Heat Transfer and Thermodynamics Conceptual Problems

Chapter 18
Heat and the First Law of Thermodynamics
Conceptual Problems
3 • The specific heat of aluminum is more than twice the specific heat of
copper. A block of copper and a block of aluminum have the same mass and
temperature (20ºC). The blocks are simultaneously dropped into a single
calorimeter containing water at 40ºC. Which statement is true when thermal
equilibrium is reached? (a) The aluminum block is at a higher temperature than
the copper block. (b) The aluminum block has absorbed less energy than the
copper block. (c) The aluminum block has absorbed more energy than the copper
block. (d) Both (a) and (c) are correct statements.
Picture the Problem We can use the relationship TmcQ Δ= to relate the amount
of energy absorbed by the aluminum and copper blocks to their masses, specific
heats, and temperature changes.
Express the energy absorbed by the
aluminum block:
TcmQ Δ= AlAlAl
Express the energy absorbed by the
copper block:
TcmQ Δ= CuCuCu
Divide the second of these equations
by the first to obtain: Tcm
Tcm
Q
Q
Δ
Δ
=
AlAl
CuCu
Al
Cu
Because the block’s masses are the
same and they experience the same
change in temperature:
1
Al
Cu
Al
Cu
<=
c
c
Q
Q
or
AlCu QQ < and )(c is correct.
11 • A real gas cools during a free expansion, while an ideal gas does not
cool during a free expansion. Explain the reason for this difference.
Determine the Concept Particles that attract each other have more potential
energy the farther apart they are. In a real gas the molecules exert weak attractive
forces on each other. These forces increase the internal potential energy during an
expansion. An increase in potential energy means a decrease in kinetic energy,
and a decrease in kinetic energy means a decrease in translational kinetic energy.
Thus, there is a decrease in temperature.
363

Chapter 18364
21 •• An ideal gas in a cylinder is at pressure P and volume V. During a
quasi-static adiabatic process, the gas is compressed until its volume has
decreased to V/2. Then, in a quasi-static isothermal process, the gas is allowed to
expand until its volume again has a value of V. What kind of process will return
the system to its original state? Sketch the cycle on a graph.
Determine the Concept During a
reversible adiabatic process, is
constant, where γ < 1 and during an
isothermal processes is constant.
Thus, the pressure rise during the
compression is greater than the pressure
drop during the expansion. The final
process could be a constant volume
process during which heat is absorbed
from the system. A constant-volume
cooling will decrease the pressure and
return the gas to its original state.
γ
PV
PV
V
P
adiabatic
isothermal
iVi2
1
f VV =
constant volume
Estimation and Approximation
25 •• A ″typical″ microwave oven has a power consumption of about 1200
W. Estimate how long it should take to boil a cup of water in the microwave
assuming that 50% of the electrical power consumption goes into heating the
water. How does this estimate correspond to everyday experience?
Picture the Problem Assume that the water is initially at 30°C and that the cup
contains 200 g of water. We can use the definition of power to express the
required time to bring the water to a boil in terms of its mass, heat capacity,
change in temperature, and the rate at which energy is supplied to the water by the
microwave oven.
Use the definition of power to relate
the energy needed to warm the water
to the elapsed time:
t
Tmc
t
W
P
Δ
Δ
=
Δ
Δ
= ⇒
P
Tmc
t
Δ
=Δ
Substitute numerical values and evaluate Δt:
( ) ( )
min1.6s63.97
W600
C30C100
Kkg
kJ
184.4kg200.0
Δ ≈=
°−°⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
=t ,
an elapsed time that seems to be consistent with experience.

