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7/27/2019 05 Chapter 5 Notes First Law of Thermodynamics
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Adiabatic work
SystemSystem
Work Transfer
SurroundingsSurroundings
Adiabatic Boundary
4
Equivalence of adiabatic processes
First LawFirst Law-- AllAlladiabatic processesadiabatic processesbetween twobetween two
equilibrium statesequilibrium statesrequire the samerequire the sameadiabatic workadiabatic work to beto be
done on the systemdone on the systemregardless of theregardless of thepathpath
State 1State 1
State 2State 2
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First Law Statement
The adiabatic work done onThe adiabatic work done on
a closed system to connecta closed system to connect
two equilibrium statestwo equilibrium states
is independent of path, thereforeis independent of path, therefore
Identifies a change in a thermodynamicIdentifies a change in a thermodynamic
property called energy.property called energy.
6
The First Law
,1 2 2 1adbyW E E =
Work doneWork doneby (out of)by (out of)the systemthe system
isis
positive by conventionpositive by conventionminusminus
work by is work on (in)work by is work on (in)..
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For a non adiabatic process, the Heat
transfer to the system, Q, is defined by
,1 2 ,1 2
,1 2 2
,1 2
,1 2 1
H eat transfer is def ined asthe d i ffe rence be tw een the w ork byand the ad iaba t ic w ork byfor the sam e chang e in s ta te
ato b yy
o
d b
b yt
Q W W
Q W E E o r
=
= +
8
Define Internal energy, U change between two states at
at the same extrinsic state , i.e. the same velocity andposition but different intrinsic properties.
,1 2 2 1adbyW U U
Work done by the system is positive byWork done by the system is positive by
convention.convention. KEKE11--22 = 0, and= 0, andPEPE11--22= 0= 0
Which are the adiabatic work on the system atWhich are the adiabatic work on the system at
same intrinsic state but different extrinsic state.same intrinsic state but different extrinsic state.
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MotorinW
fromQ
inmass sys out motorsystem
m
W W
PE
=
=
mm
Example 1Example 1
For steady operation, thereFor steady operation, there
is no change of the state ofis no change of the state of
the system, and for the motor systemthe system, and for the motor system
0in out loss
W W Q =
12
Example 2 - unsteady
MechanicalMechanicalWork to RaiseWork to Raise
the Mass, mthe Mass, m
Heat Loss fromHeat Loss fromfrom Motorfrom Motor
Force = mgForce = mgBattery has storedBattery has storedenergy. It is releasedenergy. It is releasedvia electrical current.via electrical current.
MBB
mm
System BoundarySystem Boundary
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Example 3: Q5 Q1 W=Ebattery + EmotorQQ55
To correctly analyze this arrangement, all energy & work transfersTo correctly analyze this arrangement, all energy & work transfers
must be counted. Where is the primary energy source? What aremust be counted. Where is the primary energy source? What are
the possible systems for analysis?the possible systems for analysis?
B
W
Motor
QQ11
QQ22
QQ33
QQ44
Fan
Battery
14
Cyclic Processes
P1
P2
SS11
SS22
21,2121 = byWQdU
1 2 1 1 2 1 1 2 1
0by
dU Q W
= =
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Initial SystemInitial System
State 1State 1
Deformed SystemDeformed System
BoundaryBoundary
State 2
xx
yy
zz P1
1
2
P3
P2
Process PathProcess Path
dW=PdAdN
16
MOVING BOUNDARY WORKReversible-Moving boundary work(P dVwork): The expansion andcompression work in a piston-cylinderdevice. Observer moving with the system
The work associatedwith a moving
boundary is calledboundary work.A gas does a
differentialamount of workWbas it forcesthe piston tomove by adifferentialamount ds.
Quasi-equilibrium process:A process during which the systemremains nearly in equilibrium at alltimes.
Wbis positive for expansionWbis negative for compression
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The area under the processcurve on a P-Vdiagramrepresents the boundary work.
The boundarywork done
during a processdepends on the
path followed aswell as the end
states.
The net work doneduring a cycle is thedifference between
the work done bythe system and the
work done on thesystem.
18
Polytropic, Isothermal, and Isobaric processes
Polytropic process: C, n(polytropic exponent) constants
N=1Polytropicprocess
Polytropic - for ideal gas
When n= 1(represents anisothermal process
for-ig )
Schematic andP-Vdiagram for
a polytropicprocess.
