entropy and the second law of thermodynamics heat engine and refrigerators a heat engine is a device...
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Entropy and the Second Law of
Thermodynamics
Heat Engine and Refrigerators
A heat engine is a device that carries a working substance through a cyclic process, during which it
— Absorbs thermal energy from a high-temperature source (H)
— The engine does work and
— Expels thermal energy to a low-temperature source (L)
CQ
HQ
Eint Q W1st Law
Cyclic Process
Heat added tothe substance
Work done bythe substance
H CW Q Q
Thermal efficiency: W
QH
QH QC
QH
Eint 0
1 QC
QH
H CQ Q
A refrigerator is a device that carries a working substance through a cyclic process, during which it
— Absorbs thermal energy from a low-temperature source
— by doing Work
— Expels thermal energy (the amount work done and the heat absorbed) to a high-temperature source
A refrigerator is a heat engine in reverse.
CQ
HQ
Eint Q Won
1st Law
Cyclic Process
Heat absorbed bythe substance
Work done onthe substance
C H onQ Q W Eint 0
Coefficient of performance:
K QC
Won
QC
QH QCH CQ Q
Reversible and Irreversible Processes
Reversible Process: a process that at the conclusion, the system and its surrounding return to the exact initial conditions.
All natural processes are irreversible, at best “almost” reversible.
Irreversible Processes
Heat flows from the high-temperature to the low-temperature objects.
The bouncing ball finally stops
The oscillating pendulum finally stops
Carnot Engine (an Ideal Engine)
Carnot engine: a heat engine operating in a
Carnot cycle.
Carnot cycle: a cycle consisting of four REVERSIBLE (therefore Ideal) processes: two adiabatic processes B C; D A and
two isothermal processes. A B;C D
Carnot engine is the most efficient heat engine possible.
Carnot
Efficiency of a Carnot Engine
Eint Q W
Eint 0Isothermal
Adiabatic
Isothermal:
Adiabatic:
A B
C D
B C
D A
THVB 1 TCVC
1
THVA 1 TCVD
1
VB
VA
VC
VD
dW pdV
TV 1 = const.
pV nRT
1 QC
QH
ln /H AB H B AQ W nRT V V
ln /C CD C C DQ W nRT V V
Efficiency of an Ideal Carnot Engine
Thermal efficiency is then
1 QC
QH1
TC
TH
and hence C C
H H
Q T
Q T
Thus ln / ln /B A C DV V V V
QH nRTH ln VB / VA
QC nRTC ln VC / VD
Performance of an Ideal Carnot Refrigerator
Coefficient of performance
C C
H C H C
Q TK
Q Q T T
(a)
(a) W
QH
J 1049.1
J 1030.6236.04
4
HQW
HRW 38E (5th ed.). An ideal heat engine operates in a Carnot cycle between 235˚C and 115˚C. It absorbs 6.30x104 J per cycle at the higher temperature. (a) What is the efficiency of the engine? (b) How much work per cycle is this engine capable of performing?
1TC
TH
W
QH
TH TCTH
(235 115) K(235273) K
23.6%
(a) W Area p0V0 2.27 103 J
(b) Eint Q W 3
2nRT
3
2 pV
3
2pV p0V0 4.5p0V0
HRW 45P (5th ed.). One mole of an ideal monatomic gas is taken through the cycle shown. Assume that p =2p0, V =2V0, p0 = 1.01x105 Pa, and V0 = 0.0225 m3. Calculate (a) the work done during the cycle, (b) the heat added during stroke abc, and (c) the efficiency of the cycle. (d) What is the efficiency of an ideal engine operating between the highest and lowest temperatures that occur in the cycle? How does this compare to the efficiency calculated in (c)?
a
c
d
VolumePr
essu
re
b
V0, p0
V, p
PV nRT
(b) Q 4.5p0V0 W 6.5p0V0
1.48 104 J
(c) W
Qabc
p0V0
6.5p0V015.4%
HRW 45P (5th ed.). One mole of an ideal monatomic gas is taken through the cycle shown. Assume that p =2p0, V =2V0, p0 = 1.01x105 Pa, and V0 = 0.0225 m3. Calculate (a) the work done during the cycle, (b) the heat added during stroke abc, and (c) the efficiency of the cycle. (d) What is the efficiency of an ideal engine operating between the highest and lowest temperatures that occur in the cycle? How does this compare to the efficiency calculated in (c)?
Eint Q W
a
c
d
VolumePr
essu
re
b
V0, p0
V, p
1 Ta
Tc1
paVa
pcVc
1 p0V0
2p0 2V0 75%
(d)
Larger than 15%.
HRW 45P (5th ed.). One mole of an ideal monatomic gas is taken through the cycle shown. Assume that p =2p0, V =2V0, p0 = 1.01x105 Pa, and V0 = 0.0225 m3. Calculate (a) the work done during the cycle, (b) the heat added during stroke abc, and (c) the efficiency of the cycle. (d) What is the efficiency of an ideal engine operating between the highest and lowest temperatures that occur in the cycle? How does this compare to the efficiency calculated in (c)?