Heat and the First Law of Thermodynamics 365
Heat Capacity, Specific Heat, Latent Heat
29 • How much heat must be absorbed by 60.0 g of ice at –10.0ºC to
transform it into 60.0 g of water at 40.0ºC?
Picture the Problem We can find the amount of heat that must be absorbed by
adding the heat required to warm the ice from −10.0°C to 0°C, the heat required to
melt the ice, and the heat required to warm the water formed from the ice to
40.0°C.
Express the total heat required: waterwarmicemelticewarm QQQQ ++=
Substitute for each term to obtain:
( )waterwaterficeice
waterwaterficeice
TcLTcm
TmcmLTmcQ
Δ++Δ=
Δ++Δ=
Substitute numerical values (See Tables 18-1 and 18-2) and evaluate Q:
( ) ( )( )
( )
kJ3.31
C0C0.04
Kkg
kJ
184.4
kg
kJ
5.333C0.01C0
Kkg
kJ
05.2kg0.0600
=
⎥
⎦
⎤
°−°⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
+
+⎢
⎣
⎡
°−−°⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
=Q
Calorimetry
33 • While spending the summer on your uncle’s horse farm, you spend a
week apprenticing with his farrier (a person who makes and fits horseshoes). You
observe the way he cools a shoe after pounding the hot, pliable shoe into the
correct size and shape. Suppose a 750-g iron horseshoe is taken from the farrier’s
fire, shaped, and at a temperature of 650°C, dropped into a 25.0-L bucket of water
at 10.0°C. What is the final temperature of the water after the horseshoe and
water arrive at equilibrium? Neglect any heating of the bucket and assume the
specific heat of iron is K)J/(kg460 ⋅ .
Picture the Problem During this process the water will gain energy at the
expense of the horseshoe. We can use conservation of energy to find the
equilibrium temperature. See Table 18-1 for the specific heat of water.

Chapter 18366
Apply conservation of energy to obtain:
0horseshoethecoolwaterthewarm
i
i =+=∑ QQQ
or
( ) ( ) 0C650C0.10 fFeFefwaterwater =°−+°− tcmtcm
Solve for tf to obtain:
( ) ( )
FeFewaterwater
FeFewaterwater
f
C650C0.10
cmcm
cmcm
t
+
°+°
=
Substitute numerical values and evaluate tf:
( ) ( ) ( ) ( )
( ) ( )
C1.12
Kkg
kJ
460.0kg750.0
Kkg
kJ
184.4kg0.25
C650
Kkg
kJ
460.0kg750.0C0.10
Kkg
kJ
184.4kg0.25
f
°=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
°⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
+°⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
=t
37 •• A 200-g piece of ice at 0ºC is placed in 500 g of water at 20ºC. This
system is in a container of negligible heat capacity and is insulated from its
surroundings. (a) What is the final equilibrium temperature of the system?
(b) How much of the ice melts?
Picture the Problem Because we can not tell, without performing a couple of
calculations, whether there is enough heat available in the 500 g of water to melt
all of the ice, we’ll need to resolve this question first. See Tables 18-1 and 18-2
for specific heats and the latent heat of fusion of water.
(a) Determine the energy required to
melt 200 g of ice: ( )
kJ70.66
kg
kJ
5.333kg0.200ficeicemelt
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
== LmQ
The energy available from 500 g of water at 20ºC is:
( ) ( )
kJ84.14
C02C0
Kkg
kJ
184.4kg0.500Δ waterwaterwatermaxavailable,
−=
°−°⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
== TcmQ
Because maxavailable,Q < :icemeltQ C.0isretemperatufinalThe °

(b) Equate the energy available from
the water maxavailable,Q to miceLf and
solve for mice to obtain:
f
maxavailable,
ice
L
Q
m =
Substitute numerical values and
evaluate mice:
g125
kg
kJ
5.333
kJ84.14
ice ==m
43 •• A 100-g piece of copper is heated in a furnace to a temperature tC. The
copper is then inserted into a 150-g copper calorimeter containing 200 g of water.
The initial temperature of the water and calorimeter is 16.0ºC, and the
temperature after equilibrium is established is 38.0ºC. When the calorimeter and
its contents are weighed, 1.20 g of water are found to have evaporated. What was
the temperature tC?
Picture the Problem We can find the temperature t by applying conservation of
energy to this calorimetry problem. See Tables 18-1 and 18-2 for specific heats
and the heat of vaporization of water.
Use conservation of energy to obtain:
0
sampleCu
thecool
rcalorimete
thewarm
waterthe
warm
water
vaporize
i
i =+++=∑ QQQQQ
or
0ΔΔΔ CuCuCuwcalcalOHOHOHwf,vaporizedO,H 2222
=+++ TcmTcmTcmLm
Substituting numerical values yields:
( ) ( ) ( )
( ) ( ) ( ) ( ) 0C0.38
Kkg
kJ
386.0g100C0.16C0.38
Kkg
kJ
386.0g150
C0.16C0.38
Kkg
kJ
184.4g200
Kkg
kJ
2257g1.20
C =−°⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
+°−°⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
+
°−°⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
t
Solving for tC yields: C618C °=t
Work and the PV Diagram for a Gas
51 • The gas is first cooled at constant volume until it reaches its final
pressure. It is then allowed to expand at constant pressure until it reaches its final
volume. (a) Illustrate this process on a PV diagram and calculate the work done
by the gas. (b) Find the heat absorbed by the gas during this process.