Constant pressure process
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ENERGY BALANCE FOR CLOSED SYSTEMS
Energy balance for any systemundergoing any process
Energy balancein the rate form
The total quantities are related to the quantities per unit time is
Energy balance per unit mass basiswhere represents an amount of
rather than a change of.Energy balance indifferential form
Energy balancefor a cycle
in by systemq w de =
20
Equations
In solving thermodynamics problems
Define the problem
Examine data. Is the substance known? Are two intensive propertiesidentified at the each end state?
Can the work be evaluated or is it known ?
Can heat transfer be evaluated or is it known?
Write the energy equation in form appropriate to the system. What modeling approximations can be made or need to be made?
Write the conservation of mass equation, and if possible relate to thegiven data.
Are there a sufficient number of equations for the # of unknowns?
In some cases additional fundamental laws or particular equations mustbe applied- i.e. equation of state ( or equivalent use of tables) andfurther modeling assumptions made.
Verify or check validity of assumptions, i.e. iterate.
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Scientific Method- eight step format
1. Define the problem
2. Collect the data.
3. Analyze the data
4. Form an hypothesis
5. Test the hypothesis
6. Analyze the results
7. Accept or reject the hypothesis
8. If necessary, iterate
22
net-in net-out system
Net energy transfer Change in internal, kinetic,by heat, work potential, etc. energies
in ,ou 2 1
in ,ou 2 1
( ) (since KE=PE=0)( )
b t
b t
Q W E
Q W U m u uQ W m u u
=
= =
= +
P
v
1
2300 kPa
75 kPa
45 A saturated water mixture at 75 kPa , 13 % quality with an initial volume of 2 m3 is contained in a
linear spring-loaded piston-cylinder device It is heated until the pressure is 300 kPa and volumeincreases to 5 m3. The heat transfer and the work done are to be determined.
Assumptions1 The cylinder is stationary and thus the kinetic and potential energy changes are
zero. 2 There are no work interactions involved other than the boundary work. 3 The thermal energystored in the cylinder itself is negligible. 4 The compression or expansion process is quasi-
equilibrium.
Analysis We take the contents of the cylinder as the system. This is a closed
system since no mass enters or leaves. The energy balance for this stationary
closed system can be expressed as
Peq = A + B Vol
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The initial state is saturated mixture at 75 kPa. The specific volume and internal energy at this state are (Table B1.2),
kJ/kg89.658)8.2111)(13.0(36.384
/kgm28914.0)001037.02172.2)(13.0(001037.0
1
31
=+=+=
=+=+=
fgf
fgf
xuuu
xvvv
The mass of water is
kg9170.6/kgm28914.0
m23
3
1
1===
v
Vm The final specific volume is
/kgm72285.0kg9170.6
m5 33
22 ===
m
Vv
The final state is now fixed. v> v g The internal energy at this specific volume and 300 kPa pressure is(Table B1.3)
kJ/kg2.26572 =u
1 2 3,out 2 1
3
With a linear spring, the quasi-equilibrium P varies linearly with the volume, thus(75 300)kPa 1 kJ
( ) (5 2)m2 2 1 kPa m
b
P PW PdVol Vol Vol
+ + = = = =
562.5 kJ
Substituting into energy balance equation gives
kJ14,385=+=+= kJ/kg)89.658kg)(2657.29170.6(kJ5.562)( 12out,in uumWQ b
24
Energy balance when sign convention is used (i.e., heat input and workoutput are positive; heat output and work input are negative).
Various forms of the first-law relationfor closed systems when signconvention is used.
For a cycle E =0, thus Q = W.
The first law cannot be proven mathematically, but no process in nature is knownto have violated the first law, and thus until a violation of the law is observed, itshould be taken as valid.
Energy EQ Using Sign General Convention
of 1st LawGeneral finite -period-
: /
FormsQ W E
Rate form Eq Q W DE Dt
NeglectingKE PE Q W Umass base q w eDifferentialmassbase q w de
=
=
=
=
=
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TANK A2 kg
1 MPa300C
TANK B3 kg
150Cx=0.5
Q
42 Two tanks initially separated by a partition contain steam at different states. Now the partition isremoved and they are allowed to mix until equilibrium is established. If the final pressure is 300 kPa ,
the temperature and quality of the steam at the final state and the amount of heat lost from the tanks
are to be determined.
