thermodynamics ap physics chapter 15. thermodynamics 13.3 zeroth law of thermodynamics

Thermodynamics

AP Physics

Chapter 15

Thermodynamics

13.3 Zeroth Law of Thermodynamics


If two objects of different temperatures are placed in thermal contact, they will eventually reach the same temperature.

They reach Thermal Equilibrium

Energy flowing in equals the energy flowing out

13.3

Thermal Eq Animation

http://www.yenka.com/freecontent/item.action?quick=zs


Zeroth Law of Thermodynamics – if two systems are in thermal equilibrium with a third system, then they are in thermal equilibrium with each other

Allows for a definition of temperature

(two objects have the same temperature when they are in thermal equilibirum)

13.3

Thermodynamics

13.4 Thermal Expansion


Objects usually expand when heated and contract when cooled.

This can lead to some problems

So we include expansion joints13.4


Change in length is proportional to temperature

So the equation is

is called the coefficient of linear expansion13.4

T0

T

L0

L0

L

TLL 0


Common MisconceptionWhen a object with a hole in it is heated, does the hole get larger or smaller?Imagine an infinitely thin ring

If it is heated length (circumference) of the ring increase

13.4


Volume follows the same relationship

is the coefficient of volume expansion is usually equal to approximately 3

13.4

TVV 0

Thermodynamics

13.7 The Ideal Gas Law


P – pressure in Pa (absolute pressure)

V – volume in m3

R = 8.314J/molK (in standard units)

T – temperature in K

n – quantity in mol

13.7

nRTPV


If the equation is written in terms of molecules

The number of molecules (or atoms) in one mole is

So if

And the number of molecules is

The ideal gas law can be written13.7

nRTPV

2310022.6 xN A

AnNN

RTN

NPV

A


The quantity

This is known as Boltzmann’s constant

Our equation becomes

The constant has a value 13.7

RTN

NPV

A

kN

R

A

NkTPV

KJxk /1038.1 23

Thermodynamics

14.1 Heat as Energy Transfer

14.1 Heat as Energy Transfer

Heat – energy transferred from one object to another because of a difference in temperature

Unit – joule

14.1

Thermodynamics

14.2 Internal Energy


Temperature (K)– measurement of average kinetic energy of the particles

Internal Energy – total energy of all the particles in the object

Heat - transfer

14.2

Temperature

http://www.explorelearning.com/index.cfm?method=cResource.dspView&ResourceID=555


Internal Energy equation

First internal Energy (U) is equal to the number of particles (N) times average kinetic energy

Since

The equation changes to

And since14.2

)( 221 vmNU

kTvmK 232

21

NkTU 23

nRNk

nRTU 23

Thermodynamics

14.3 Specific Heat and Calorimetry


Remember from chemistry

Q – heat transfer in Joules

m – mass in kg

C – specific heat (Cp – constant pressure, Cv – constant volume)

T – change in temperature (oC or K)

14.3

TmCQ


Also when energy is transfered

Expands to

14.3

gainedlost QQ

BBBAAA TCmTCm

Thermodynamics

14.6 Heat Transfer: Conduction


Conduction – by molecular collisions

If heat is transferred through a substance

The rate of heat transfer (Q/t) depends on

14.6

A

l

Tc

TH

l

TTkA

t

Q CH


Q = heat (J)

t = time (s)

k = thermal conductivity (J/smCo)

T = temperature (K or Co)

l = length

14.6

l

TTkA

t

Q CH

Thermodynamics

15.1 The First Law of Thermodynamics


The change in internal energy of a closed system will be equal to the energy added to the system by heat minus the work done by the system on the surroundings.

