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Chapter 20 Chapter 20 -- ThermodynamicsThermodynamicsA PowerPoint Presentation byA PowerPoint Presentation by
Paul E. Paul E. TippensTippens, Professor of Physics, Professor of Physics
Southern Polytechnic State UniversitySouthern Polytechnic State University
© 2007
THERMODYNAMICSTHERMODYNAMICSThermodynamics is Thermodynamics is the study of energy the study of energy relationships that relationships that involve heat, involve heat, mechanical work, mechanical work, and other aspects and other aspects of energy and heat of energy and heat transfer.transfer.
Central Heating
Objectives: After finishing this Objectives: After finishing this unit, you should be able to:unit, you should be 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. (For which heat transfer can take place. (For example, the gas, walls, and cylinder of example, the gas, walls, and cylinder of an automobile engine.)an automobile engine.)
Work done on Work done on gas or work gas or work done by gasdone by gas
INTERNAL ENERGY OF SYSTEMINTERNAL ENERGY OF SYSTEM
•• 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 BY EXPANDING GAS: W is positive
WORK DONE WORK DONE BYBY EXPANDING GAS: EXPANDING GAS: W is W is positivepositive
-UDecrease
--UUDecreaseDecrease
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
QQ outout
hot
Wout WW outout
hot
THERMODYNAMIC STATETHERMODYNAMIC STATE
The STATE of a thermodynamic The STATE of a thermodynamic system is determined by four system is determined by four factors:factors:
•• Absolute Pressure Absolute Pressure PP in in PascalsPascals
•• Temperature Temperature TT in in KelvinsKelvins•• Volume Volume VV in cubic metersin cubic meters•• Number of moles,Number of moles, nn, of working gas, of working gas
THERMODYNAMIC PROCESSTHERMODYNAMIC PROCESSIncrease in Internal Energy, U.
Initial State:
P1 V1 T1 n1
Final State:
P2 V2 T2 n2
Heat inputHeat input
QQ inin
WW outout
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 heatLoss of heat
QQ outout
WW inin
THE FIRST LAW OF THE FIRST LAW OF THERMODYAMICS:THERMODYAMICS:
•• The net heat put into a system is equal to The net heat put into a system is equal to the change in internal energy of the the change in internal energy of the system plus the work done system plus the work done BYBY the system.the system.
Q = U + W final - initial)
•• Conversely, the work done Conversely, the work done ONON a system is a system is equal to the change in internal energy plus equal to the change in internal energy plus the heat lost in the process.the heat lost in the process.
SIGN CONVENTIONS SIGN CONVENTIONS FOR FIRST LAWFOR FIRST LAW
•• Heat Q input is positiveHeat Q input is positive
Q = U + W final - initial)
•• Heat OUT is negativeHeat OUT is negative
• Work BY a gas is positive
• Work ON a gas is negative
++QQ inin++WW outout
U
--WW inin
--QQ outout
U
APPLICATION OF FIRST APPLICATION OF FIRST LAW OF THERMODYNAMICSLAW OF THERMODYNAMICS
Example 1:Example 1: In the figure, the In the figure, the gas absorbsgas absorbs 400 J400 J of heat and of heat and at the same time doesat the same time does 120 J120 J of work on the piston. What of work on the piston. What is the change in internal is the change in internal energy of the system?energy of the system?
Q = U + W
Apply First Law:
QQ inin400 J400 J
WW outout =120 J=120 J
Example 1 (Cont.): Example 1 (Cont.): Apply First LawApply First Law
U = +280 J
QQinin
400 J400 J
WWoutout =120 J=120 J
UU = = Q Q -- W W
= (+400 J) = (+400 J) -- (+120 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 LawApply First Law
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: V = 0, W = 0
• Isobaric Process: P = 0
• Isothermal Process: T = 0, U = 0
• Adiabatic Process: 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 = 0W = 0
00
+U -U
QQININ QQOUTOUT
HEAT IN = INCREASE IN INTERNAL ENERGYHEAT IN = INCREASE IN INTERNAL ENERGY
HEAT OUT = DECREASE IN INTERNAL ENERGYHEAT OUT = DECREASE IN INTERNAL ENERGY
No Work No Work DoneDone
ISOCHORIC EXAMPLE:ISOCHORIC EXAMPLE:
Heat input Heat input increases P increases P with const. Vwith const. V
400 J400 J heat input increases heat input increases internal energy by internal energy by 400 J400 J and zero work is done.and zero work is done.
