chapter 20 -thermodynamics
DESCRIPTION
Chatper 20 about Heat Work EngineTRANSCRIPT
Chapter 20 - Chapter 20 - ThermodynamicsThermodynamics
A PowerPoint Presentation byA PowerPoint Presentation by
Paul E. Tippens, Professor of Paul E. Tippens, Professor of PhysicsPhysics
Southern Polytechnic State Southern Polytechnic State UniversityUniversity© 2007
THERMODYNAMICSTHERMODYNAMICSThermodynamics Thermodynamics is the study of is the study of energy energy relationships that relationships that involve heat, involve heat, mechanical work, mechanical work, and other aspects and other aspects of energy and of energy and heat transfer.heat transfer. Central Heating
Objectives: After finishing Objectives: After finishing this unit, you should be this unit, you should be able to:able to:
• State and apply theState and apply the first and and second laws ofof thermodynamics.
• Demonstrate your understanding Demonstrate your understanding ofof adiabatic, isochoric, isothermal, and isobaric processes.processes.
• Write and apply a relationship for determining Write and apply a relationship for determining thethe ideal efficiency of a heat engine.of a heat engine.
• Write and apply a relationship for determiningWrite and apply a relationship for determining coefficient of performance for a refrigeratior.for a refrigeratior.
A THERMODYNAMIC SYSTEMA THERMODYNAMIC SYSTEM
• A system is a closed environment in A system is a closed environment in which heat transfer can take place. which heat transfer can take place. (For example, the gas, walls, and (For example, the gas, walls, and cylinder of an automobile engine.)cylinder of an automobile engine.)
Work done Work done on gas or on gas or work done work done by gasby gas
INTERNAL ENERGY OF INTERNAL ENERGY OF SYSTEMSYSTEM• The internal energy The internal energy UU of a system is the of a system is the
total of all kinds of energy possessed by total of all kinds of energy possessed by the particles that make up the system.the particles that make up the system.
Usually the internal energy consists of the sum of the potential and kinetic energies of the working gas molecules.
TWO WAYS TO TWO WAYS TO INCREASEINCREASE THE THE INTERNAL ENERGY, INTERNAL ENERGY, U.U.
HEAT PUT HEAT PUT INTOINTO A SYSTEM A SYSTEM
(Positive)(Positive)
++UU
WORK DONE WORK DONE ONON A GAS A GAS (Positive)(Positive)
WORK DONE WORK DONE BYBY EXPANDING EXPANDING
GAS: GAS: W is W is positivepositive
WORK DONE WORK DONE BYBY EXPANDING EXPANDING
GAS: GAS: W is W is positivepositive
--UUDecreasDecreas
ee
--UUDecreasDecreas
ee
TWO WAYS TO TWO WAYS TO DECREASEDECREASE THE THE INTERNAL ENERGY, INTERNAL ENERGY, U.U.
HEAT HEAT LEAVESLEAVES A A SYSTEMSYSTEM Q is Q is
negativenegative
QQoutout
hot
WWoutoutWWoutout
hot
THERMODYNAMIC STATETHERMODYNAMIC STATE
The STATE of a The STATE of a thermodynamic system is thermodynamic system is determined by four factors:determined by four factors:
• Absolute Pressure Absolute Pressure PP in in PascalsPascals
• Temperature Temperature TT in in KelvinsKelvins
• Volume Volume VV in cubic in cubic metersmeters
• Number of moles,Number of moles, n n, of working , of working gasgas
THERMODYNAMIC PROCESSTHERMODYNAMIC PROCESSIncrease in Internal Energy,
U.
Initial State:
P1 V1 T1 n1
Final State:
P2 V2 T2 n2
Heat inputHeat input
QQinin
WWoutout
Work by gasWork by gas
The Reverse ProcessThe Reverse ProcessDecrease in Internal Energy, U.