1TC
TH
a
c
d
VolumePr
essu
re
b
V0, p0
V, p
K QC
Won
QH Won
Won
For a complete cycle, ∆Eint = 0
Rate = 1.57 x 106 J/3600 s = 440 W
QC QH Won 0
Won QH
K 1
In one hour Won 7.54 MJ
3.8 11.57 MJ
HRW 60P (5th ed.). An ideal heat pump is used to heat a building. The outside temperature is -5.0˚C and the temperature inside the building is to be maintained at 22˚C. The coefficient of performance is 3.8 and the heat pump delivers 7.54 MJ of heat to the building each hour. At what rate must work be done to run the heat pump?
Entropy and the Second Law of Thermodynamics
The Second Law of Thermodynamics
It is impossible to construct a heat engine with100% efficiency.
It is impossible to construct a refrigerator thatdoes not require work.
1 QC
QH
K QC
Won
Absolute Temperature Scale
Carnot engine — the most efficient heat engine:
1 TC
TH
TC = 0 = 1
Second Law Temperature cannot be
equal or lower than T = 0 K
Entropy
Entropy is a state function like p, T, and Eint.A state function describes the thermodynamicstate of a system, which is independent of thehistory.
The change in entropy S for an infinitesimal process:
dS dQr
TThe subscript “r” stands for reversible process
For a finite process from an initial state “i” to a final state “f”, the change in entropy is
S dS dQr
Ti
f
i
f
The integration is along a path that represents a reversible process
Unit: J/K
From the statistical mechanical point of view,
entropy is a measure of disorder.
The Second Law of Thermodynamics
S 0= : reversible process
> : irreversible process
In a CLOSED system:
No energy exchange with other systems
Carnot Engine
For the gas: the Carnot cycle is a reversible closed cycle — back to the same state:
Entropyf
r
i
dQS
T
0rdQS
T
Carnot Engine
For the hot reservoir: heat is lost
A B
QH nRTH ln VB / VA
Isothermal
For the cold reservoir: heat is gained
C D
QC nRTC ln VC /VD
Eint 0
Eint Q W
pV nRT
ln /HAB B A
H
QS nR V V
T
ln /CCD C D
C
QS nR V V
T
Carnot Engine
B C and D A: Adiabatic
VB
VA
VC
VD
Carnot cycle
00)/ln()/ln( DCAB VVVV
0
0BC DAS S
CD AB BC DAS S S S S
Isothermal ∆Eint = 0 Q = W
W pdV nRTdV
Vi
f
i
f
nRT lnVf
Vi
S Q
T
W
T
nRT
Tln
Vf
Vi
nR ln2
Change of entropy of the ideal gas:
Reversible process: ∆Stotal = 0
Change of entropy of the reservoir: ∆S’ = -∆S = -nRln2
HRW 5E (5th ed.). An ideal gas in contact with a constant-temperature reservoir undergoes a reversible isothermal expansion to twice its initial volume. Show that the reservoir’s change in entropy is independent of its temperature.
S dQr
Ti
f
a. Q TdSi
f
Area under the T - S curve 4.5 103 J.
c. W Q Eint
9.5 103 J
Eint nCVT 2 3
2nRT
5.0 103 J
b. Monatomic
HRW 15P (5th ed.). A 2.0 mol sample of an ideal monatomic gas undergoes the reversible process shown. (a) How much heat is absorbed by the gas? (b) What is the change in the internal energy of the gas? (c) How much work is done by the gas?
Entropy (J/K)
5 10
Tem
pera
ture
(K
)200
400
15 20
S dQr
Ti
f
(1) S1 dQ
T mcIdT
TTi
T f
mcI lnTf
Ti
10 g 2220 J/kgK ln 273 K
263 K
0.828 J/K
The entropy change of the ice:
The ice warms from -10˚C to 0˚C (1), then melts (2), then the water warms to the lake temperature of 15˚C (3).
HRW 25P (5th ed.). A 10 g ice cube at -10˚C is placed in a lake whose temperature is 15˚C. Calculate the change in entropy of the cube-lake system as the ice cube comes to thermal equilibrium with the lake. The specific heat of ice is 2220 J/kg·K.
(2) S2 Q
T
mLF
T
10 g 333 kJ/kg 273 K
12.16 J/K
(3) S3 mcW lnTf
Ti
10g 4180J/kgK ln 288 K
273 K
2.26 J/K
S S1 S2 S3 15.25 J/K
HRW 25P (5th ed.). A 10 g ice cube at -10˚C is placed in a lake whose temperature is 15˚C. Calculate the change in entropy of the cube-lake system as the ice cube comes to thermal equilibrium with the lake. The specific heat of ice is 2220 J/kg·K.
The entropy change of the lake: assume its temperature does not change: ∆SL = Q/T
(1) Q1 mcI Tf Ti 222 J
(2) Q2 mLF 3330 J
(3) Q3 mcW Tf Ti 627 J
SL Q1 Q2 Q3
T
4180J
288 K 14.51 J/K
HRW 25P (5th ed.). A 10 g ice cube at -10˚C is placed in a lake whose temperature is 15˚C. Calculate the change in entropy of the cube-lake system as the ice cube comes to thermal equilibrium with the lake. The specific heat of ice is 2220 J/kg·K.
S SI SL 15.25 J/K 14.51 J/K
= 0.74 J/K
HRW 25P (5th ed.). A 10 g ice cube at -10˚C is placed in a lake whose temperature is 15˚C. Calculate the change in entropy of the cube-lake system as the ice cube comes to thermal equilibrium with the lake. The specific heat of ice is 2220 J/kg·K.