Assumptions1 The tank is stationary and thus the kinetic and potential energy changes arezero. 2 There are no work interactions.
Analysis(a) We take the contents of both tanks as the system. This is a closed system since no
mass enters or leaves. Noting that the volume of the system is constant and thus there is noboundary work, the energy balance for this stationary closed system can be expressed as
The properties of steam in both tanks at the initial state are (Tables B1.1 through B1.3)kJ/kg7.2793
/kgm25799.0
C300
kPa1000
,1
3
,1
,1
,1
=
=
=
=
A
A
A
A
uT
P v
( )[ ]
( ) kJ/kg4.15954.19270.50.66631
/kgm0.196790.0010910.392480.500.001091
kJ/kg4.1927,66.631
/kgm.392480,001091.0
50.0
C150
1,1
31,1
3
1
,1
=+=+=
=+=+=
==
==
=
=
fgfB
fgfB
fgf
gfB
uxuu
x
uux
T
vvv
vv
[ ] [ ]
in out system system
Net energy transfer Change in internal, kinetic,by heat, work potential, etc. energies
2
out 2 1 2 1
[ ) ) ]( ) ( )
(with
A B
A B
A B A B
Q W E U
m m mU mu mu muQ U U U m u u m u u
W
= =
= +
= +
= = + = +
KE=PE=0)=
26
Two intensive properties are required to define final state. Assume the fina l state is an equilibrium state.The total volume and total mass of the system are
kg523
m106.1/kg)m19679.0kg)(3(/kg)m25799.0kg)(2( 333,1,1
=+=+=
=+=+=+=
BA
BBAABA
mmm
mm vvVVV
Now, the specific volume at the final state may be determined
/kgm22127.0kg5
m106.1 33
2 ===m
Vv
which fixes the final state and we can determine other properties. Table B1.2 SinceVf
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Energy balance for a constant-pressure
expansion or compression process
HWU b =+
For a constant-pressure expansionor compression process:
An example of constant-pressure, Pequ,bdy condition-not reversible-no piston-friction
General analysis for a closed systemundergoing a quasi-equilibriumconstant-pressure process. Qis tothesystem and Wis fromthe system.
Wb
Weee0.3A
Since frictional effects neg
28
Net out system
Net energy transfer Change in internal, kinetic,by heat, work potential, etc. energies
e,in pw,in
e,in pw,in 2 1
(since =KE=PE=0)
( )
(
NetoutQ W E
W W PAdh U Q
W W m h h
I t
=
+ =
+ =
Volt pw,in 2 1) ( )W m h h+ =
H2OPequil.
WpwWe
v
P
1 2
----39393939 An insulated cylinder is initially filled with 5 L ofsaturated liquid water at a pressure of175 kPa. The water is heated electrically with a
current of8A for 45 min through a resistor as it is stirred by a paddle-wheel with constant pressureboundary condition. Ifone half of the
liquid is evaporated during this process and the paddle wheel work is 400 kJ, determine the voltage of the current source . Show the
process on a P-vdiagram.
AssumptionsAssumptionsAssumptionsAssumptions1111 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2222 The cylinder is well-insulated and
thus heat transfer is negligible. 3333 The thermal energy stored in the cylinder itself is negligible. 4444 The piston moves with no friction and
negligible acceleration.
AnalysisAnalysisAnalysisAnalysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance
for this stationary closed system can be expressed as
U+ Wb = HSince the initial and final P is the same a nd the endstates are-equilibrium states. The properties of water ar e (Tables
B1.1through B1.3)
( )
kg4.731/kgm0.001057
m0.005
kJ/kg1593.61.22135.001.4875.0
kPa175
/kgm0.001057
kJ/kg487.01
liquidsat.
kPa175
3
3
1
1
222
2
3kPa175@1
kPa175@11
===
=+=+=
=
=
==
==
=
v
V
vv
m
hxhhx
P
hhP
fgf
f
f
Substituting,
(400kJ) (4.731 kg)(1593.6 487.01)kJ/kg
4835 kJ
4835 kJ 1000 VA
(8 A)(45 60 s) 1 kJ/s
elect
elect
W
I t
+ =
=
= =
W = V
V 223.9 V
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Steam0.15 kg3.5 MPa Q
154 A piston-cylinder device contains 0.15 kg steam at 3.5 MPa , superheated by 5 C in a cylinder with a set of stops.
As Heat is lost from the steam, the piston moves down and hit the stops at which point the state is saturated liquid. It is
cooled further until a temperature of200 C is reached. The final pressure and quality (if mixture), the boundary work,
and the heat transfer until the piston first hits the stops and the total heat transfer are to be determined.