Initial State:
P1 V1 T1 n1
Final State:
P2 V2 T2 n2
Work on gasWork on gas
Loss of Loss of heatheat
QQoutout
WWinin
THE FIRST LAW OF THE FIRST LAW OF THERMODYAMICS:THERMODYAMICS:
• The net heat put into a system is The net heat put into a system is equal to the change in internal equal to the change in internal energy of the system plus the work energy of the system plus the work done done BYBY the system. the system.
Q = U + W final - initial)
• Conversely, the work done Conversely, the work done ONON a a system is equal to the change in system is equal to the change in internal energy plus the heat lost in internal energy plus the heat lost in the process.the process.
SIGN SIGN CONVENTIONS FOR CONVENTIONS FOR
FIRST LAWFIRST LAW• Heat Q input is Heat Q input is positivepositive
Q = U + W final - initial)
• Heat OUT is negativeHeat OUT is negative
• Work BY a gas is positive• Work ON a gas is negative
+Q+Qinin
+W+Woutout
U
-W-Winin
-Q-Qoutout
U
APPLICATION OF FIRST APPLICATION OF FIRST LAW OF LAW OF
THERMODYNAMICSTHERMODYNAMICSExample 1:Example 1: In the figure, In the figure, the gas absorbsthe gas absorbs 400 J400 J of of heat and at the same time heat and at the same time doesdoes 120 J120 J of work on the of work on the piston. What is the piston. What is the change in internal energy change in internal energy of the system?of the system?
Q = U + W
Apply First Law:
QQinin
400 J400 J
WWoutout =120 =120 JJ
Example 1 (Cont.): Example 1 (Cont.): Apply First Apply First LawLaw
U = +280 J
QQinin
400 J400 J
WWoutout =120 J =120 J
UU = = Q - Q - W W
= (+400 J) - (+120 J)= (+400 J) - (+120 J)
= +280 J= +280 J
W is positive: +120 J (Work OUT)
Q = Q = U + U + WW
UU = = Q - Q - WW
Q is positive: +400 J (Heat IN)
Example 1 (Cont.): Example 1 (Cont.): Apply First Apply First LawLaw
U = +280 J
The The 400 J400 J of input thermal of input thermal energy is used to perform energy is used to perform 120 J120 J of external work, of external work, increasingincreasing the internal the internal energy of the system by energy of the system by 280 J280 J
QQinin
400 J400 J
WWoutout =120 J =120 J
The increase in internal energy is:
Energy is conserved:
FOUR THERMODYNAMIC FOUR THERMODYNAMIC PROCESSES:PROCESSES:
• Isochoric Process: Isochoric Process: V = 0, V = 0, W = 0 W = 0
• Isobaric Process: Isobaric Process: P = 0 P = 0
• Isothermal Process: Isothermal Process: T = 0, T = 0, U = 0 U = 0
• Adiabatic Process: Adiabatic Process: Q = 0 Q = 0
• Isochoric Process: Isochoric Process: V = 0, V = 0, W = 0 W = 0
• Isobaric Process: Isobaric Process: P = 0 P = 0
• Isothermal Process: Isothermal Process: T = 0, T = 0, U = 0 U = 0
• Adiabatic Process: Adiabatic Process: Q = 0 Q = 0
Q = U + W
Q = Q = U + U + W so that W so that Q = Q = UU
ISOCHORIC PROCESS: ISOCHORIC PROCESS: CONSTANT VOLUME, CONSTANT VOLUME, V = 0, V = 0, W = W =
0000
+U -U
QQININ QQOUTOUT
HEAT IN = INCREASE IN INTERNAL ENERGYHEAT IN = INCREASE IN INTERNAL ENERGY
HEAT OUT = DECREASE IN INTERNAL HEAT OUT = DECREASE IN INTERNAL ENERGYENERGY
No Work No Work DoneDone
ISOCHORIC EXAMPLE:ISOCHORIC EXAMPLE:
Heat input Heat input increases P increases P with const. with const. VV
400 J400 J heat input heat input increases internal increases internal energy by energy by 400 J400 J and and zero work is done.zero work is done.