BB
AA
PP 22
VV 11 = V= V 22
PP1
PP A A PP BB
TT A A TT BB=
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 OutWork Out Work Work InIn
HEAT OUT = HEAT OUT = WWoutout + 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 of work, increasing the work, increasing the internal energy by internal energy 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 = 0U = 0
NET HEAT INPUT = WORK OUTPUTNET HEAT INPUT = WORK OUTPUT
Q = Q = U + U + W ANDW AND Q = Q = WW
U = 0 U = 0
QQ OUTOUT
Work Work InIn
Work OutWork Out
QQ ININ
WORK INPUT = NET HEAT OUTWORK INPUT = NET HEAT OUT
ISOTHERMAL EXAMPLE ISOTHERMAL EXAMPLE (Constant T):(Constant T):
PA VA = PBVB
Slow compression at constant temperature: ----- No change in UNo change in U.
U = U = TT = 0= 0
B
APA
V2 V1
PB
ISOTHERMAL EXPANSION (ISOTHERMAL EXPANSION (Constant T)Constant 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
PA VA = PBVB
TA = TB
ln B
A
VW nRTV
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
APP AA
VV 11 VV 22
PP BB
Expanding gas does Expanding gas does work with zero heat work with zero heat loss. 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
APP AA
VV AA VV BB
PP BB
PP AA VV A A PP BB VV BB
TT A A TT BB=
A A B BP V P V
MOLAR HEAT CAPACITYMOLAR HEAT CAPACITYOPTIONAL TREATMENT
The The molar heat capacity Cmolar heat capacity C is defined as is defined as the heat per unit mole per Celsius degree.the heat per unit mole per Celsius 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 CAPACITY
Remember the definition of specific heat Remember the definition of specific heat capacity as the heat per unit mass capacity as the heat per unit mass required to change the temperature?required to change the temperature?
For example, copper: c = 390 J/kgFor example, copper: c = 390 J/kgKK
Qcm t
MOLAR SPECIFIC HEAT CAPACITYMOLAR SPECIFIC HEAT CAPACITYThe The ““molemole”” is a better reference for gases is a better reference for gases than is the than is the ““kilogram.kilogram.”” Thus the molar Thus the molar specific heat capacity is defined by:specific heat capacity is defined by:
For example, a constant volume of oxygen For example, a constant volume of oxygen requires requires 21.1 J21.1 J to raise the temperature of to raise the temperature of one moleone mole by one by one kelvin degreekelvin degree..
C =C =QQ
n n TT
SPECIFIC HEAT CAPACITYSPECIFIC HEAT CAPACITY CONSTANT VOLUMECONSTANT VOLUME
How much heat is required to How much heat is required to raise the temperature of 2 moles raise the temperature of 2 moles of Oof O 22 from 0from 0ooC to 100C to 100ooC?C?
QQ = (2 mol)(21.1 J/mol K)(373 K = (2 mol)(21.1 J/mol K)(373 K -- 273 K)273 K)
Q = nCv T
Q = +4220 J
SPECIFIC HEAT CAPACITYSPECIFIC HEAT CAPACITY CONSTANT VOLUME (Cont.)CONSTANT VOLUME (Cont.)
Since the volume has not Since the volume has not changed, changed, no workno work is done. The is done. The entire entire 4220 J4220 J goes to increase goes to increase the internal energy,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 CAPACITY CONSTANT PRESSURECONSTANT PRESSURE
We have just seen that We have just seen that 4220 J4220 J of of heat were needed at heat were needed at constant constant volumevolume. Suppose we want to also . Suppose we want to also dodo 1000 J1000 J of work at of work at constant constant pressurepressure??