Assumptions1 The kinetic and potential energy changes are negligible,. 2 The friction between the piston and the cylinder is negligible.
Analysis(a) We take the steam in the cy linder to be the system. This is a closed system since no masscrosses the system boundary. The energy balance for this stationary closed system can be ex pressed as
net-in net-out system system
Net energy transfer Change in internal, kinetic,by heat, work potential, etc. energies
b,in out (since KE PE 0)
Q W E U
W Q U
= =
= = =
Denoting when piston first hits the stops as state (2) and the final state as (3), the energy balance relations may be written as
30
)u-(
)u-(
133-out,1inb,
122-out,1inb,
umQW
umQW
=
=
The properties of steam at various states are (Tables B1.1 through B 1.3)[email protected] MPa
1
242.56 C
242.56 5 247.56 Csat
T
T T dSH
=
= + = + =
kJ/kg3.2617
/kgm05821.0
C56.247
MPa5.3
1
31
1
1
=
=
=
=
uT
P v
kJ/kg4.1045
/kgm001235.0
0
MPa5.3
2
32
2
12
=
=
=
==
ux
PP v
kJ/kg55.851C200
/kgm001235.0
3
3
3
3
323
=
=
=
=
==
u
P
x
TkPa1555
0.00062vv
(b) Noting that the pressure is constant until the piston hits the stops during which the boundary work is done, it can be determined from
its definition as
kJ29.91=== 3211inb, 0.001235)m21kPa)(0.058kg)(350015.0()( vvmPW
(c) Substituting into energy balance relations, yields Q out 1-2 = m (h1h2 ) or
kJ265.7== kJ/kg)3.26174.1045(kg)15.0(kJ91.292-out,1Q
(d) kJ294.8== kJ/kg)3.261755.851(kg)15.0(kJ91.293-out,1Q
Specific volume comparison
out,1-2 2 1 b,in,1-3
out,1-3 3 1 b,in,1-3 out,1-2 out,2-3
( -u )+( -u )+
Q m u W
Q m u W Q Q=
= = +
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SPECIFIC HEATS- properties of a substance
Specific heat at constant volume, cv: Informal definition-The energyrequired to raise the temperature of the unit mass of a substance byone degree as the volume is maintained constant.
Specific heat at constant pressure, cp: Informal definition-The energyrequired to raise the temperature of the unit mass of a substance byone degree as the pressure is maintained constant.
Specific heat is a propertythat can be qrev/dT as the
energy required to raise thetemperature of a unit massof a substance by onedegree in a specified way.
Constant-
volume andconstant-
pressure specificheats cvand cp(values are for
helium gas).
32
The equations in the figure are valid for anysubstance undergoing anyprocess.
cvand cpare properties.
cvis related to the changes in internal energyand cpto the changes inenthalpy.
A common unit for specific heats is kJ/kg C or k J/kg K. Are these unitsidentical?
The specific heat of a substancechanges with temperature.
cpis always greater orequal to cv.
Formal definitions of cvand cp.
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If the heat transfer results in changes of the translational ke of the molecule, thespecific heat will remain constant; but if some of the input energy is diverted to the
rotational or viberational modes than the specific heat will increase, pjf
33
v1
v2
U(T,v)
T
T
u
34
INTERNAL ENERGY, ENTHALPY,AND SPECIFIC HEATS OF IDEAL GASES
For ideal gases, u,h, cv, and cpvarywith temperatureonly(Tables--.
Internal energy andenthalpy change of
an ideal gas
General internal energy relation for apure , simple compressible substance
( , )
/ [ / ]v v
u T v
du u T dT T P T P dv
=
= +
Molecular-kinetic + potential
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Specific heats
Quantities are defined that are related to the variables applicable toreversible work.
The specific heat at constant volume for a simple compressible sub isdefined to be cv = u/T)v at constant displacement (volume).
For a reversible constant volume process this is equivalent to c v = qrev/dT)v
Another specific heat is defined in terms of the Thermodynamic Force
36
Specific heat at const force
The specific heat at constant pressure for a scs is defined as cP = h/T)P ,at const
Thermo force. [h=u +p v>u]
For a reversible process,
this is equivalent to cP = qrev /T)P .
