3b Thermodynamics

•Specific heat capacity, c

•Latent heat capacity, L

•Change of phase, evaporation

•First law of thermodynamics,

byin WQU

Real Life Application of specific heat capacity, c

Thinking QuestionCheese is still hot a long time, after being taken out of the oven. Why?

Heat capacity, Cnote: C = m.c

m = Mass of the substance, c = Specific heat capacity

The heat capacity, C, of a body is defined as

the amount of heat Q required to raise its

temperature by 1 Kelvin without going

through a change in state.

The S.I. unit is J K–1

QC

Why is C worth learning?

• Demo video on Heat capacity, C

• Water Balloon Heat Capacity.MPG

Nice explanation

• http://www.asknlearn.com/contentpackaging/14297/RS14/Container.html

Specific Heat Capacity

The specific heat capacity, c, of a body is defined as the amount of heat required to raise the temperature of 1 kg of the body by 1 Kelvin.

The S.I. unit is J kg–1 K–1

m

Qc

Substance J/ kg oC

Aluminum 900Bismuth 123

Copper 386

Brass 380

Gold 126

Lead 128Silver 233

Tungsten 134

Zinc 387

Mercury 140

Alcohol (ethyl) 2400

Water 4186Ice (-10 0C) 2050

Specific Heat Capacity, c

of some substances at 25 oC and atmospheric pressure

What does larger specific heat capacity mean?

• Specific Heat with Rods and Wax.MPG• Hint:

• For the same mass, m, • cAl > c steel> c lead means• QAl > Q steel> Q lead for the same ∆θ

mcQ

To solve problems mcQ

if

0gainQ

Example 1• How much energy is required when a

piece of copper of mass 0.275 kg is heated from 14.0 C to 100.0 C? (Specific heat capacity of copper = 380 J kg–1 K–1)

J 31099.8

14100380275.0

,

mcQcopperbygainedHeat

m = 0.275 kgc = 380 J/(kg.K)

θf - θi = (100 – 14)Q

Example 2

• An electric heater of 5 kW is used to heat up a piece of copper of mass 0.5 kg from 10 C to 90 C. Calculate the time taken. (ccopper = 380 J kg–1 K–1)

m = 0.5 kgc = 380 J/(kg.K)

θf - θi = (90 – 10)

P= 5x103 W

heater

• Q = m c (θf - θi )

• (5x103)(t)= (0.5)(380)(90-10)

• t = 3.0 s

m = 0.5 kgc = 380 J/(kg.K)

θf - θi = (90 – 10)

P= Q/t = 5x103 W

heater

Example 3• An ethnic restaurant serves coffee in copper mugs. A waiter

fills a cup having a mass of 0.10 kg, initially at 20 °C, with 0.20 kg of coffee initially at 70 °C. What is the final temperature after the coffee and the cup has attained thermal equilibrium? Assume that there is no heat loss to the surroundings.

• (ccoffee = 4200 J kg–1 K–1; ccopper = 380 J kg–1 K–1)

H2Om = 0.2 kg

c = 4200 J/(kg.K)θf - θi = (θ – 70)

Q= ?cupm = 0.1 kg

c = 380 J/(kg.K)θf - θi = (θ – 20)

• ∑Q = 0

• mcup.ccup (θ – 20) + mH20.cH20 (θ – 70) = 0

• (0.1)(380)(θ – 20) + (0.2)(4200)(θ – 70)=0

• Solve for θ

• θ = 67.8 0C = 68 0C

H2Om = 0.2 kg

c = 4200 J/(kg.K)θf - θi = (θ – 70)

Q= ?cupm = 0.1 kg

c = 380 J/(kg.K)θf - θi = (θ – 20)

Flash Simulation• heat_metal.swf

• Thinking and reflection• The random fluctuation of the thermometer

instrument, relate back to chapter 1 called uncertainty or random error

• the assumption no heat loss to surrounding is shown in simulation but not easily explain with clarity with words

• The concept of thermal equilibrium is vividly show in simulation when calculations alone fail to show the process of thermal equilibrium

Electrical method to determine c

AV

Heater & material under

test

Connect to power supply

Remember PWS6 electrical setup for heater power?Term 1 practical YJC.

• Example 4• Hand write

Phase Changes

Q= mcsolid∆θ

Q= mcliquid∆θ

Q= mcgas∆θ

Change of phase-melting

• Latent heat of fusion, Lf

• Heat is supplied, Q

• Temperature remains constant. T=constant

Q= mLf

Change of phase-boiling• Latent heat of vaporisation, Lv

• Heat is supplied, Q• Energy (Work done by system) is needed to

expand against atmosphere Wby

• Temperature remains constant, T =constantQ= mLv

Change of phaseSpecific Latent heat of vaporisation >

Specific latent heat of fusion. Lv > Lf

Energy is needed to work against atmosphere, Wby

• Difference in PE between molecules in liquid to gaseous > in solid to liquid

Q= mLv

Q= mLf

Select 2

• http://www.asknlearn.com/contentpackaging/14297/RS14/Container.html

cooling effect during evaporation

Molecules with higher KE and are near to surface, are able to overcome intermolecular force and escape from liquid surface.

