (12 x 10-6 oC-1)(0.1000 m)(580 oC) = 6.96 x 10-4 m
L = Lo + L = 0.1000 m + 0.0007 m = 0.1007 m
m1cT1 = m4cT  (1 kg)(100 – T) = (4 kg)(T – 0)
 T = 20oC
23.
The final temperature will be closer to water because
water has a higher value of c and will have a smaller T.
24.
Tile feels cooler so it must be conducting heat away
from your feet faster than rug.
25.
Run hot water over the outer glass will make it expand
more than the cooler inner glass.
26. a.
Q = (mAlcAl + malcoholcalcohol)T
Q = [(2)(900) + (20)(2400)]65 = 3.2 x 106 J
7.
A
8.
B
9.
C
10.
A
Q = Pt = mcT
(750 J/s)t = (0.500 kg)(4200 J/kg•K)(78 K)  t = 218 s
22.
6.
B
Qalloy = QH2O + QAl malloycalloyT = (mH2OcH2O + mAlcAl)T
(.120)c(200 – 31.5) = [(.500)(4200) + (.200)(900)](31.5 – 25)
calloy = 733 J/kg•K
mH2OcH2OTH2O = mAlcAlTAl
(0.150)(4200)(75 – T) = (0.050)(900)(T – 25)  T = 72oC
U = Qin + Win = 2000 – 1000 = 1000 J
Q = 3/2nRT = 3/2(1)(8.31)(200) = 2500 J
The hole expands the same as if it was filled with
material, which expands upon heating.
b.
Internal Energy
is the total energy of a body
c.
Temperature
measures the "warmth" of an object
Heat energy
is transferred between two bodies at
different temperatures
11.
Q = 5/2nRT = 5/2(1)(8.31)(200) = 4200 J
Q = -W = -(1000 J) = -1000 J
12.
No
Internal energy
Work
Hotter
Colder
When two objects are the same
temperature __ heat is transferred
When heat is transferred to a system, it
either increases the systems __ or the
system performs __
When two objects are at different
temperatures, heat naturally flows from
the __object to the __ object.
13.
Conduction
Energy transfer through collisions
Radiation
Photons are emitted from hot objects
Convection
Hot fluid moves to cooler location
14.
The nature of glass, surface area and thickness of the
glass, and the difference in temperature.
15.
16.
Conduction
Double-paned windows
Radiation
Overhangs along the south side
Convection
Weather striping around windows and doors
d.
.
Q=0J
27. a.
?
0.0010 m3 x 0.179 kg/m3 x 1 mol/.004 kg = .045 mol
b.
Q = 3/2nRT
Q = 3/2(0.045 mol)(8.31 J/mol•K)(500 K) = 280 J
c.
P = Q/t = 280 J/1 s = 280 W
28. Which process (isobaric, isothermic, or adiabatic) would
use the least amount of work to compress a gas? Explain.
Isobaric has the least area under its curve.
29. a.
PV = nRT
(1 x 105 Pa)(.1 m3) = (1 mol)(8.31)T  T = 1200 K
b. (1)
P1V1/T1 = P2V2/T2
T2 = T1(P2/P1) = 1200 K(2/1) = 2400 K
(2)
No work is done because the volume is unchanged
(3)
L = LoT
Al: (26 x 10-6 oC-1)(20 cm)(128oC) = 0.0666 cm
Brass: (19 x 10-6 oC-1)(20 cm)(128oC) = 0.0486 cm
L = 0.0666 cm – 0.0486 cm = 0.0180 cm
Qin = 3/2PV = 3/2(1.0 x 105 Pa)(0.10 m3) = 1.5 x 104 J
(4)
U = Qin + Win = 1.5 x 104 J + 0 J = 1.5 x 104 J
c. (1)
P1V1/T1 = P2V2/T2
T2 = T1(V2/V1) = 1200 K(2/1) = 2400 K
(2)
Win = -PV
Win = -(1 x 105 Pa)(0.1 m3) = -1 x 104 J
(3)
D
c.
Step
U
=
Qin
+
Win
x 105)(.01) Qin = U – Win
0
AB
= 1500 J
= 1500 J
3/ (2 x 105)(.01)
Qin = U – Win -(2 x 105)(.01)
2
BC
= 3000 J
= -2000 J
= 5000 J
3/ (-1 x 105)(.02)
Qin = U – Win
2
0
CD
= -3000 J
= -3000 J
3/ (1 x 105)(-.01)
Qin = U – Win -(1 x 105)(-.01)
2
DA
= -1500 J
= 1000 J
= -2500 J
0J
1000 J
-1000 J
Change
d.
Qin + Win = 0 because T per cycle = 0 U = Qin + Win = 0.
e.
3/
Qin = 5/2PV = 5/2(1 x 105 Pa)(0.1 m3) = 2.5 x 104 J
(4)
U = Qin + Win = 2.5 x 104 J + -1 x 104 J = 1.5 x 104 J
d.