For an ideal gas u and h are functions only of temperature. Thus, so to are the specific
heats.
For an ideal gas, cp = cv + R. (important)
For a gas, at any T the energy is distributed between the translational ke, rotational
ke and vibrational energy and electronic. Only the first is directly proportional to
the temperature,T.
For an ideal gas undergoing a polytropic process the equivalent specific heat is
Cn = Cv + 1/(1-N)
For a given substance rotation of the molecule begins at a critical velocity of impact . As
more molecules collide at or above the critical speed the more will be put into rotation.
When the velocities become high enough to cause the molecules to vibration and result in
an increase in the specific heat. Molecules shake , rotate while they move. Rotational
energy mostly transform to translational energy on cooling, while vibration converts to
thermal radiation.
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Diatomic molecule lower level Energy
Translation, rotation,vibration
At higher T the heat in part goes to increase the translational ke, resulting in T increase,part to rotation, then part to vibration; thus for a given amount of Q the more that goes
into other than translation results in a smaller increase in T, the ratio of dQ/dT increases due to smaller increase in T.
Translation - T
rotation not proportional to T
vibrationnot propto T
Q (sameamount)
38
Specific heats for ideal gases. ~(f+2)/2
(Ideal gas tables)
tri
di
mon
rotation
vibration
translation
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39
Real gas Pressure effect on cp, low P ~ig as P increase deviation from ig
40
Ideal-gas constant-
pressure specificheats for somegases (see TableA6 for cpequations).
At low pressures, all real gases approachideal-gas behavior, and therefore theirspecific heats depend on temperature only.
The specific heats of real gases at lowpressures are called ideal-gas specificheats, or zero-pressure specific heats, andare often denoted cp0 and cv0.
uand hdata for a number ofgases have been tabulated, A-7.
These tables are obtained bychoosing an arbitrary referencepoint and performing theintegrations by treating state 1as the reference state.
In the preparation of ideal-gastables, 0 K is chosen as thereference temperature.
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(kJ/kg)
For small temperature intervals, thespecific heats may be assumed to vary
linearly with temperature, for such acase the average value is at the
average T.
Internal energy and enthalpy change and
use the mean value theorem for specificheat average value
The relation u = cvTis valid for anykind ofprocess, constant-volume or not.
Area=h2h1
42
1. By using the tabulated uand h, A7 -8data. This is the easiest and mostaccurate way when tables arereadily available.
2. By using the cvor cprelations (TableA-6) as a function of temperatureand performing the integrations. Thisis very inconvenient for hand
calculations but quite desirable forcomputerized calculations. Theresults obtained are accurate.
3. By using average specific heats.This is very simple and certainly veryconvenient when property tables arenot available. The results obtainedare reasonably accurate if thetemperature interval is not verylarge. A.5 at 25 C
Three ways of calculating u andh
Three ways of calculating uWith varying accuracy.
For a linear variation, cp,avg =(cp(T2) + cp(T1))/2
cp
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43
Specific Heat Relations of Ideal Gases
The cpof an ideal gas can bedetermined from a knowledge ofcvand R.
On a molar basis
The relationship between cp, cvand R
Specificheat ratio
The specific ratio varies withtemperature, but this variation isvery mild.
For monatomic gases (helium,argon, etc.), its value is essentiallyconstant at 1.667.
Many diatomic gases, including air,have a specific heat ratio of about1.4 at room temperature.
dh= cpdT and du= cvdT
44
0peke
QO2
T1 = 25CT2 = 300C
QO2
T1 = 25CT2 = 300C
71 1-kg of Oxygen is heated from 25 to 300 C to experience a specified temperature change. The heat transfer
is to be determined for two cases., a) a constant volume process and b) a constant pressure process for both
cases assume the initial volume is 1 unit ofvolume.
Assumptions1 Oxygen is an ideal gas s ince it is at a high temperature and low pressure relative to its critical
point values of 154.8 K and 5.08 MPa. 2 The kinetic and potential energy changes are negligible,
. 3 a. Constant specific heats can be used for oxygen, b. Variable specific heats..
PropertiesThe specific heats of oxygen at the average temperature of (25+300)/2=162.5C=436 K are cp=
0.952 kJ/kgK (Table A-6).