Q = U + W
Q = 4220 J + J
Q =Q = 5220 J5220 J CC pp > C> C vv
Same
HEAT CAPACITY (Cont.)HEAT CAPACITY (Cont.)
CC pp > C> C vvFor constant pressureFor constant pressure
Q = Q = U + U + WW
nCnC pp T = nCT = nCvv T + P T + P VV
U = nCv T
Heat to raise temperature Heat to raise temperature of an ideal gas, of an ideal gas, UU,, is the is the same for any process.same for any process.
Cp
Cv
REMEMBER, FOR REMEMBER, FOR ANYANY PROCESS PROCESS INVOLVING AN IDEAL GAS:INVOLVING AN IDEAL GAS:
PV = nRTPV = nRT
U = nCU = nCv v TTQ = Q = U + U + WW
PP AA VV A A PP BB VV BBTT A A TT BB
==
Example Problem:Example Problem:
•• AB: Heated at constant V to 400 K.AB: Heated at constant V to 400 K.
A A 22--L L sample of Oxygen gas has an initial tempsample of Oxygen gas has an initial temp-- erature and pressure of erature and pressure of 200 K200 K and and 1 atm1 atm. The . The gas undergoes four processes: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 atmatm..
•• DA: Cooled at constant P back to 200 K.DA: Cooled at constant P back to 200 K.
PVPV--DIAGRAM FOR PROBLEMDIAGRAM FOR PROBLEMBB
A
PP BB
2 L2 L
1 atm1 atm200 K
400 K 800 KHow many moles How many moles of Oof O22 are present?are present?
Consider point A: Consider point A: PV = nRTPV = nRT
3(101, 300Pa)(0.002m ) 0.122 mol(8.314J/mol K)(200K)
PVnRT
PROCESS AB: ISOCHORICPROCESS AB: ISOCHORIC
What is the pressure What is the pressure at point B?at point B?
PP A A PP BBTA T B
==
1 atm1 atm PP BB
200 K200 K 400 K400 K==
P B = 2 atm
or 203 kPa
BB
AA
PP BB
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 (0.122 mol)(21.1 J/mol K)(400 K -- 200 K)200 K)
BB
AA
PP BB
2 L2 L
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)?
VV B B VV CC
TT B B TT CC==
2 L2 L VV CC
400 K400 K 800 K800 K==
BBCCPP BB
2 L2 L
1 atm1 atm200 K
400 K 800 K
DD
4 L4 L
V C = V D = 4 L
FINDING FINDING U FOR PROCESS BC. U FOR PROCESS 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 = (0.122 mol)(21.1 J/mol K)(800 K -- 400 K)400 K)
U = +1028 J
BBCC
2 L2 L
1 atm1 atm200 K
400 K 800 K
4 L4 L
2 atm2 atm
FINDING FINDING W FOR PROCESS BC. W FOR PROCESS BC.
Work depends on change in V.
P = 0
Work = PV
WW = (2 atm)(4 L = (2 atm)(4 L -- 2 L) = 4 atm L = 405 J2 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 BC. Q FOR PROCESS 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 L2 L
1 atm1 atm200 K
400 K 800 K
4 L4 L
2 atm2 atm
U = 1028 J
PROCESS CD: ISOCHORICPROCESS CD: ISOCHORIC
What is temperature at point D?
PP C C PP DD
TT C C TT DD==
2 atm2 atm 1 atm1 atm
800 K TT DD== T D = 400 K
B
A
PB
2 L
1 atm200 K
400 K 800 KC
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 L2 L
1 atm200 K
400 K 800 K
400 K
FINDING FINDING U FOR PROCESS DA. U FOR PROCESS 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 (0.122 mol)(21.1 J/mol K)(400 K -- 200 K)200 K)
U = -514 J
AA DD
2 L2 L
1 atm1 atm200 K
400 K 800 K
4 L
2 atm2 atm
400 K
FINDING FINDING W FOR PROCESS DA. W FOR PROCESS DA.
Work depends Work depends on change inon change in VV.