Chapter 18368
Picture the Problem We can find the work done by the gas during this process
from the area under the curve. Because no work is done along the constant
volume (vertical) part of the path, the work done by the gas is done during its
isobaric expansion. We can then use the first law of thermodynamics to find the
heat absorbed by the gas during this process
(a) The path from the initial state (1)
to the final state (2) is shown on the
PV diagram.
2
0
1
0
P, atm
3.00
2.00
1.00
1.00 2.00 3.00
V, L
The work done by the gas equals the area under the curve:
( )( )
J054
L
m10
L2.00
atm
kPa101.325
atm2.00L2.00atm2.00Δ
33
gasby
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×⎟
⎠
⎞
⎜
⎝
⎛
×===
−
VPW
(b) The work done by the gas is the
negative of the work done on the
gas. Apply the first law of
thermodynamics to the system to
obtain:
( ) ( )
( ) gasbyint,1int,2
gasbyint,1int,2
onintin
WEE
WEE
WEQ
+−=
−−−=
−Δ=
evaluate Qin:
( ) J618J054J456J912in =+−=Q
57 •• An ideal gas initially at 20ºC and 200 kPa has a volume of 4.00 L. It
undergoes a quasi-static, isothermal expansion until its pressure is reduced to
100 kPa. Find (a) the work done by the gas, and (b) the heat absorbed by the gas
during the expansion.

Picture the Problem The PV diagram
shows the isothermal expansion of the
ideal gas from its initial state 1 to its
final state 2. We can use the ideal-gas
law for a fixed amount of gas to find
V2 and then evaluate for an
isothermal process to find the work
done by the gas. In Part (b) of the
problem we can apply the first law of
thermodynamics to find the heat added
to the gas during the expansion.
∫PdV
200
100
2
1
V2
P, kPa
293 K
4.00
V, L
(a) Express the work done by a gas
during an isothermal process: ∫∫∫ ===
2
1
2
1
2
1
11gasby
V
V
V
V
V
V
V
dV
VP
V
dV
nRTdVPW
Apply the ideal-gas law for a fixed
amount of gas undergoing an
isothermal process:
2211 VPVP = ⇒ 1
2
1
2 V
P
P
V =
evaluate V2:
( ) L00.8L4.00
kPa100
kPa200
2 ==V
Substitute numerical values and evaluate W:
( )( ) ( )[ ]
( )
J555
L
m10
LkPa5.554
L4.00
L00.8
lnLkPa800
lnLkPa800L00.4kPa200
33
L8.00
L00.4
L8.00
L00.4
gasby
=
×⋅=⎟
⎠
⎞
⎜
⎝
⎛
⋅=
⋅==
−
∫ V
V
dV
W
(b) Apply the first law of
thermodynamics to the system to
obtain:
onintin WEQ −Δ=
or, because ΔEint = 0 for an isothermal
process,
onin WQ −=
Because the work done by the gas is
the negative of the work done on the
gas:
( ) J555gasbygasbyin ==−−= WWQ

Chapter 18370
Remarks: in an isothermal expansion the heat added to the gas is always
equal to the work done by the gas (ΔEint = 0).
Heat Capacities of Gases and the Equipartition Theorem
59 •• The heat capacity at constant pressure of a certain amount of a
diatomic gas is 14.4 J/K. (a) Find the number of moles of the gas. (b) What is the
internal energy of the gas at T = 300 K? (c) What is the molar heat capacity of this
gas at constant volume? (d) What is the heat capacity of this gas at constant
volume?
Picture the Problem (a) The number of moles of the gas is related to its heat
capacity at constant pressure and its molar heat capacity at constant pressure
according to . For a diatomic gas, the molar heat capacity at constant
pressure is
PP nc'C =
Rc' 2
7
P = . (b) The internal energy of a gas depends on its number of
degrees of freedom and, for a diatomic gas, is given by nRTE 2
5
int = . (c) The
molar heat capacity of this gas at constant volume is related to its molar heat
capacity at constant pressure according to Rc'c' −= PV . (d) The heat capacity of
this gas at constant volume is the product of the number of moles in the gas and
its molar heat capacity at constant volume.
(a) The number of moles of the gas
is the ratio of its heat capacity at
constant pressure to its molar heat
capacity at constant pressure:
P
P
c'
C
n =
For a diatomic gas, the molar heat
capacity is given by: Kmol
J
1.292
7
P
⋅
== Rc'
evaluate n:
mol495.0
mol4948.0
Kmol
J
1.29
K
J
4.14
=
=
⋅
=n
(b) With 5 degrees of freedom at this
temperature:
nRTE 2
5
int =