BB
AA
PP
22
VV11= V= V22
PP1
PPA A PP B B
TTA A TT B B
=
400 J400 J
No Change in No Change in volume:volume:
Q = Q = U + U + W But W But W = P W = P VV
ISOBARIC PROCESS: ISOBARIC PROCESS: CONSTANT PRESSURE, CONSTANT PRESSURE, P = 0P = 0
+U -U
QQININ QQOUTOUT
HEAT IN = WHEAT IN = Woutout + INCREASE IN INTERNAL ENERGY + INCREASE IN INTERNAL ENERGY
Work Work OutOut
Work Work InIn
HEAT OUT = WHEAT OUT = Woutout + DECREASE IN INTERNAL ENERGY + DECREASE IN INTERNAL ENERGY
ISOBARIC EXAMPLE (ISOBARIC EXAMPLE (Constant Pressure):
Heat input increases V with const. P
400 J400 J heat does heat does 120 J120 J of work, increasing of work, increasing the internal energy the internal energy by by 280 J280 J.
400 J400 J
BAP
V1 V2
VA VB
TA T B
=
ISOBARIC WORKISOBARIC WORK
400 J400 J
Work = Area under PV curve
Work P V
BAP
V1 V2
VA VB
TA T B
=
PPA A = P= PBB
ISOTHERMAL PROCESS: ISOTHERMAL PROCESS: CONST. TEMPERATURE, CONST. TEMPERATURE, T = 0, T = 0, U U
= 0= 0
NET HEAT INPUT = WORK OUTPUTNET HEAT INPUT = WORK OUTPUT
Q = Q = U + U + W ANDW ANDQ Q = = WW
U = 0
U = 0
QQOUTOUT
Work Work InIn
Work Work OutOut
QQININ
WORK INPUT = NET HEAT OUTWORK INPUT = NET HEAT OUT
ISOTHERMAL EXAMPLE ISOTHERMAL EXAMPLE (Constant (Constant T):T):
PAVA =
PBVB
Slow compression at constant temperature: ----- No change in UNo change in U.
U = U = TT = = 00
B
APA
V2 V1
PB
ISOTHERMAL EXPANSION (ISOTHERMAL EXPANSION (Constant Constant T)T)::
400 J of energy is absorbed by gas as 400 J of work is done on gas.
T = U = 0
U = T = 0
BB
AAPA
VA VB
PB
PAVA = PBVB
TA = TB
ln B
A
VW nRT
V
Isothermal Work
Q = Q = U + U + W ; W ; W = -W = -U or U or U = -U = -WW
ADIABATIC PROCESS: ADIABATIC PROCESS: NO HEAT EXCHANGE, NO HEAT EXCHANGE, Q = 0Q = 0
Work done at EXPENSE of internal energy
INPUT Work INCREASES internal energy
Work Out
Work InU +U
Q = 0
W = -U U = -W
ADIABATIC EXAMPLE:ADIABATIC EXAMPLE:
Insulated Walls: Q =
0
B
APPAA
VV11 V V22
PPBB
Expanding gas Expanding gas does work with does work with zero heat loss. zero heat loss. Work = -Work = -UU
ADIABATIC EXPANSION:ADIABATIC EXPANSION:
400 J of WORK is done, DECREASING the internal energy by 400 J: Net heat exchange is ZERO. Q = 0Q = 0
Q = 0
B
APPAA
VVAA VVBB
PPBB
PPAAVVA A PPBBVVBB
TTA A TT B B
=
A A B BP V P V
MOLAR HEAT CAPACITYMOLAR HEAT CAPACITYOPTIONAL TREATMENT
The The molar heat capacity Cmolar heat capacity C is defined is defined as the heat per unit mole per Celsius as the heat per unit mole per Celsius degree.degree.
Check with your instructor to see if this more thorough treatment of thermodynamic processes is required.
Check with your instructor to see if this more thorough treatment of thermodynamic processes is required.
SPECIFIC HEAT CAPACITYSPECIFIC HEAT CAPACITYRemember the definition of specific Remember the definition of specific heat capacity as the heat per unit heat capacity as the heat per unit mass required to change the mass required to change the temperature?temperature?