P = 0 P = 0
Work = PWork = P VV
WW = (1 atm)(2 L = (1 atm)(2 L -- 4 L) = 4 L) = --2 atm L = 2 atm L = --203 J203 J
W = -203 J
AD
2 L2 L
1 atm1 atm200 K
400 K 800 K
4 L4 L
2 atm2 atm
400 K
FINDING FINDING Q FOR PROCESS DA. Q FOR PROCESS DA.
Analyze first law for DA.
Q = Q = U + U + WW
Q Q = = --514 J 514 J -- 203 J203 J
Q = Q = --717 J717 J
Q = -717 J W = -203 JU = -514 J
AA DD
2 L2 L
1 atm1 atm200 K
400 K 800 K
4 L4 L
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 COMPLETE NET WORK FOR COMPLETE CYCLE IS ENCLOSED AREACYCLE IS ENCLOSED AREA
BB C
2 L
1 atm1 atm
4 L4 L
2 atm2 atm
+404 J+404 J B CC
2 L2 L
1 atm1 atm
4 L4 L
2 atm2 atmNegNeg
-202 J
Area = (1 atm)(2 L)
Net Work = 2 atm L = 202 J2 L2 L 4 L4 L
BB CC
1 atm1 atm
2 atm2 atm
ADIABATIC EXAMPLE:
Q = 0
AA
BBPP BB
VV BB VV AA
PP AA PA VA PBVB
TT A A TT BB=
PP AA VV AA = P= P BB VV BB
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 PV
1.4(1 atm)(12)BP
PP AA VV AA = P= P BB VV BB
AA
BBPP BB
VV BB 12VV BB
1 atm1 atm300 K Solve for PSolve for P BB ::
AB A
B
VP PV
ADIABATIC (Cont.): FIND TADIABATIC (Cont.): FIND TBB
Q = 0
TB = 810 K
(1 atm)(12V(1 atm)(12V BB )) (32.4 atm)(1 V(32.4 atm)(1 V BB ))
(300 K)(300 K) TT BB==
AA
BB32.4 atm32.4 atm
VV BB 1212VV BB
1 atm1 atm300 K
Solve for TSolve for T BB
TT BB =?=?A A B B
A B
P V P VT T
ADIABATIC (Cont.): ADIABATIC (Cont.): If VIf V AA = 96 cm= 96 cm33
and Vand V AA = 8 cm= 8 cm33, FIND , FIND WW
Q = 0
W = W = -- U = U = -- nCnC VV TT & & CC VV == 21.1 j/mol K21.1 j/mol K
AA
B32.4 atm32.4 atm
1 atm1 atm300 K
810 KSince Since Q = 0,Q = 0,
W = W = -- UU
8 cm8 cm3 3 96 cm96 cm3 3
Find n from Find n from point Apoint A PV = nRTPV = nRT
PVPV
RTRTn =n =
ADIABATIC (Cont.): ADIABATIC (Cont.): If VIf V AA = 96 cm= 96 cm33
and Vand V AA = 8 cm= 8 cm33, FIND , FIND WW
AA
BB32.4 atm32.4 atm
1 atm300 K
810 K
8 cm8 cm3 3 96 cm96 cm33
PVPV
RTRTn =n = = =
(101,300 Pa)(8 x10(101,300 Pa)(8 x10--66 mm33))
(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 = 810 -- 300 = 510 K300 = 510 K
W = W = -- U = U = -- nCnCVV TT
W = - 3.50 J
•• Absorbs heat Absorbs heat QQ hothot
•• Performs work Performs work WW outout
•• Rejects heat Rejects heat QQ coldcold
A heat engine is any device which through a cyclic process:
Cold Res. TC
Engine
Hot Res. TH
Qhot 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. TCold Res. TCC
Engine
Hot Res. THot Res. THH
QH W
QC
The efficiency of a heat engine The efficiency of a heat engine is the ratio of the net work is the ratio of the net work done W to the heat input Qdone W to the heat input Q HH ..