Substitute numerical values and evaluate Eint:
( ) ( ) kJ09.3K300
Kmol
J
314.8mol4948.02
5
int =⎟
⎠
⎞
⎜
⎝
⎛
⋅
=E
(c) The molar heat capacity of this
gas at constant volume is the
difference between the molar heat
capacity at constant pressure and the
gas constant R:
Rc'c' −= PV
Because Rc' 2
7
P = for a diatomic gas: RRRc' 2
5
2
7
V =−=
Substitute the numerical value of R
to obtain:
Kmol
J
8.20
Kmol
J
79.20
Kmol
J
314.82
5
V
⋅
=
⋅
=⎟
⎠
⎞
⎜
⎝
⎛
⋅
=c'
(d) The heat capacity of this gas at
constant volume is given by:
VV nc'C' =
evaluate :VC' ( )
K
J
3.10
Kmol
J
79.20mol4948.0V
=
⎟
⎠
⎞
⎜
⎝
⎛
⋅
=C'
65 •• Carbon dioxide (CO2) at a pressure of 1.00 atm and a temperature of
–78.5ºC sublimates directly from a solid to a gaseous state without going through
a liquid phase. What is the change in the heat capacity at constant pressure per
mole of CO2 when it undergoes sublimation? (Assume that the gas molecules can
rotate but do not vibrate.) Is the change in the heat capacity positive or negative
during sublimation? The CO2 molecule is pictured in Figure 18-22.
Picture the Problem We can find the change in the heat capacity at constant
pressure as CO2 undergoes sublimation from the energy per molecule of CO2 in
the solid and gaseous states.
Express the change in the heat
capacity (at constant pressure) per
mole as the CO2 undergoes
sublimation:
solidP,gasP,P CCC −=Δ (1)

Chapter 18372
Express Cp,gas in terms of the number
of degrees of freedom per molecule:
( ) NkNkfC 2
5
2
1
gasP, ==
because each molecule has three
translational and two rotational degrees
of freedom in the gaseous state.
We know, from the Dulong-Petit
Law, that the molar specific heat of
most solids is 3R = 3Nk. This result
is essentially a per-atom result as it
was obtained for a monatomic solid
with six degrees of freedom. Use this
result and the fact CO2 is triatomic to
express CP,solid:
Nk
Nk
C 9atoms3
atom
3
solidP, =×=
Substitute in equation (1) to obtain: NkNkNkC 2
13
2
18
2
5
PΔ −=−=
Quasi-Static Adiabatic Expansion of a Gas
69 •• A 0.500-mol sample of an ideal monatomic gas at 400 kPa and 300 K,
expands quasi-statically until the pressure decreases to 160 kPa. Find the final
temperature and volume of the gas, the work done by the gas, and the heat
absorbed by the gas if the expansion is (a) isothermal and (b) adiabatic.
Picture the Problem We can use the ideal-gas law to find the initial volume of
the gas. In Part (a) we can apply the ideal-gas law for a fixed amount of gas to
find the final volume and the expression for the work done in an isothermal
process. Application of the first law of thermodynamics will allow us to find the
heat absorbed by the gas during this process. In Part (b) we can use the
relationship between the pressures and volumes for a quasi-static adiabatic
process to find the final volume of the gas. We can apply the ideal-gas law to find
the final temperature and, as in (a), apply the first law of thermodynamics, this
time to find the work done by the gas.
Use the ideal-gas law to express the
initial volume of the gas: i
i
i
P
nRT
V =
Substitute numerical values and evaluate Vi:
( ) ( )
33
i m103.118
kPa400
K300
Kmol
J
8.314mol0.500
−
×=
⎟
⎠
⎞
⎜
⎝
⎛
⋅
=V
(a) Because the process is isothermal: K300if == TT

