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)
QH
—
—
The engine does work and
Expels thermal energy to a lowtemperature source (L) Q
—
C
1st Law
Eint  Q  W
Cyclic Process
W  QH  QC
Work done by
the substance
Eint  0
Heat added to
the substance
Thermal efficiency:
QH  QC
W


QH
QH
QC
  1
QH
QH  QC
A refrigerator is a heat engine in reverse.
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 Q
—
—
by doing Work
C
Expels thermal energy (the
amount work done and the heat
absorbed) to a high-temperature
source
—
QH
1st Law
Eint  Q  Won
Cyclic Process
QC  QH  Won
Heat absorbed by
the substance
Eint  0
Work done on
the substance
Coefficient of performance:
QC
QC
K

Won QH  QC
QH  QC
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
Carnot engine is the most efficient heat
engine possible.
QC
  1
QH
Efficiency of a Carnot EngineIsothermal
Eint  0
Eint  Q  W
Isothermal:
dW  pdV
A  B QH  WAB  nRTH ln VB / VA 
C  D QC  WCD  nRTC ln VC / VD 
pV  nRT
Adiabatic
 1
TV
= const.
Adiabatic:
 1
B  C TH VB
 1
D  A TH VA
 1
 TCVC
 1
 TCVD
VB VC

VA VD
Efficiency of an Ideal Carnot Engine
Thus ln VB / VA   ln VC / VD 
QC
TC
and hence

QH TH
QC  nRTC ln VC / VD 
QH  nRTH ln VB / VA 
Thermal efficiency is then
QC
TC
  1
1
QH
TH
Performance of an Ideal Carnot
Refrigerator
Coefficient of performance
QC
TC
K

QH  QC TH  TC
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?
(a)
(a)
TH  TC (235  115) K


 23.6%
TH
(235  273) K
W

QH

W   QH  0.236 6.30 10 J
 1.49 10 4 J
4

W

QH
TC
  1
TH
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)?
c
V, p
(a) W  Area  p0V0  2.27  103 J
(b)
Eint  Q  W 
3
nRT
2
3
  pV 
2
3
  pV  p0V0   4.5p0 V0
2
Pressure
b
V0, p0
d
a
Volume
PV  nRT
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)?
b
 1.48 10 4 J
W
p0V0
(c)  

 15.4%
Qabc 6.5 p0V0
Pressure
(b) Q  4.5p0V0  W  6.5p0V0
V0, p0
a
c
V, p
d
Volume
Eint  Q  W
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)?
p0 V0
 1
 75%
2 p0 2V0 
Pressure
Ta
pa Va
 1
(d)   1 
Tc
pc Vc
b
V0, p0
a
c
V, p
d
Volume
Larger than 15%.
  1
TC
TH
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?
For a complete cycle, ∆Eint = 0
QC  QH  Won  0
QC
QH  Won
K

Won
Won
In one hour
Won
Won
7.54 MJ

 1.57 MJ
3.8  1
Rate = 1.57 x 106 J/3600 s = 440 W
QH

K 1
Entropy and the Second
Law of Thermodynamics
The Second Law of
Thermodynamics
QC
  1
QH
It is impossible to construct a heat engine with
100% efficiency.
It is impossible to construct a refrigerator that
does not require work.
QC
K
Won
Absolute Temperature Scale
Carnot engine — the most efficient heat
engine:
TC
  1
TH
TC = 0

Second Law
=1

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 thermodynamic
state of a system, which is independent of the
history.
The change in entropy S for an infinitesimal
process:
dQr
dS 
T
The 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
f
f
dQr
S   dS  
T
i
i
Unit: J/K
The integration is along a path that
represents a reversible process
From the statistical mechanical
point of view,
entropy
is a measure of disorder.
The Second Law of
Thermodynamics
In a CLOSED system:
No energy
exchange
with other
systems
S  0
= : reversible process
> : irreversible process
Entropy
f
Carnot Engine
For the gas: the Carnot cycle is a reversible
closed cycle — back to the same state:
dQr
S  
 T 0
S  
i
dQr
T
Isothermal
Eint  0
Carnot Engine
Eint  Q  W
pV  nRT
For the hot reservoir: heat is lost
QH  nRTH lnVB / VA 
A  B S   QH  nR ln V / V 
AB
B
A
TH
QC  nRTC ln VC / VD 
For the cold reservoir: heat is gained
C  D SCD
QC

 nR ln VC / VD 
TC
Carnot cycle
Carnot Engine
B  C and D  A: Adiabatic
SBC  SDA  0
S  SCD  S AB  S BC  SDA
 ln( VB / VA )  ln( VC / VD )  0  0
0
VB VC

VA VD
HRW 5E (5th ed.). An ideal gas in contact with a constanttemperature reservoir undergoes a reversible isothermal expansion to
twice its initial volume. Show that the reservoir’s change in entropy
is independent of its temperature.
Isothermal
∆Eint = 0

Q=W
 Vf 
dV
W   pdV  nRT 
 nRT ln  
V
 Vi 
i
i
f
f
f
dQr
S  
i T
Change of entropy of the ideal gas:
Q W nRT  V f 
S   
ln
 nR ln 2
T T
T
 Vi 
Reversible process: ∆Stotal = 0
Change of entropy of the reservoir: ∆S’ = -∆S = -nRln2
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?
f
a.
Q   TdS  Area under the T - S curve  4.5  10 J.
3
i
f
dQr
i T
S  
3
Eint  nCV T  2  nRT
2
 5.0  103 J
c.
W  Q  Eint
Temperature (K)
b. Monatomic
400
200
 9.5  10 3 J
5
10
15
Entropy (J/K)
20
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 ice warms from -10˚C to 0˚C (1), then melts (2),
then the water warms to the lake temperature of 15˚C (3).
The entropy change of the ice:
Tf
 Tf 
dQ
dT
(1) S1 
 T  mcI  T  mcI ln  T 
i
T
i
 273 K
 10 g2220 J/kg  K ln
 0.828 J/K
 263 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.
Q mLF 10 g333 kJ/kg
(2) S2  

 12.16 J/K
T
T
273 K
 Tf 
288
K




(3) S3  mcW ln
 10g4180J/kg  K ln
 273 K 
 Ti 
 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 T f  Ti  222 J
(2)
Q2  mLF  3330 J
(3)
Q3  mcW T f  Ti  627 J


Q1  Q2  Q3 4180J
SL 

 14.51 J/K
T
288 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