Plan for Today (AP Physics Plan for Today (AP Physics 2)2)

• Lecture/Notes on Heat Engines, Lecture/Notes on Heat Engines, Carnot Engines, and EntropyCarnot Engines, and Entropy

Energy Transfer by Energy Transfer by ConductionConduction

• Kinetic energy exchange between Kinetic energy exchange between microscopic particlesmicroscopic particles

• Collisions take place between Collisions take place between particles, energy transfer, and particles, energy transfer, and gradually it works through a gradually it works through a materialmaterial

ConductionConduction

• Rate depends on properties of the Rate depends on properties of the substancesubstance

• Metals are good thermal conductors Metals are good thermal conductors (free electrons to transfer energy)(free electrons to transfer energy)

ConductionConduction

• Only occurs if there is a difference in Only occurs if there is a difference in temperaturetemperature

• Temperature difference drives the Temperature difference drives the flow in energyflow in energy

ConductionConduction

• Rate of energy transfer is Rate of energy transfer is proportional to the cross sectional proportional to the cross sectional area of the slab and the temperature area of the slab and the temperature differencedifference

• H = Q/tH = Q/t

• And is inversely proportional to the And is inversely proportional to the thickness of the slabthickness of the slab

Conduction EquationConduction Equation

• k is proportionality constant, k is proportionality constant, depending on the materialdepending on the material

Heat engineHeat engine• A device that takes in heat energy A device that takes in heat energy

and converts some of it to other and converts some of it to other forms of energy (like mechanical)forms of energy (like mechanical)

• Work done by a heat engine Work done by a heat engine through a cyclic process isthrough a cyclic process is

• W = |Qh| - |Qc|W = |Qh| - |Qc|

• Qh is energy absorbed from hot Qh is energy absorbed from hot reservoir, Qc is energy expelled to reservoir, Qc is energy expelled to coldcold

• Absorbs heat Absorbs heat QQhothot

• Performs work Performs work WWoutout

• Rejects heat Rejects heat QQcoldcold

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. Cold Res. TTCC

Engine

Hot Res. THot Res. THH

QH W

QC

The efficiency of a heat The efficiency of a heat engine is the ratio of the engine is the ratio of the net work done W to the net work done W to the heat input Qheat input QHH..

e = 1 - QC

QH

e = = W

QH

QH- QC

QH

EFFICIENCY EXAMPLEEFFICIENCY EXAMPLE

Cold Res. Cold Res. TTCC

EnginEnginee

Hot Res. Hot Res. TTHH

800 J W

600 J

An engine absorbs 800 J An engine absorbs 800 J and wastes 600 J every and wastes 600 J every cycle. What is the cycle. What is the efficiency?efficiency?

e = 1 - 600 J

800 J

e = 1 - QC

QH

e = 25%

Question: How many joules of work is done?

Carnot EngineCarnot Engine

• No real engine operating between No real engine operating between two reservoirs can be more efficient two reservoirs can be more efficient than a Carnot enginethan a Carnot engine

Carnot EngineCarnot Engine

• Substance whose temperature Substance whose temperature varies between Tc and Thvaries between Tc and Th

• Ideal gas in a cylinder with a Ideal gas in a cylinder with a movable pistonmovable piston

• Cycle consists of 2 adiabatic and 2 Cycle consists of 2 adiabatic and 2 isothermal processesisothermal processes

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. Cold Res.

TTCC

EnginEnginee

Hot Res. Hot Res. TTHH

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 is only half of the ideal efficiency, how much efficiency, how much workwork is done is done during each cycle?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 A refrigerator is an engine operating in engine operating in reverse: Work is done reverse: Work is done onon gas extracting heat gas extracting heat fromfrom cold reservoir and cold reservoir and depositing heat depositing heat intointo hot reservoir.hot reservoir.Win + Qcold = Qhot

WIN = Qhot - Qcold

Cold Res. Cold Res. TTCC

Engine

Hot Res. Hot Res. TTHH

Qhot

Qcold

Win

THE SECOND LAW FOR THE SECOND LAW FOR REFRIGERATORSREFRIGERATORS

It is impossible to It is impossible to construct a refrigerator construct a refrigerator that absorbs heat from a that absorbs heat from a cold reservoir and cold reservoir and deposits equal heat to a deposits equal heat to a hot reservoir with hot reservoir with W = W = 0.0.If this were possible, we could establish perpetual motion!