For example, copper: c = 390 J/kgFor example, copper: c = 390 J/kgKK
Qc
m t
MOLAR SPECIFIC HEAT MOLAR SPECIFIC HEAT CAPACITYCAPACITY
The “mole” is a better reference for The “mole” is a better reference for gases than is the “kilogram.” Thus gases than is the “kilogram.” Thus the molar specific heat capacity is the molar specific heat capacity is defined by:defined by:
For example, a constant volume of For example, a constant volume of oxygen requires oxygen requires 21.1 J21.1 J to raise the to raise the temperature of temperature of one moleone mole by one by one kelvin kelvin degreedegree..
C =C = QQ
n n TT
SPECIFIC HEAT CAPACITYSPECIFIC HEAT CAPACITYCONSTANT VOLUMECONSTANT VOLUME
How much heat is required How much heat is required to raise the temperature of 2 to raise the temperature of 2 moles of Omoles of O22 from 0 from 0ooC to C to 100100ooC?C?
QQ = (2 mol)(21.1 J/mol K)(373 K - 273 K) = (2 mol)(21.1 J/mol K)(373 K - 273 K)
Q = nCv T
Q = +4220 J
SPECIFIC HEAT CAPACITYSPECIFIC HEAT CAPACITYCONSTANT VOLUME (Cont.)CONSTANT VOLUME (Cont.)
Since the volume has not Since the volume has not changed, changed, no workno work is done. is done. The entire The entire 4220 J4220 J goes to goes to increase the internal energy,increase the internal energy, UU.
QQ = = U = nCU = nCv v TT = 4220 J= 4220 J
U = nCv TThus, U is determined by the change of temperature and the specific heat at constant volume.
SPECIFIC HEAT CAPACITYSPECIFIC HEAT CAPACITYCONSTANT PRESSURECONSTANT PRESSURE
We have just seen that We have just seen that 4220 J4220 J of heat were needed at of heat were needed at constant volumeconstant volume. Suppose we . Suppose we want to also dowant to also do 1000 J1000 J of work of work at at constant pressureconstant pressure??
Q = U + W
Q = 4220 J + J
Q =Q = 5220 J5220 JCCpp > > CCvv
Same
HEAT CAPACITY (Cont.)HEAT CAPACITY (Cont.)
CCpp > C > Cvv
For constant pressureFor constant pressure
Q = Q = U + U + WW
nCnCppT = nCT = nCvvT + P T + P VV
U = nCvT
Heat to raise Heat to raise temperature of an ideal temperature of an ideal gas, gas, UU,, is the same is the same for any process.for any process.
Cp
Cv
REMEMBER, FOR REMEMBER, FOR ANYANY PROCESS INVOLVING AN PROCESS INVOLVING AN
IDEAL GAS:IDEAL GAS:
PV = nRTPV = nRT
U = nCU = nCv v TTQ = Q = U + U + WW
PPAAVVA A PPBBVVBB
TTA A TT B B
==
Example Problem:Example Problem:
• AB: Heated at constant V to 400 K.AB: Heated at constant V to 400 K.
A A 2-L 2-L sample of Oxygen gas has an initial sample of Oxygen gas has an initial temp-erature and pressure of temp-erature and pressure of 200 K200 K and and 1 1 atmatm. The gas undergoes four processes:. The gas undergoes four processes:
• BC: Heated at constant P to 800 K.BC: Heated at constant P to 800 K.
• CD: Cooled at constant V back to 1 CD: Cooled at constant V back to 1 atm.atm.
• DA: Cooled at constant P back to 200 DA: Cooled at constant P back to 200 K.K.
PV-DIAGRAM FOR PV-DIAGRAM FOR PROBLEMPROBLEM
BB
A
PPBB
2 2
LL
1 atm1 atm200 K
400 K
800 KHow many How many
moles of Omoles of O22 are are present?present?Consider point Consider point A: PV = nRTA: PV = nRT
3(101,300Pa)(0.002m )0.122 mol
(8.314J/mol K)(200K)PV
nRT
PROCESS AB: ISOCHORICPROCESS AB: ISOCHORIC
What is the What is the pressure at point pressure at point B?B?