A broad statement of the law of conservation of energy

15.1

WQU


Internal energy, U, is a property of the system

Work and heat are not

First Law proven by Joule in an experiment

The work done by

The weight as if fell

(done by gravity)

Equaled the energy increase of the liquid in the sealed chamber

15.1


The first law can be expanded

If the system is moving and has potential energy, then

Remember

Q is positive when work flows in

W is positive when the system does work

15.1

WQUUK g

Thermodynamics

15.2 Thermodynamic Processes & the First Law


You need to remember the names of these processes

Isothermal – constant temperature

If temperature is held constant, then

A graph would look like

The curves are called

isotherms15.2

nRTPV .constPV


If T is zero, then U is zero because

We can then show

The work done by the gas in an isothermal process equals the heat added to the gas

15.2

nRTU 23 TnRU 23 )0(0 23 nR

WQU WQ 0 WQ


Adiabatic – no heat is allowed to flow into or out of the system

That leaves First Law as

If the gas expands, the internal energy decreases, and so does the temperature

15.2

0Q WU


Isobaric – pressure is constant

Work = PV

Isovolumetric – volume is

Constant

Work = 0

15.2


Example 1: Isobaric Process

A gas is placed in a piston with an area of .1m2. Pressure is maintained at a constant 8000 Pa while heat energy is added. The piston moves upward 4 cm. If 42 J of energy is added to the system what is the change in internal energy?

15.2


Example 1: Isobaric Process

A gas is placed in a piston with an area of .1m2. Pressure is maintained at a constant 8000 Pa while heat energy is added. The piston moves upward 4 cm. If 42 J of energy is added to the system what is the change in internal energy?

15.2

WQU

VPW xPAW )04)(.1)(.8000(W JW 32

JU 103242


Example 2: Adiabatic Expansion

How much work is done the adiabatic expansion of a car piston if it contains 0.10 mole an ideal monatomic gas that goes from 1200 K to 400 K?

15.2


Example 2: Adiabatic Expansion

How much work is done the adiabatic expansion of a car piston if it contains 0.10 mole an ideal monatomic gas that goes from 1200 K to 400 K?

Adiabatic Q=0

15.2

WQU WU nRTU 2

3 TnRU 23 )1200400)(314.8)(1(.2

3 U JU 998 JW 998


Example 3: Isovolumetric Process

Water with a mass of 2 kg is held at a constant volume in a container, while 10 kJ of energy is slowly added. 2 kJ of energy leaks out to the surroundings. What is the temperature change of the water?

15.2


Example 3: Isovolumetric Process

Water with a mass of 2 kg is held at a constant volume in a container, while 10 kJ of energy is slowly added. 2 kJ of energy leaks out to the surroundings. What is the temperature change of the water?

Constant volume and we don’t have Cv

15.2

WQU TnRU 2

3

QU 8000U

molnmolemm 11118

2000

T )314.8)(111(8000 23 KT 8.5


When a process is cyclical

And the work done is the area bound by the curves

15.2

0U


Example 4: First law in a Cyclic ProcessAn ideal monatomic gas is confined in a cylinder by a movable piston. The gas starts at A with P = 101.3 kPa, V = .005 m3 and T = 300 K.The cycle is

A B is isovolumetric and raises P to 3 atm.

BC Isothermal Expansion (Pave = 172.2 kPa)CA IsobaricCalculate U, Q, and W for each step and for the entire cycle

15.2


15.2

A

B

C



A B is isovolumetric and raises P to 3 atm.

15.2

0)0( PVPW BA

WQU 0 QU QU QTnR 23

nRTPV )300)(314.8()005)(.101300( n nmol 203.0

QT )300)(314.8)(203.0(23

nRTPV T)314.8)(203.0()005)(.303900( T900

Q )300900)(314.8)(203.0(23 JQ 1519

JU 1519



BC Isothermal Expansion (Pave = 172.2 kPa)

15.2

0U

WQU WQ 0 WQ

nRTPV

VPQ

)900)(314.8)(203.0()101300( V 3015.0 mV

)005.0015.0)(172200( Q JQ 1722

JW 1722



CA Isobaric

15.2

WQU

VPW )015.0005.0)(101300( W JW 1013

TnRU 23 )900300)(314.8)(203.0(2

3 U JU 1519

)1013(1591 Q JQ 2532



Totals

15.2

JW 709101317220

JU 0151901519

JQ 709253217221519

Thermodynamics

15.4 The Second Law of Thermodynamics-Intro


The first law deals with conservation of energy.However there are situations that would

conserve energy, but do not occur.1. Falling objects convert from Ug to K to Q

Never Q to K to Ug

2. Heat flows from TH to TC

Never TC to TH

15.4

Falling and Energy

http://www.regentsprep.org/Regents/physics/phys02/falleng/default.htm


The second law explains why some processes occur and some don’t

In terms of heat, the second law could be stated-Heat can flow spontaneously from a hot object

to a cold object; heat will not flow spontaneously from a cold object to a hot object