e = 1 -QC
QH
e = = W
QH
QH- QC
QH
EFFICIENCY EXAMPLEEFFICIENCY EXAMPLE
Cold Res. TCold Res. TCC
EngineEngine
Hot Res. THot Res. THH
800 J W
600 J
An engine absorbs 800 J and An engine absorbs 800 J and wastes 600 J every cycle. What wastes 600 J every cycle. What is the efficiency?is the 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. TCold Res. TCC
EngineEngine
Hot Res. THot Res. THH
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, efficiency is only half of the ideal efficiency, how much how much workwork is done during each cycle?is done 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 engine A refrigerator is an engine operating in reverse: operating in reverse: Work is done Work is done onon gas gas extracting heat extracting heat fromfrom cold cold reservoir and depositing reservoir and depositing heat heat intointo hot reservoir.hot reservoir.
Win + Qcold = Qhot
WIN = Qhot - QcoldCold Res. TCold Res. T CC
Engine
Hot Res. THot Res. T HH
Qhot
Qcold
Win
THE SECOND LAW FOR THE SECOND LAW FOR REFRIGERATORSREFRIGERATORS
It is impossible to construct a It is impossible to construct a refrigerator that absorbs refrigerator that absorbs heat from a cold reservoir heat from a cold reservoir and deposits equal heat to a and deposits equal heat to a hot reservoir with hot reservoir with W = 0.W = 0.
If this were possible, we could establish perpetual motion!
Cold Res. TC
EngineEngine
Hot Res. TH
Qhot
Qcold
COEFFICIENT OF PERFORMANCECOEFFICIENT OF PERFORMANCE
Cold Res. TCold Res. TCC
EngineEngine
Hot Res. TH
QH W
QC
The The COP (K)COP (K) of a heat of a heat engine is the ratio of the engine is the ratio of the HEATHEAT QQ cc extracted to the extracted to the net net WORKWORK done done WW..
K =TH
TH- TC
For an IDEAL For an IDEAL refrigerator:refrigerator:
QC
WK = =
QH
QH- QC
COP EXAMPLECOP EXAMPLE
A 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. TCold Res. TCC
Eng ine
Hot Res. THot Res. THH
800 J
WQH
500 K
400 K
K =400 K400 K
500 K 500 K -- 400 K400 K
TC
TH- TC
=
C.O.P. (K) = 4.0
COP EXAMPLE (Cont.)COP EXAMPLE (Cont.)
Next we will find QNext we will find Q HH by by assuming same K for actual assuming same K for actual refrigerator (Carnot).refrigerator (Carnot).
Cold Res. TCold Res. TCC
Eng ine
Hot Res. THot Res. THH
800 J
WQH
500 K
400 K
K =K =QC
QH- QC
QH = 1000 J
800 J800 J
QQ HH -- 800 J800 J=4.0
COP EXAMPLE (Cont.)COP EXAMPLE (Cont.)
Now, can you say how much Now, can you say how much work is done in each cycle?work is done in each cycle?
Cold Res. TCold Res. TCC
EngineEngine
Hot Res. THot Res. THH
800 J
W1000 J
500 K
400 K
Work = 1000 J Work = 1000 J -- 800 J800 J
Work = 200 J
SummarySummary
Q = U + W final - initial)
TheThe First Law of ThermodynamicsFirst Law of Thermodynamics:: The net The net heat taken in by a system is equal to the heat taken in by a system is equal to the sum of the change in internal energy and sum of the change in internal energy and the work done by the system.the work done by the system.
•• 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
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 VT T
Summary (Cont.)Summary (Cont.)
TheThe Second Law of Thermo:Second Law of Thermo: It is It is impossible to construct an engine impossible to construct an engine that, operating in a cycle, that, operating in a cycle, produces no effect other than the produces no effect other than the extraction of heat from a reservoir extraction of heat from a reservoir and the performance of an and the performance of an equivalent amount of work.equivalent amount of work.Cold Res. TCold Res. TCC
EngineEngine
Hot Res. THot Res. THH
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 -QCQH
e = 1 -TC
TH
The coefficient of performance of a refrigerator:
C C
in H C
Q QKW Q Q
C
H C
TKT T
CONCLUSION: Chapter 20CONCLUSION: Chapter 20 ThermodynamicsThermodynamics