PPA A PP B B
TA T B
==
1 atm1 atm PP B B
200 K200 K 400 K400 K==
P B = 2 atm or 203 kPa
BB
AA
PPBB
2 L
1 atm1 atm200 K
400 K
800 K
PROCESS AB: PROCESS AB: Q = Q = U + U + WW
Analyze first law for ISOCHORIC process AB.W = 0 W = 0
Q = Q = U = nCU = nCv v TT
U =U = (0.122 mol)(21.1 J/mol K)(400 K - 200 K) (0.122 mol)(21.1 J/mol K)(400 K - 200 K)
BB
AA
PPBB
2 2
LL
1 atm1 atm200 K
400 K
800 K
Q = +514 J W = 0U = +514 J
PROCESS BC: ISOBARICPROCESS BC: ISOBARIC
What is the volume at point C (& D)?
VVB B VV C C
TTB B TT C C
==
2 L2 L VV C C
400 K400 K 800 K800 K==
BBCCPPBB
2 2
LL
1 atm1 atm200 K
400 K
800 K
DD
4 4
LL
V C = V D = 4 L
FINDING FINDING U FOR PROCESS U FOR PROCESS BC. BC.
Process BC is ISOBARIC.P = 0 P = 0
U = nCU = nCv v TT
UU = (0.122 mol)(21.1 J/mol K)(800 K - 400 K)= (0.122 mol)(21.1 J/mol K)(800 K - 400 K)
U = +1028 J
BBCC
2 2
LL
1 atm1 atm200 K
400 K
800 K
4 4
LL
2 atm2 atm
FINDING FINDING W FOR PROCESS W FOR PROCESS BC. BC.
Work depends on change in V.
P = 0
Work = P V
WW = (2 atm)(4 L - 2 L) = 4 atm L = 405 J = (2 atm)(4 L - 2 L) = 4 atm L = 405 J
W = +405 J
BBCC
2 L
1 atm200 K
400 K
800 K
4 L
2 atm
FINDING FINDING Q FOR PROCESS Q FOR PROCESS BC. BC.
Analyze first law for BC.
Q = Q = U + U + WW
Q = Q = +1028 J + 405 J+1028 J + 405 J
Q = Q = +1433 J+1433 J
Q = 1433 J W = +405 J
BBCC
2 2
LL
1 atm1 atm200 K
400 K
800 K
4 4
LL
2 atm2 atm
U = 1028 J
PROCESS CD: ISOCHORICPROCESS CD: ISOCHORIC
What is temperature at point D?
PPC C PP D D
TTC C TT D D
==
2 atm2 atm 1 atm1 atm
800 K TTDD
== T D = 400 K
B
A
PB
2 L
1 atm200 K
400 K
800 K C
D
PROCESS CD: PROCESS CD: Q = Q = U + U + WW
Analyze first law for ISOCHORIC process CD.W = 0 W = 0
Q = Q = U = nCU = nCv v TT
U = (0.122 mol)(21.1 J/mol K)(400 K - 800 K)
Q = -1028 J W = 0U = -1028 J
CC
DD
PB
2 2
LL
1 atm200 K
400 K
800 K
400 K
FINDING FINDING U FOR PROCESS U FOR PROCESS DA. DA.
Process DA is ISOBARIC.
P = 0 P = 0
U = nCU = nCv v TT
U = U = (0.122 mol)(21.1 J/mol K)(400 K - 200 K)(0.122 mol)(21.1 J/mol K)(400 K - 200 K)
U = -514 J
AADD
2 2
LL
1 atm1 atm200 K
400 K
800 K
4 L
2 atm2 atm
400 K
FINDING FINDING W FOR PROCESS W FOR PROCESS DA. DA.
Work Work depends on depends on change inchange in VV.