AnalysisWe take the oxygen as the system. This is a closed systemsince no mass crosses the boundaries of
the system. The energy balance for a constant-volume process can be expressed as
net-in net-out system 2 1
Net energy transfer Change in internal, kinetic,by heat, work potential, etc. energies
in 2 1
( )
( )v
Q W E m u u
Q U m u mc T T
= =
= = =
The energy balance during a constant-pressure process (such as in a piston-cylinder device) can be ex pressed as
in out system
Net energy transfer Change in internal, kinetic,by heat, work, potential, etc. energies
in ,out
in ,out
in , 2 1
( )( )
b
b
p avg
Q W E
Q W U
Q W P Vol U Q H m h mc T T
=
=
= = +
= = =
since U+ Wb = Hduring a constant p ressure quasi-equilibrium process. Substituting for both cases,
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45
in, const , 2 1( ) (1 kg)(0.952 kJ/kg K)(300 25)K
P p avgQ mc T T
== = = 261.8 kJ
in, const 2 1( ) (1 kg)(0.692 kJ/kg K)(300 25)KQ mc T T = = = = 190.3 kJV v,avg
46
INTERNAL ENERGY, ENTHALPY, ANDSPECIFIC HEATS OF SOLIDS AND LIQUIDS
The specific volumes ofincompressible substancesremain constant during aprocess.
The cvand cpvalues ofincompressible substances areidentical and are denoted by c.
Incompressible substance: A substance whose specific volume(or density) is constant. Solids and liquids are approximatelyincompressible substances ; (T). dhp = du(T)
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47
Internal Energy Changes for an incompressiblesubstance u(T,v) ; du=
Enthalpy Changes; cv=cp=c
The enthalpy of acompressed liquid
Which is more accurate relation than
/ ) / )v T
u T dT u v dv +
cv=cp=cdhp = dup or cp = cv =c
48
in out system in
Net energy transfer Change in internal, kinetic,by heat, work, and mass potential, etc. energies
waterE E E Q U = =
Food
Reactionchamber
Water
T = 3.2C
141 A 2 g sample of a food is burned in a bomb calorimeter, that contains 3 kg of water. The
water temperature rises by 3.2C when equilibrium is established in a reaction chamber that
contains and 100 g of air.. The energy content of the food is to be determined in kJ/kg by
neglecting the energy supplied to the mixer and the thermal storage in the reaction chamber.
Assumptions1 Water is an incompressible substance with constant specific heats. 2 Air is an ideal gas with constantspecific heats. 3 The energy stored in the reaction chamber is negligible relative to the energy stored in water. 4 Theenergy supplied by the mixer is negligible.PropertiesThe specific heat of water at room temperature is c= 4.18 kJ/kgC (Table A-4). The constant volume specificheat of air at room temperature is c
v= 0.718 kJ/kgC (Table A-5).
AnalysisThe chemical energy released during the combustion of the sample is transferred to the water as heat.Therefore, disregarding the change in the sensible energy of the reaction chamber, the energy content of the food issimply the heat transferred to the water. Taking the water as our system, the energy balance can be written as
( )( ) ( )
food
2 1water water
[outofreaction ChemicalE in water
in Water
Q E Q
Q U mc T T
= =
= = Substituting,
Qin = (3 kg)(4.18 kJ/kgC)(3.2C) = 40.13 kJfor a 2-g sample. Then the energy content of the food per unit
mass is
40.13 kJ 1000 g
2 g 1 kg
=
20,060 kJ/kg- food
To make a rough estimate of the error inv olved in neglecting the thermal energy stored in the contents of the reactionchamber, we treat the entire mass within the chamber as air and determine the change in sensible internal energy:
( ) ( )[ ] ( )( ( kJ0.23C3.2CkJ/kg0.718kg0.102chamber12chamber
===
TTmcU v
which is less than 1% of the internal energy change of water. Therefore, it isreasonable to disregard the change in the sensible energy content of the reactionchamber in the analysis.
&
Watersys
For reaction chamber
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Summary
Moving boundary work
Wbfor an isothermal process
Wbfor a constant-pressure process
Wbfor a polytropic process
Energy balance for closed systems
Energy balance for a constant-pressure expansion orcompression process
Specific heats
Constant-pressure specific heat, cp Constant-volume specific heat, cv
Internal energy, enthalpy, and specific heats of ideal gases Specific heat relations of ideal gases
Internal energy, enthalpy, and specific heats ofincompressible substances (solids and liquids)
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