15.4

Thermodynamics

15.5 Heat Engines

15.5 Heat Engines

The Heat Engine1. Heat flows into

the engine2. Energy is convert-

ed to work3. Remaining heat is

exhausted to cold

15.5

15.5 Heat Engines

Steps in an internal combustion engine1. The intake valve is open, and fuel and air are

drawn past the valve and into the combustion chamber and cylinder from the intake manifold located on top of the combustion chamber (Intake Stroke)

15.5

15.5 Heat Engines

Steps in an internal combustion engine2. With both valves closed, the combination of the

cylinder and combustion chamber form a completely closed vessel containing the fuel/air mixture. As the piston is pushed to the right, volume is reduced & the fuel/air mixture is compressed (Compression Stroke)

15.5

15.5 Heat Engines

Steps in an internal combustion engine3. the electrical contact is opened. The sudden opening

of the contact produces a spark in the combustion chamber which ignites the fuel/air mixture. Rapid combustion of fuel releases heat, produces exhaust gases in the combustion chamber. (Power Stroke)

15.5

15.5 Heat Engines

Steps in an internal combustion engine4. The purpose of the exhaust stroke is to clear

the cylinder of the spent exhaust in preparation for another ignition cycle. (Exhaust Stroke)

15.5

15.5 Heat Engines

Complete cycle

15.5

15.5 Heat Engines

Looking at a steam engineIf the steam were the same temperature throughout-exhaust pressurewould be the sameas the intake pressure-then exhaust work would be the same as intake

workFor net work there must be a T

15.5

15.5 Heat Engines

Actual Efficiency of an engine is defined as

The ratio of work to heat inputSinceWe can write the efficiency as

H

WeQ

15.5

H CQ W Q

H C

H

Q Qe

Q

1 C

H

Qe

Q

15.5 Heat Engines

Carnot Engine (ideal) – no actual Carnot engineA four cycle engine1. Isothermal expansion (=0, Q=W)2. Adiabatic expansion (Q=0, U=-W)3. Isothermal compression 4. Adiabatic compressionEach process was considered reversibleThat is that each step is done very slowlyReal reactions occur quickly – there would be

turbulence, friction - irreversible

15.5

15.5 Heat Engines

The Carnot efficiency is defined as

15.5

1 Cideal

H

Te

T

Thermodynamics

15.6 Refrigerators, Air Conditioners

15.6 Refrigerators, Air Conditioners

The reverse of Heat Engines

Work must be done – because heat flows from hot to cold

15.6

Thermodynamics

15.7 Entropy and the 2nd Law of Thermodynamics


Entropy – a measure of the order or disorder of a system

Change in entropy is defined as

Q must be added as a reversible process at a constant temperature

15.7

QS

T


The Second Law of Thermodynamics – the entropy of an isolated system never decreases. It can only stay the same or increase.

Only idealized processes have a S=0Or – the total entropy of any system plus that of

its environment increases as a result of any natural process

15.7


Example – A sample of 50 kg of water at 20oC is mixed with 50 kg at 24oC. The final temperature is 22oC. Estimate the change in Entropy.

The reaction does not occur at constant temperature, so use the average temperature to estimate the entropy change.

15.7


Example – A sample of 50 kg of water at 20oC is mixed with 50 kg at 24oC. The final

temperature is 22oC. Estimate the change in Entropy. Cold Water

Hot Water

15.7

PQ mC T (50)(4186)(22 20)Q 418,600Q J418600294S 1424 /CS J K

(50)(4186)(22 24)Q 418600Q J418600296S 1414 /HS J K 1414 1424S 10 /S J K

Thermodynamics

15.8 Order to Disorder

15.8 Order to Disorder

2nd Law can be stated – natural processes tend to move toward a state of greater disorder

15.7

thermodynamics ap physics chapter 15. thermodynamics 13.3 zeroth law of thermodynamics

Documents

temperature slide

internal energy slide

energy transfer slide

heat energy

conduction slide

calorimetry slide

thermal expansion slide

ideal gas law slide