Use the ideal-gas law for a fixed
amount of gas with T constant to
express Vf:
ffi VPVP = ⇒
f
i
if
P
P
VV =
evaluate Vf:
( )
L7.80
L795.7
kPa160
kPa400
L3.118f
=
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=V
Express the work done by the gas
during the isothermal expansion: i
f
gasby ln
V
V
nRTW =
evaluate Wby gas:
( )
( )
kJ14.1
L3.118
L7.795
lnK300
Kmol
J
8.314mol0.500gasby
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×
⎟
⎠
⎞
⎜
⎝
⎛
⋅
=W
Noting that the work done by the gas
during the process equals the
negative of the work done on the
gas, apply the first law of
thermodynamics to find the heat
absorbed by the gas:
( )
kJ1.14
kJ1.140onintin
=
−−=−Δ= WEQ
(b) Using γ = 5/3 and the relationship
between the pressures and volumes
for a quasi-static adiabatic process,
express Vf:
γγ
ffii VPVP = ⇒
γ1
f
i
if ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
P
P
VV
evaluate Vf: ( )
L5.40
L403.5
kPa160
kPa400
L118.3
53
f
=
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=V
Apply the ideal-gas law to find the
final temperature of the gas: nR
VP
T ff
f =

Chapter 18374
evaluate Tf:
( )( )
( )
K208
Kmol
J
8.314mol0.500
m105.403kPa160 33
f
=
⎟
⎠
⎞
⎜
⎝
⎛
⋅
×
=
−
T
For an adiabatic process: 0in =Q
Apply the first law of thermodynamics
to express the work done on the gas
during the adiabatic process:
TnRTCQEW Δ=−Δ=−Δ= 2
3
Vininton 0
evaluate Won:
( )( )
( )
J745
K300K208
KJ/mol8.314mol0.5002
3
on
−=
−×
⋅=W
Because the work done by the gas
equals the negative of the work done
on the gas:
( ) J745J574gasby =−−=W
Cyclic Processes
73 •• A 1.00-mol sample of an ideal diatomic gas is allowed to expand. This
expansion is represented by the straight line from 1 to 2 in the PV diagram (Figure
18-23). The gas is then compressed isothermally. This compression is represented
by the straight line from 2 to 1 in the PV diagram. Calculate the work per cycle
done by the gas.
Picture the Problem The total work done as the gas is taken through this cycle is
the area bounded by the two processes. Because the process from 1→2 is linear,
we can use the formula for the area of a trapezoid to find the work done during
this expansion. We can use ( )ifprocessisothermal ln VVnRTW = to find the work done
on the gas during the process 2→1. The net work done during this cycle is then
the sum of these two terms.
Express the net work done per cycle:
1221
gason thegasby thenet
→→ +=
+=
WW
WWW
(1)

Work is done by the gas during its
expansion from 1 to 2 and hence is
equal to the negative of the area of
the trapezoid defined by this path
and the vertical lines at V1 = 11.5 L
and V2 = 23 L. Use the formula for
the area of a trapezoid to express
W1→2:
( )
( )
atmL3.17
atm1.0atm2.0
L11.5L232
1
trap21
⋅−=
+×
−−=
−=→ AW
Work is done on the gas during the
isothermal compression from V2 to
V1 and hence is equal to the area
under the curve representing this
process. Use the expression for the
work done during an isothermal
process to express W2→1:
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=→
i
f
12 ln
V
V
nRTW
Apply the ideal-gas law at point 1 to find the temperature along the isotherm
2→1:
( )( )
( )( ) K280
Katm/molL10206.8mol1.00
L5.11atm0.2
2
=
⋅⋅×
== −
nR
PV
T
Substitute numerical values and evaluate W2→1:
( )( )( ) atmL9.15
L23
L5.11
lnK280Katm/molL10206.8mol00.1 2
12 ⋅=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅⋅×= −
→W
Substitute in equation (1) and
evaluate Wnet:
kJ14.0
atmL
J101.325
atmL40.1
atmL15.9atmL3.17net
−=
⋅
×⋅−=
⋅+⋅−=W
75 ••• At point D in Figure 18-24 the pressure and temperature of 2.00 mol of
an ideal monatomic gas are 2.00 atm and 360 K, respectively. The volume of the
gas at point B on the PV diagram is three times that at point D and its pressure is
twice that at point C. Paths AB and CD represent isothermal processes. The gas is
carried through a complete cycle along the path DABCD. Determine the total
amount of work done by the gas and the heat absorbed by the gas along each
portion of the cycle.

