entropy and the second law of thermodynamics heat engine and refrigerators a heat engine is a device...

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.