T = PV/R = (1 x 105)(0.02)/8.31 = 240 K
(1)
T = 0 for an isothermic process 1200 K
(2)
Win = -Qin = -1 x 104 J
(3)
2(1
A = Bh = (0.01)(1 x 105) = 1000 N/m2•m3 = N•m = J
U = 0 for an isothermic process.
f.
e.
e = Wnet/Qin = 1000/6500 = 0.15
(1)
Q = 0 J for an adiabatic process
g.
(2)
ec = (TH – TL)/TH = (480 – 120)/480 = 0.75
Uin = Win = 1 x
104
J
32.
g.
U
+
+
+
–
BC
+
+
+
–
CD
–
–
?
–
+
DE
–
–
–
+
EA
–
–
–
+
0
0
+
–
+
isobaric (c)
+
+
isothermic (d)
0
+
+
0
T

U
+
Win
0
adiabatic (e)
30. a.
Process
T
AB
Qin
+
Process
isometric (b)
Process
U = Qin + W in
Change
33. a.
=
Qin
Win
1.
Isothermal expansion
0
0
+
–
2.
Adiabatic expansion
–
–
0
–
1 hr x 3600 s/1 hr x 1.2 x 108 J/1 s = 4.3 x 1011 J
3.
Isothermal compression
0
0
–
+
4.
Adiabatic compression
+
+
0
+
Change
0
0
+
–
b.
e = |W|/Qin = 0.40
Qin = W/0.40 = 4.3 x 1011 J/0.40 = 1.1 x 1012 J
c.
b.
generate power for the heat engine
heat gained to the system
heat lost by the system
c.
The net work done by the system.
31. a.
P (kPa)
B
200
100
1
X
X
2
X
3
0
–
Qout = Qin – W = 1.1 x 1012 J – 4.3 x 1011 J = 6.6 x 1011 J
4
d.
Q = mcT
T = 6.6 x 1011 J/(1 x 108 kg)(4200 J/kg•K) = 1.6 K
X
34.
U = 0 for an isothermic process.
35.
Work is equal to the area under a PV curve.
C
36.
Isobaric expansion would require the most work.
A
0.01
D
0.02
b.
A
T = PV/R = (1 x 105)(0.01)/8.31 = 120 K
B
T = PV/R = (2 x 105)(0.01)/8.31 = 240 K
C
T = PV/R = (2 x 105)(0.02)/8.31 = 480 K
37.
V (m3)
W = U = 3/2nRT = 3/2(0.30)(8.31)(-750) = -2800 J
38. a.
PV = nRT
(2 x 105)V = (1)(8.31)(273)V = 0.01 m3
b.
Win = -PV = -(2 x 105)(0.01) = -2000 J
ec = (TH – TC)/TH
0.3 = (1000 – TC)/1000 TC = 700 K
c.
U =
3/
2PV
=
3/
2(2
x
105)(0.01)
= 3000 J
d.
Practice Multiple Choice
Q = U – Win = 3000 + 2000 = 5000 J
1.
39. a.
D
P (kPa)
2.
B
100
L = LoT, Since the plate's side increased by 1 %,
then the whole diameter increase by 1 %  0.101 m
D
Collisions are elastic otherwise molecules cool down.
3.
50
C
A
4.
C
B
0.01
U = Qin + Win
U = 275 J + 125 J – 50 J = 350 J
V (m3)
0.02
b.
Q/t = kA(TH – TL)/L  increase rate by decrease A,
T2 – T1 and/or increase d
5.
C
A
T = PV/R = (5 x 104)(0.01)/8.31 = 60 K
B
T = PV/R = (2 x 105)(0.01)/8.31 = 120 K
B
C
T = PV/R = (2 x 105)(0.02)/8.31 = 120 K
7.
(1.5 kg)(200 J/kg•K)(80 K) = (3.0 kg)(1,000 J/kg•K)Tl
24,000 J = (3,000 J/kg•K)Tl Tl = 8 K (8oC)  Tl = 8 – 0
6.
A
c.
Step
AB
U
=
Qin
+
3/ (5 x 104)(.01)
2
+750 J
= 750 J
BC
CA
Change
d.
CA
e.
0
3/
x 104)(-.01)
= -750 J
0J
-1250 J
+200 J
0
-700 J
D
-(5 x 104)(-0.01)
= 500 J
-200 J
10.
PV = nRT  T  PV  greatest PV product, which is
farthest from the origin.
A
Adiabatic process occurs when there is no exchange
of heat with the environment.
Qin = W + Qout  100 J = W + 60 J  W = 40 J
e = W/Qin = (40 J/100 J) x 100 = 40 %
11.
C
ec = (TH – TL)/TH
ec = (1500 K – 600 K)/1500 K = 0.6  60 %
12.
C
f.