Chapter 18376
Picture the Problem We can find the temperatures, pressures, and volumes at all
points for this ideal monatomic gas (3 degrees of freedom) using the ideal-gas law
and the work for each process by finding the areas under each curve. We can find
the heat exchanged for each process from the heat capacities and the initial and
final temperatures for each process.
Express the total work done by the
gas per cycle:
DCCBBAADtotgas,by →→→→ +++= WWWWW
1. Use the ideal-gas law to find the
volume of the gas at point D:
( ) ( )
( )
L29.54
atm
kPa
101.325atm2.00
K360
Kmol
J
8.314mol2.00
D
D
D
=
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⋅
=
=
P
nRT
V
2. We’re given that the volume of
the gas at point B is three times
that at point D:
( )
L62.88
L54.2933 DCB
=
=== VVV
Use the ideal-gas law to find the pressure of the gas at point C:
( ) ( )
atm6667.0
L62.88
K360
Kmol
atmL
10206.8mol2.00 2
C
C
C =
⎟
⎠
⎞
⎜
⎝
⎛
⋅
⋅
×
==
−
V
nRT
P
We’re given that the pressure at
point B is twice that at point C:
( ) atm333.1atm6667.022 CB === PP
3. Because path DC represents an
isothermal process:
K360CD == TT
Use the ideal-gas law to find the
temperatures at points B and A:
( )( )
( )
K719.8
Kmol
atmL
108.206mol2.00
L88.62atm1.333
2
BB
BA
=
⎟
⎠
⎞
⎜
⎝
⎛
⋅
⋅
×
=
==
−
nR
VP
TT

Because the temperature at point A is
twice that at D and the volumes are
the same, we can conclude that:
atm00.42 DA == PP
The pressure, volume, and
temperature at points A, B, C, and D
are summarized in the table to the
right.
Point P V T
(atm) (L) (K)
A 4.00 29.5 720
B 1.33 88.6 720
C 0.667 88.6 360
D 2.00 29.5 360
4. For the path D→A, 0AD =→W
and: ( )DA2
3
AD2
3
ADint,AD
TTnR
TnREQ
−=
Δ=Δ= →→→
Substitute numerical values and evaluate QD→A:
( ) ( ) kJ979.8K360K720
Kmol
J
8.314mol2.002
3
AD =−⎟
⎠
⎞
⎜
⎝
⎛
⋅
=→Q
For the path A→B:
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
== →→
A
B
BA,BABA ln
V
V
nRTQW
Substitute numerical values and evaluate WA→B:
( ) ( ) kJ15.13
L29.54
L88.62
lnK720
Kmol
J
8.314mol2.00BA =⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⋅
=→W
and, because process A→B is isothermal, 0BAint, =Δ →E
For the path B→C, , and:0CB =→W
( )BC2
3
VCBCB ΔΔ
TTnR
TCUQ
−=
== →→
Substitute numerical values and evaluate QB→C:
( )( )( ) kJ979.8K720K360KJ/mol8.314mol2.002
3
CB −=−⋅=→Q
For the path C→D:
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=→
C
D
DC,DC ln
V
V
nRTW

Chapter 18378
Substitute numerical values and evaluate WC→D:
( ) ( ) kJ576.6
L62.88
L54.92
lnK603
Kmol
J
8.314mol2.00DC −=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⋅
=→W
Also, because process A→B is isothermal, 0BAint, =Δ →E , and
kJ58.6DCDC −== →→ WQ
Qin, Won, and ΔEint are summarized
for each of the processes in the table
to the right.
Process Qin Won ΔEint
(kJ) (kJ) (kJ)
D→A 98.8 0 8.98
A→B 2.13 −13.2 0
B→C 98.8− 0 −8.98
C→D 58.6− 6.58 0
Referring to the table, find the total work done by the gas per cycle:
kJ6.6
kJ6.580kJ13.20
DCCBBAADtotgas,by
=
−++=
+++= →→→→ WWWWW
Remarks: Note that, as it should be, ΔEint is zero for the complete cycle.
General Problems
79 • The PV diagram in Figure 18-25 represents 3.00 mol of an ideal
monatomic gas. The gas is initially at point A. The paths AD and BC represent
isothermal changes. If the system is brought to point C along the path AEC, find
(a) the initial and final temperatures of the gas, (b) the work done by the gas, and
(c) the heat absorbed by the gas.
Picture the Problem We can use the ideal-gas law to find the temperatures TA
and TC. Because the process EDC is isobaric, we can find the area under this line
geometrically and then use the 1st law of thermodynamics to find QAEC.