Cold Res. TC

EnginEnginee

Hot Res. TH

Qhot

Qcold

Second Law of Second Law of ThermodynamicsThermodynamics

• Energy will not flow spontaneously Energy will not flow spontaneously by heat from a cold object to a hot by heat from a cold object to a hot objectobject

• No heat engine operating in a cycle No heat engine operating in a cycle can absorb energy from a reservoir can absorb energy from a reservoir and perform an equal amount of and perform an equal amount of workwork

EntropyEntropy

• Systems tend toward disorderSystems tend toward disorder

• A disorderly arrangement is more A disorderly arrangement is more probable than an orderly oneprobable than an orderly one

• Entropy is a measure of the disorderEntropy is a measure of the disorder

• The entropy of the Universe increases The entropy of the Universe increases in all natural processes (alternate 2in all natural processes (alternate 2ndnd law)law)

EntropyEntropy

• The change in entropy is equal to The change in entropy is equal to the energy flowing into a system the energy flowing into a system (while the system changes from one (while the system changes from one state to another) divided by the state to another) divided by the absolute temperatureabsolute temperature

• Change in S = Qr/TChange in S = Qr/T

COEFFICIENT OF COEFFICIENT OF PERFORMANCEPERFORMANCE

Cold Res. Cold Res. TTCC

EnginEnginee

Hot Res. TH

QH W

QC

The The COP (K)COP (K) of a heat of a heat engine is the ratio of engine is the ratio of the the HEATHEAT QQcc extracted extracted to the net to the net WORKWORK done done WW..

K =

TH

TH- TC

For an For an IDEAL IDEAL

refrigerator:refrigerator:

QC

WK = =

QH

QH- QC

COP EXAMPLECOP EXAMPLEA 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. Cold Res. TTCC

Engine

Hot Res. Hot Res. TTHH

800 J

WQH

500 K

400 K

K = 400 K400 K

500 K - 400 K500 K - 400 K

TC

TH- TC

=

C.O.P. (K) = 4.0

COP EXAMPLE (Cont.)COP EXAMPLE (Cont.)Next we will find QNext we will find QHH by by assuming same K for assuming same K for actual refrigerator actual refrigerator (Carnot).(Carnot).

Cold Res. Cold Res. TTCC

Engine

Hot Res. Hot Res. TTHH

800 J

WQH

500 K

400 K

K =K = QC

QH- QC

QH = 1000 J

800 J800 J

QQHH - 800 J - 800 J=4.0

COP EXAMPLE (Cont.)COP EXAMPLE (Cont.)

Now, can you say how Now, can you say how much work is done in much work is done in each cycle?each cycle?

Cold Res. Cold Res. TTCC

EnginEnginee

Hot Res. THot Res. THH

800 J

W1000 J

500 K

400 K

Work = 1000 J - 800 JWork = 1000 J - 800 J

Work = 200 J

SummarySummary

Q = U + W final - initial)

TheThe First Law of ThermodynamicsFirst Law of Thermodynamics:: The The net heat taken in by a system is equal net heat taken in by a system is equal to the sum of the change in internal to the sum of the change in internal energy and the work done by the energy and the work done by the system.system.

• Isochoric Process: Isochoric Process: V = 0, V = 0, W = W = 0 0

• Isobaric Process: Isobaric Process: P = 0 P = 0

• Isothermal Process: Isothermal Process: T = 0, T = 0, U = U = 0 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 V

T T

Summary (Cont.)Summary (Cont.)

TheThe Second Law of Thermo:Second Law of Thermo: It It is impossible to construct an is impossible to construct an engine that, operating in a engine that, operating in a cycle, produces no effect other cycle, produces no effect other than the extraction of heat than the extraction of heat from a reservoir and the from a reservoir and the performance of an equivalent performance of an equivalent amount of work.amount of work.

Cold Res. Cold Res. TTCC

EnginEnginee

Hot Res. Hot Res. TTHH

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 - QC

QHe = 1 -

TC

TH

The coefficient of performance of a refrigerator:

C C

in H C

Q QK

W Q Q

C

H C

TK

T T