P = 0 P = 0
Work = PWork = P VV
WW = (1 atm)(2 L - 4 L) = -2 atm L = -203 J= (1 atm)(2 L - 4 L) = -2 atm L = -203 J
W = -203 J
AD
2 2
LL
1 atm1 atm200 K
400 K
800 K
4 4
LL
2 atm2 atm
400 K
FINDING FINDING Q FOR PROCESS Q FOR PROCESS DA. DA.
Analyze first law for DA.
Q = Q = U + U + WW
Q Q = -514 J - 203 J= -514 J - 203 J
Q = Q = -717 J-717 J
Q = -717 J W = -203 JU = -514 J
AADD
2 2
LL
1 atm1 atm200 K
400 K
800 K
4 4
LL
2 atm2 atm
400 K
PROBLEM SUMMARYPROBLEM SUMMARY
Q = Q = U + U + WWFor all For all
processes:processes:
Process Q U W
AB 514 J 514 J 0
BC 1433 J 1028 J 405 J
CD -1028 J -1028 J 0
DA -717 J -514 J -203 J
Totals 202 J 0 202 J
NET WORK FOR NET WORK FOR COMPLETE CYCLE IS COMPLETE CYCLE IS
ENCLOSED AREAENCLOSED AREABB C
2 L
1 atm1 atm
4 4
LL
2 atm2 atm
+404 J+404 JB CC
2 2
LL
1 atm1 atm
4 4
LL
2 atm2 atmNegNeg
-202 J
Area = (1 atm)(2 L)
Net Work = 2 atm L = 202 J2 2
LL 4 4
LL
BB CC
1 1 atmatm
2 2 atmatm
ADIABATIC EXAMPLE:
Q = 0
AA
BBPPBB
VVBB V VAA
PPAA PAVA PBVB
TTA A TT B B
=
PPAAVVAA = P = PBBVVBB
Example 2: A diatomic gas at 300 K and 1 atm is compressed adiabatically, decreasing its volume by 1/12. (VA = 12VB). What is the new pressure and temperature? ( = 1.4)
ADIABATIC (Cont.): FIND PADIABATIC (Cont.): FIND PBB
Q = 0
PB = 32.4 atm or 3284 kPa
1.412 B
B AB
VP P
V
1.4(1 atm)(12)BP
PPAAVVAA = P = PBBVVBB
AA
BBPPBB
VVBB 12VVBB
1 1 atmatm
300 K Solve for Solve for PPBB::
AB A
B
VP P
V
ADIABATIC (Cont.): FIND TADIABATIC (Cont.): FIND TBB
Q = 0
TB = 810 K
(1 atm)(12V(1 atm)(12VBB)) (32.4 atm)(1 V(32.4 atm)(1 VBB))
(300 K)(300 K) TT B B
==
AA
BB32.4 32.4 atmatm
VVBB 12 12VVBB
1 1 atmatm
300 K
Solve for Solve for TTBB
TTBB=?=?A A B B
A B
P V P V
T T
ADIABATIC (Cont.): ADIABATIC (Cont.): If VIf VAA= 96 cm= 96 cm33 and Vand VAA= 8 cm= 8 cm33, FIND , FIND WW
Q = 0
W = - W = - U = - nCU = - nCVV TT & & CCVV== 21.1 j/mol 21.1 j/mol KK
AA
B32.4 32.4 atmatm
1 1 atmatm
300 K
810 KSince Since Q = Q =
0,0,
W = - W = - UU 8 cm8 cm3 3 96 cm96 cm3 3
Find n Find n from point from point
AAPV = nRTPV = nRT
PVPV
RTRT n =n =
ADIABATIC (Cont.): ADIABATIC (Cont.): If VIf VAA= 96 cm= 96 cm33 and Vand VAA= 8 cm= 8 cm33, FIND , FIND WW
AA
BB32.4 32.4 atmatm
1 atm
300 K
810 K
8 cm8 cm3 3 96 cm96 cm33
PVPV
RTRT n =n = = =
(101,300 Pa)(8 x10(101,300 Pa)(8 x10-6-6 m m33))
(8.314 J/mol K)(300 K)(8.314 J/mol K)(300 K)
nn = 0.000325 mol = 0.000325 mol & & CCVV= 21.1 j/mol K= 21.1 j/mol K
TT = 810 - 300 = 510 K = 810 - 300 = 510 K
W = - W = - U = - nCU = - nCVV TT
W = - 3.50 J
• Absorbs heat Absorbs heat QQhothot
• Performs work Performs work WWoutout
• Rejects heat Rejects heat QQcoldcold
A heat engine is any device which through a cyclic process:
Cold Res. TC
Engine
Hot Res. THQhot Wout
Qcold
HEAT ENGINESHEAT ENGINES
THE SECOND LAW OF THE SECOND LAW OF THERMODYNAMICSTHERMODYNAMICS
It is impossible to construct an engine that, operating in a cycle, produces no effect other than the extraction of heat from a reservoir and the performance of an equivalent amount of work.
