chapter 18 the second law of thermodynamics. irreversible processes irreversible processes: always...

Chapter 18

The Second Law of Thermodynamics

Irreversible Processes

• Irreversible Processes:always found to proceed in one direction

• Examples:

• free expansion of ideal gas

• heat flow (for finite T): from hot to cold

• friction: converts mechanical energy to heat

Reversible Processes

• Reverisble processes: just an idealization, but we can come close

• put system in equilibrium (with itself and its surroundings)

• make slow, infinitesimal changes in order to reverse a process

Heat Engine

• Heat engine:transforms heat into mechanical energy, W

• Reservoirs:

• hot reservoir at TH

• cold reservoir at TC

Heat Engine

• engine absorbs heat QH from hot reservoir

• engine discards heat QC to cold reservoir

• engine does work W

Heat Engine

• Working substance:material in engine undergoing the heat transfer

• Cyclic process: reuse same substance, return it to initial state at end of each cycle

Q = net heat input

• For each cycle:

• heat QH enters engine

• heat QC leaves engine

• net heat in per cycle:

Q = QH + QC

= QH – |QC|

Heat Engine

• cyclic process:U = 0 for engine during one cycle

• 1st Law: Q = W + U

• Thus for each cycle:Q = W

Heat Engine

• For each cycle:

Q = QH – |QC|

• So by 1st Law:

W = Q = QH – |QC|

EngineEfficiency

W= QH – |QC|

Perfect engine: |QC| = 0

All experiments: |QC| > 0

Efficiency: e• measures how close an

engine gets to |QC| = 0

EngineEfficiency

H

C

H

CH

H

||1

||

Q

Q

Q

QQ

Q

We

Ideal Engine

= limiting behavior of real engines

• engine: converts heat to mechanical energy

• recall: converting mechanical energy to heat is ‘irreversible’

Ideal Engine

• avoid irreversibility

• carry out engine cycle slowly and with reversible heat flow

• the reversibility of heat flow motivates the allowed processes

Ideal Engine

• For reversible heat flow, there can be no finite T between engine and reservoir

• so during heat transfer we need an isothermal process

Ideal Engine

• If there is a finite T between engine and reservoir, any heat flow would be irreversible

• So when there is a finite T we need an adiabatic process

Ideal Engine

• so we are motivated to consider a cycle with only two processes:

• isothermal • adiabatic

• called a ‘Carnot cycle’

Ideal Engine

• efficiency: e = ?• working substance = ?

• We’ll calculate e for substance = ideal gas

• Actually, e is independent of the working substance

Carnot Cycle

Example engine: internal combustion engine

• TH is provided by combustion of air-fuel mixture

• TC is provided by exhaust gases venting to outside air

• fuel = gasoline• working substance = mixture of air and burned fuel• this engine doesn’t actually recycle the same mixture in

each cycle, but we can idealize with simple models

Problem 18-44

Refrigerator

• Refrigerator:like a heat engine, but operating in reverse

• Reservoirs:

• hot reservoir at TH

• cold reservoir at TC

Refrigerator

• refrigerator absorbs heat QC

• refrigerator discards heat QH

• work W done on refrigerator

Refrigerator

• cyclic process:U = 0 during one cycle

• 1st Law: Q = W + U

• Thus for each cycle:Q = W

Refrigerator

• So by 1st Law:

W = Q

= QH + QC

– |W| = – |QH| + QC

|QH| = QC + |W|

Announcements

• Today: finish Chapter 18

• Wednesday: review

• Thursday: no class (office hours 12:30-2:30)

• HW 6 (Ch. 15): returned at front

• Midterms: returned at front (midterm scores: classweb, solutions: E-Res)

The Second Law of Thermodynamics

2nd Law of Thermodynamics

Four equivalent approaches, in terms of:

• Engines

• Refrigerators

• Carnot Cycle

• Entropy

Heat Engine

• engine absorbs heat QH

• engine discards heat QC

• engine does work W

• engine efficiency :

H

C

H

||1

Q

Q

Q

We

2nd Law:Engines

• There are no perfect engines (real engines have e < 1)

• It is impossible for a system to change heat completely into work, with no other change to the system taking place

2nd Law: Engine Version

If 2nd Law weren’t true:

• ships could move by cooling the ocean

• cars could move by cooling surrounding air

Refrigerator

• refrigerator absorbs heat QC

• refrigerator discards heat QH

• work W done on refrigerator

2nd Law: RefrigeratorVersion

• There are no perfect refrigerators (for real refrigerators |W|> 0).

• It is impossible for heat to flow from hot to cold with no other change to the system taking place

‘Engine’ and ‘Refrigerator’ versions of 2nd Law are equivalent

(a) If you can build a perfect refrigerator, then you can build a perfect engine

(perfect refrigerator) + (real engine) = perfect engine

See notesSee notes

(b) If you can build a perfect engine, then you can build a perfect refrigerator

(perfect engine) + (real refrigerator) = perfect refrigerator

See notesSee notes

Engine Efficiency Revisited

2nd Law:Real engines have efficiency e < 1.

• How close can we get to e = 1?

• Is there a maximum possible efficiency?

• Yes!

Carnot Cycle

2nd Law: Carnot Cycle Version

e < eCarnot

The efficiency e of a real engine operating between temperatures TH and TC can never exceed that of a Carnot engine operating between the same TH and TC:

H

CCarnot 1

T

Te

ProofProof

Problem 18-44, Revisited

Defining Entropy

• For Carnot cycle:

0

0C

C

H

H

H

C

H

C

T

Q

T

Q

T

Q

T

T

Q

Q

See notesSee notes

• Any reversible cycle = sum of many Carnot cycles• For cycle, in limit of infinitesimally close isotherms:

0T

dQ

Entropy (S)

• Result for a reversible cycle:

• This defines a new state variable, entropy (S):

0T

dQ

T

dQdS

dS

0

• Entropy measures the disorder of a system

• The more heat dQ you add, the higher the entropy increase dS that results

• The smaller the temperature T, the larger the entropy increase dS that results

process) reversible a(for T

dQdS

Entropy (S)

• For any reversible path from state 1 to 2:

• This must also equal S for an irreversible path from 1 to 2, since S is a state variable

2

1

2

1 T

dQdSS

Exercises 18-20, 18-26Exercises 18-20, 18-26 Example 18-8, Problem 18-50Example 18-8, Problem 18-50

0T

dQ

T

dQdS

dS

0

Example 18-8: Free Expansion of Ideal Gas

Entropy (S)

• For any reversible path from state 1 to 2:

• This must also equal S for an irreversible path from 1 to 2, since S is a state variable

2

1

2

1 T

dQdSS

Exercises 18-20, 18-26Exercises 18-20, 18-26 Example 18-8, Problem 18-50Example 18-8, Problem 18-50

2nd Law: Entropy Version

If we consider all systems taking part in a process,

For a reversible process:

Stotal = 0

For an irreversible process:

Stotal > 0

Laws of Thermodynamics

Law state variable

Zeroth Law temperature: T

1st Law U = Q – W internal energy: U

2nd Law entropy: S0S

Announcements

• Today: finish Chapter 18

• Wednesday: review

• Thursday: no class (office hours 12:30-2:30)

• HW 6 (Ch. 15): returned at front

• Midterms: returned at front (midterm scores: classweb, solutions: E-Res)