13.
1250 J
B
g.
14.
e = Wnet/Qin = 200/1450 = 0.14
C
h.
A
40.
T
16.

U
=
Qin
XY
+
+
+
+
YZ
+
+
+
0
ZX
–
–
–
+
–
+
0
0
Change
41. a.
|W|/Qin = 0.2
(1000 J)/Qin = 0.2  Qin = 5000 J
b.
Qout = Qin – W = 5000 – 1000 = 4000 J
W and Q depend on path taken. U depends only on
I and F. (Isothermic paths are curved.)
Win = -PV, since V > 0, then Win < 0, U > 0 because
U  T (TY > TX), Qin > 0 because U = Qin + Win.
U is proportional to T. Since X is closer to the
origin than Z, then TX < TZ and U < 0
15.
ec = (TH – TL)/TH = (120 – 60)/120 = 0.50
c.
A
9.
750 J + 700 J = 1450 J
Process
Mechanical work equals area under the curve. Since
(1) has the most area, it does the most work.
8.
Win
+700 J
2(5
rate = P = Q/t = 24,000 J/5 s = 4,800 J/s
Win
C
–
Win = -PV, when V = 0, then Win = 0.
PV = nRT  highest T has the greatest PV product.
C: (3Po)(3Vo) = 9 PoVo, D: (2Po)(4Vo) = 8 PoVo  C > D
17.
B
18.
A
19.
C
½Ug = Q  ½mgh = mcT
½(10 m/s2)(100 m) = (100 J/kg•K)T T = 5 K
U = Qin + Win
U = 685 J + (-120 J) = 565 J
U = Qin + Win
-450 J = Qin + 0  Qin = -450 J 
20.
D
c d: Win = 0 (V = 0), d a: Win = Area
Win = (6 x 105 Pa)(0.25 x 10-3 m3) = 150 J
5.
Practice Free Response
1.
a.
Q = mcT = (0.2 kg)(4190 J/kg•K)(30 K) = 25,000 J
b.
Q = mcT = (0.1 kg)(840 J/kg•K)(30 K) = 2500 J
c.
Qlost = Qgained = 25,000 J + 2500 J = 27,500 J
2.
d.
Q = mcT
27,500 J = (0.6 kg)c(50 K)  c = 920 J/kg•K
a.
PV = nRT  P(0.025 m3) = (1 mol)(8.31 J/mol•K)(400 K)
P = 133,000 Pa
b.
W = -PV = -(133,000 Pa)(0.030 m3) = -4000 J
c.
U = 3/2PV = 3/2(4000) = 6000 J
d
Q = 5/2PV = 5/2(4,000) = 10,000 J
3.
4.
e.
PV = nRT
(133,000 Pa)(0.055 m3) = (1 mol)(8.31 J/mol•K)T
T = 880 K
a.
2.2 kg x 1 mol/0.018 kg = 122 mol H2O
PV = nRT = (3.0 x 105)(2.0) = (2.2/0.018)(8.31)T  T = 590 K
b.
PV = nRT
(4.0 x 105)(2.5) = (122 mol)(8.31 J/mol•K)T  T = 980 K
c.
Increase. Internal energy is proportional to
temperature, which increases from A  C.
d.
W = WA  B + WBC
W = 0 + (-PV) = -(4.0 x 105 Pa)(0.5 m3) = -2.0 x 105 J
a. (1)
PV = nRT
(5 x 105 Pa)(0.001 m3) = (0.03 mol)(8.31 J/mol•K)T
T = 2,000 K
(2)
PV = nRT
(1 x 105 Pa)VC = (0.03 mol)(8.31 J/mol•K)(2,000 K)
VC = 0.005 m3
b.
Qin
Win
U
3/ PV
2
U – Win
3/ (4E5)(1E-3)
Win = 0
2
Qin = 600 J
U = 600 J
U – Win
-800 J
0 J (T = 0)
0 – (-800)
Qin = 800 J
3/ PV
U – Win
-PV
2
3/ (1E5)(-4E-3)
-600 – (400)
-(1E5)(-4E-3)
2
Qin = -1000 J
Win = 400 J
U = -600 J
600 + 0 – 600
600 + 800 – 1,000 0 + -800 + 400
Qtot = 400 J
Win = -400 J
Utot = 0 J
a.
P = F/A 
F = PA = P(r2) = (4.0 x 105 Pa)(0.10 m)2 = 1.3 x 104 N
b.
PV = nRT 
V = nRT/P = (2 mol)(8.31 J/mol•K)(300 K)/(4.0 x 105 Pa)
V = 0.012 m3
c.
Wout = PV = PAx = Fx
Wout = (1.3 x 104 N)(0.15 m) = 1950 J
d.
I (Heat is transferred to the gas)
Q = U + Wout = 3/2PV + PV = 5/2PV
Q = 5/2(1950 J) = 4875 J