Not only can you not win (1st law); you can’t even break even (2nd law)!
Wout
Cold Res. TC
Engine
Hot Res. TH
Qhot
Qcold
THE SECOND LAW OF THE SECOND LAW OF THERMODYNAMICSTHERMODYNAMICS
Cold Res. TC
Engine
Hot Res. TH
400 J
300 J
100 J
• A possible engine. • An IMPOSSIBLE engine.
Cold Res. TCold Res. TCC
Engine
Hot Res. TH
400 J 400 J
EFFICIENCY OF AN ENGINEEFFICIENCY OF AN ENGINE
Cold Res. Cold Res. TTCC
Engine
Hot Res. THot Res. THH
QH W
QC
The efficiency of a heat The efficiency of a heat engine is the ratio of the engine is the ratio of the net work done W to the net work done W to the heat input Qheat input QHH..
e = 1 - QC
QH
e = = W
QH
QH- QC
QH
EFFICIENCY EXAMPLEEFFICIENCY EXAMPLE
Cold Res. Cold Res. TTCC
EnginEnginee
Hot Res. Hot Res. TTHH
800 J W
600 J
An engine absorbs 800 J An engine absorbs 800 J and wastes 600 J every and wastes 600 J every cycle. What is the cycle. What is the efficiency?efficiency?
e = 1 - 600 J
800 J
e = 1 - QC
QH
e = 25%
Question: How many joules of work is done?
EFFICIENCY OF AN IDEAL EFFICIENCY OF AN IDEAL ENGINE (Carnot Engine)ENGINE (Carnot Engine)
For a perfect engine, the quantities Q of heat gained and lost are proportional to the absolute temperatures T.
e = 1 - TC
TH
e = TH- TC
THCold Res. Cold Res.
TTCC
EnginEnginee
Hot Res. Hot Res. TTHH
QH W
QC
Example 3:Example 3: A steam engine absorbs A steam engine absorbs 600 J600 J of heat at of heat at 500 K500 K and the exhaust and the exhaust temperature is temperature is 300 K300 K. If the actual . If the actual efficiency is only half of the ideal efficiency is only half of the ideal efficiency, how much efficiency, how much workwork is done is done during each cycle?during each cycle?
e = 1 - TC
TH
e = 1 - 300 K
500 K
e = 40%
Actual e = 0.5ei = 20%
e = W
QH
W = eQH = 0.20 (600 J)
Work = 120 J
REFRIGERATORSREFRIGERATORSA refrigerator is an A refrigerator is an engine operating in engine operating in reverse: Work is done reverse: Work is done onon gas extracting heat gas extracting heat fromfrom cold reservoir and cold reservoir and depositing heat depositing heat intointo hot reservoir.hot reservoir.Win + Qcold = Qhot
WIN = Qhot - Qcold
Cold Res. Cold Res. TTCC
Engine
Hot Res. Hot Res. TTHH
Qhot
Qcold
Win
THE SECOND LAW FOR THE SECOND LAW FOR REFRIGERATORSREFRIGERATORS
It is impossible to It is impossible to construct a refrigerator construct a refrigerator that absorbs heat from a that absorbs heat from a cold reservoir and cold reservoir and deposits equal heat to a deposits equal heat to a hot reservoir with hot reservoir with W = W = 0.0.If this were possible, we could establish perpetual motion!