(a) Using the ideal-gas law, find the
temperature at point A:
( )( )
( )
K65K65.2
Kmol
atmL
108.206mol3.00
L4.01atm4.0
2
AA
A
==
⎟
⎠
⎞
⎜
⎝
⎛
⋅
⋅
×
=
=
−
nR
VP
T
Using the ideal-gas law, find the
temperature at point C:
( )( )
( )
K81K81.2
Kmol
atmL
108.206mol3.00
L0.02atm1.0
2
CC
C
==
⎟
⎠
⎞
⎜
⎝
⎛
⋅
⋅
×
=
=
−
nR
VP
T
(b) Express the work done by the gas
along the path AEC: ( )( )
kJ1.6kJ1.62
atmL
J101.325
atmL15.99
L4.01L20.0atm1.0
Δ0 ECECECAEAEC
==
⋅
×⋅=
−=
+=+= VPWWW
(c) Apply the first law of
thermodynamics to express QAEC:
( )ATTnRW
TnRW
TCWEWQ
−+=
Δ+=
Δ+=Δ+=
C2
3
AEC
2
3
AEC
VAECintAECAEC
Substitute numerical values and evaluate QAEC:
( )( )( ) kJ2.2K65.2K81.2KJ/mol8.314mol3.00kJ1.62 2
3
AEC =−⋅+=Q
Remarks The difference between WAEC and QAEC is the change in the internal
energy ΔEint,AEC during this process.
83 •• As part of a laboratory experiment, you test the calorie content of
various foods. Assume that when you eat these foods, 100% of the energy
released by the foods is absorbed by your body. Suppose you burn a 2.50-g
potato chip, and the resulting flame warms a small aluminum can of water. After
burning the potato chip, you measure its mass to be 2.20 g. The mass of the can is
25.0 g, and the volume of water contained in the can is 15.0 ml. If the
temperature increase in the water is 12.5°C, how many kilocalories

Chapter 18380
(1 kcal = 1 dietary calorie) per 150-g serving of these potato chips would you
estimate there are? Assume the can of water captures 50.0 percent of the heat
released during the burning of the potato chip. Note: Although the joule is the SI
unit of choice in most thermodynamic situations, the food industry in the United
States currently expresses the energy released during metabolism in terms of the
″dietary calorie,″ which is our kilocalorie.
Picture the Problem The ratio of the energy in a 150-g serving to the energy in
0.30 g of potato chip is the same as the ratio of the masses of the serving and the
amount of the chip burned while heating the aluminum can and the water in it.
The ratio of the energy in a
150-g serving to the
energy in 0.30 g of potato
chip is the same as the
ratio of the masses of the
serving and the amount of
the chip burned while
heating the aluminum can
and the water in it:
500
g0.30
g150
g0.30
servingg-150
==
Q
Q
or
g0.30servingg150 500QQ =
Letting f represent the
fraction of the heat
captured by the can of
water, express the energy
transferred to the
aluminum can and the
water in it during the
burning of the potato chip:
( ) Tcmcm
TcmTcm
QQfQ
Δ
ΔΔ
OHOHAlAl
OHOHAlAl
OHAlg0.30
22
22
2
+=
+=
+=
where ΔT is the common temperature change of
the aluminum cup and the water it contains.
Substituting for
yields and solving
for yields:
g0.30Q
servingg-150Q
( )
f
Tcmcm
Q
Δ500 OHOHAlAl
servingg-150
22
+
=
Substitute numerical values and evaluate :servingg-150Q
( ) ( ) ( )
kcal652cal10256
J4.184
cal1
J1007.1
500.0
C5.12
Kkg
kJ
184.4kg0150.0
Kkg
kJ
900.0kg0250.0500
36
servingg-150
≈×=××=
°⎥
⎦
⎤
⎢
⎣
⎡
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
=Q
89 •• A thermally insulated system consists of 1.00 mol of a diatomic gas at
100 K and 2.00 mol of a solid at 200 K that are separated by a rigid insulating