Cold Res. TC
EnginEnginee
Hot Res. TH
Qhot
Qcold
COEFFICIENT OF COEFFICIENT OF PERFORMANCEPERFORMANCE
Cold Res. Cold Res. TTCC
EnginEnginee
Hot Res. TH
QH W
QC
The The COP (K)COP (K) of a heat of a heat engine is the ratio of engine is the ratio of the the HEATHEAT QQcc extracted extracted to the net to the net WORKWORK done done WW..
K =
TH
TH- TC
For an For an IDEAL IDEAL
refrigerator:refrigerator:
QC
WK = =
QH
QH- QC
COP EXAMPLECOP EXAMPLEA Carnot refrigerator operates between 500 K and 400 K. It extracts 800 J from a cold reservoir during each cycle. What is C.O.P., W and QH ?
Cold Res. Cold Res. TTCC
Engine
Hot Res. Hot Res. TTHH
800 J
WQH
500 K
400 K
K = 400 K400 K
500 K - 400 K500 K - 400 K
TC
TH- TC
=
C.O.P. (K) = 4.0
COP EXAMPLE (Cont.)COP EXAMPLE (Cont.)Next we will find QNext we will find QHH by by assuming same K for assuming same K for actual refrigerator actual refrigerator (Carnot).(Carnot).
Cold Res. Cold Res. TTCC
Engine
Hot Res. Hot Res. TTHH
800 J
WQH
500 K
400 K
K =K = QC
QH- QC
QH = 1000 J
800 J800 J
QQHH - 800 J - 800 J=4.0
COP EXAMPLE (Cont.)COP EXAMPLE (Cont.)
Now, can you say how Now, can you say how much work is done in much work is done in each cycle?each cycle?
Cold Res. Cold Res. TTCC
EnginEnginee
Hot Res. THot Res. THH
800 J
W1000 J
500 K
400 K
Work = 1000 J - 800 JWork = 1000 J - 800 J
Work = 200 J
SummarySummary
Q = U + W final - initial)
TheThe First Law of ThermodynamicsFirst Law of Thermodynamics:: The The net heat taken in by a system is equal net heat taken in by a system is equal to the sum of the change in internal to the sum of the change in internal energy and the work done by the energy and the work done by the system.system.
• Isochoric Process: Isochoric Process: V = 0, V = 0, W = W = 0 0
• Isobaric Process: Isobaric Process: P = 0 P = 0
• Isothermal Process: Isothermal Process: T = 0, T = 0, U = U = 0 0
• Adiabatic Process: Adiabatic Process: Q = 0 Q = 0
Summary (Cont.)Summary (Cont.)
cc = = QQ
n n TT
U = nCv T
The Molar Specific Heat capacity, C:
Units are:Joules per mole per Kelvin degree
The following are true for ANY process:
Q = U + W
PV = nRT
A A B B
A B
P V P V
T T
Summary (Cont.)Summary (Cont.)
TheThe Second Law of Thermo:Second Law of Thermo: It It is impossible to construct an is impossible to construct an engine that, operating in a engine that, operating in a cycle, produces no effect other cycle, produces no effect other than the extraction of heat than the extraction of heat from a reservoir and the from a reservoir and the performance of an equivalent performance of an equivalent amount of work.amount of work.
Cold Res. Cold Res. TTCC
EnginEnginee
Hot Res. Hot Res. TTHH
Qhot
Qcold
Wout
Not only can you not win (1st law); you can’t even break even
(2nd law)!
Summary (Cont.)Summary (Cont.)The efficiency of a heat engine:
e = 1 - QC
QHe = 1 -
TC
TH
The coefficient of performance of a refrigerator:
C C
in H C
Q QK
W Q Q
C
H C
TK
T T
CONCLUSION: Chapter 20CONCLUSION: Chapter 20ThermodynamicsThermodynamics