wall. Find the equilibrium temperature of the system after the insulating wall is
removed, assuming that the gas obeys the ideal-gas law and that the solid obeys
the Dulong–Petit law.
Picture the Problem We can use conservation of energy to relate the equilibrium
temperature to the heat capacities of the gas and the solid. We can apply the
Dulong-Petit law to find the heat capacity of the solid at constant volume and use
the fact that the gas is diatomic to find its heat capacity at constant volume.
Apply conservation of energy to this
process:
0Δ solidgas =+= QQQ
Use to substitute for QTCQ ΔV= gas and Qsolid:
( ) ( ) 0K200K100 equilsolidV,equilgasV, =−+− TCTC
Solving for Tequil yields: ( )( ) ( )( )
solidV,gasV,
solidV,gasV,
equil
K200K100
CC
CC
T
+
+
=
Using the Dulong-Petit law,
determine the heat capacity of the
solid at constant volume:
RnC solidsolidV, 3=
The heat capacity of the gas at
constant volume is given by:
RnC gas2
5
gasV, =
Substitute for CV,solid and CV,gas and simplify to obtain:
( )( ) ( )( ) ( )( ) ( )( )
solidgas2
5
solidgas2
5
solidgas2
5
solidgas2
5
equil
3
3K200K100
3
3K200K100
nn
nn
RnRn
RnRn
T
+
+
=
+
+
=
Substitute numerical values for ngas and nsolid and evaluate Tequil:
( )( )( ) ( )( )( )
( ) ( )
K171
mol00.23mol00.1
mol00.23K200mol00.1K100
2
5
2
5
equil =
+
+
=T
95 ••• (a) Use the results of Problem 94 to show that in the limit that
ETT >> , the Einstein model gives the same expression for specific heat that the
Dulong–Petit law does. (b) For diamond, TE is approximately 1060 K. Integrate
numerically dEint = dT to find the increase in the internal energy if 1.00 mol of
diamond is heated from 300 to 600 K.
v
′c

Chapter 18382
Picture the Problem (a) We can rewrite our expression for by dividing its
numerator and denominator by
'
cV
TT
e E
and then using the power series for ex
to
show that, for T > TE, . In Part (b), we can use the result of Problem 94 to
obtain values for every 100 K between 300 K and 600 K and use this data to
find ΔU numerically.
Rc'
3V ≈
'
cV
(a) From Problem 94 we have:
( )2
2
E
V
1
3
E
E
−
⎟
⎠
⎞
⎜
⎝
⎛
=
TT
TT
e
e
T
T
Rc'
Divide the numerator and denominator
by TT
e E
to obtain:
TTTT
TT
TTTT
eeT
T
R
e
eeT
T
Rc'
EE
E
EE
2
1
3
12
1
3
2
E
2
2
E
V
−
+−
⎟
⎠
⎞
⎜
⎝
⎛
=
+−
⎟
⎠
⎞
⎜
⎝
⎛
=
Express the exponential terms in their power series to obtain:
E
2
E
2
EE
2
EE
for
...
2
1
12...
2
1
12 EE
TT
T
T
T
T
T
T
T
T
T
T
ee TTTT
>>⎟
⎠
⎞
⎜
⎝
⎛
≈
+⎟
⎠
⎞
⎜
⎝
⎛
+−+−+⎟
⎠
⎞
⎜
⎝
⎛
++=+− −
Substitute for TTTT
ee EE
2 −
+− to
obtain: R
T
TT
T
Rc' 3
1
3 2
E
2
E
V =
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
≈
(b) Use the result of Problem 94 to verify the following table:
T (K) 300 400 500 600
cV (J/mol⋅K) 9.65 14.33 17.38 19.35

The following graph of specific heat as a function of temperature was plotted
using a spreadsheet program:
5
7
9
11
13
15
17
19
21
300 350 400 450 500 550 600
T (K)
CV(J/mol-K)
Integrate numerically, using the formula for the area of a trapezoid, to obtain:
( )( )( )
( )( )( )
( )( )( )
kJ62.4
Kmol
J
35.1938.17K100mol00.1
Kmol
J
38.1733.14K100mol00.1
Kmol
J
33.1465.9K100mol00.1Δ
2
1
2
1
2
1
=
⋅
++
⋅
++
⋅
+=U

Heat Transfer and Thermodynamics Conceptual Problems

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