Molecules with lower KE are left behind in the liquid.

As a result, the average molecular KE decreases and hence the temperature drops.

Evaporation

• Occurs at all temperature (including boiling)

• Average KE of liquid decreases

• Cooling process

http://www.colorado.edu/physics/2000/applets/bec.html

Video on Real Life Application of Evaporation & Condensation

• Drinking Bird.MPG

• Clever synthesis of many Physics ideas

• You can read up about it by “Goggling” Dippy Bird

• http://science.howstuffworks.com/question608.htm

Now let’s test yourselvesQn 1For a given liquid at atmospheric pressure, which process can occur at any temperature?

A) BoilingB) EvaporationC) MeltingD) Solidification

Now let’s test yourselvesQn 2Latent heat of vaporisation is the energy required to

A) Separate the molecules of the liquidB) Force back the atmosphere to make space for the

vapourC) Increase the average molecular speed in the liquid

phase to that in the gas phaseD) Separate the molecules and to force back the

atmosphereE) Separate the molecules and to increase their

average molecular speed to that in the gas phase.

Latent heat, L

The S.I. unit for l is J kg–1.

mLQ

Specific latent heat of fusion, L f

Specific latent heat of vaporisation, L v

Q = m L f Q = m L v

No change in temperature

• Example 5

Total heat gained = Total heat lost / supplied

Or

∑ (heat) = 0

summary of Specific heat capacity, Specific latent heat

Specific heat capacity, c : Q=mc ∆θ

Specific latent heat of fusion, Lf : Q= m Lf

Specific latent heat of vaporisation, Lv: Q= m Lv

In a closed system,

3 most important equations:

Phase Change of water

Electrical method to determine L

AV

Heater & material under test

Connect to power supply

• Example 6

• Example 7

• Suggested to hand write, for ease of following the chain of thought

Example 6An electric kettle contains 1.5 kg of water at 100 C and is powered by a 2.0 kW electric element. If the thermostat of the kettle fails to operate, calculate the time taken for the kettle to boil dry.(lv of water = 2000 kJ kg–1)

t = 1500 s

heat gained by water (boil) = heat supplied by kettle

Example 7 (a) better to use h/t

Therefore, lv = 1.98 x 106 J Kg-1

1st set up:

Mass = 0.1 kgPower = 240 WTime taken = 15 min

2nd set up:

Mass = 0.15 kgPower = 350 WTime taken = 15 min

a) Calculate the specific latent heat of vaporisation.

Heat supplied = heat gained by liquid + heat loss

P t = m lv + h h = P t – m lv

Since time taken is the same, h will also be the same.

(240)(15 x 60) – (0.1) lv = (350)(15 x 60) – (0.15) lv

Example 7 (b)

= 8.3 %

b) Find out what percentage of the heat supplied was lost in 1st attempt.

Percentage of heat loss =

Heat loss, h = (240)(15 x 60) – (0.1)(1.98 x 106)

Using h = P t – m lv for 1st set up:

= 1.80 x 10 4 J

=

First law of thermodynamics

• Conservation of energy

• To change the internal energy of system

1. Forces to do work

2. By means of heat transfer

3. Combination of both

byin WQU

First law of Thermodynamics

The increase in internal energy of a closed system is the sum of the heat supplied to the system subtract the work done by the system.

byWQU Increase in internal energy of system

Heat supplied to system

Work done by system

(+)(+) (+)

Volume will

increase.

Temperature will

increase.

First law of Thermodynamics

The increase in internal energy of a closed system is the sum of the heat supplied to the system subtract the work done by the system.

byWQU Increase in internal energy of system

Work done by system

Heat supplied to system

↑ to

In a closed system, the mass is fixed. Therefore no mass can be added into

or removed from the system.

Recap: Internal energy

Internal energy, U of a system is the total sum of the kinetic energy, Ek and the potential energy, Ep of the molecules in the system.

Kinetic energy, Ek is associated with the temperature.

Potential energy, Ep depends on the separation of molecules.

moleculespk )EE(U

Mechanical Equivalent of Heat.MPG

• Video showing Q is a form of energy.

Nice explanation

• http://www.asknlearn.com/contentpackaging/14297/RS15/Container.html

byin WQU

onin WQU

onby WW

Remember one form

First law of Thermodynamics

byWQU Sign convention:

Descriptors Meaning Signs

Heat enters system

Heat supplied to system

+ Q

Heat leaves system

Heat given out by system

– Q

onby WW

First law of Thermodynamics

Descriptors Meaning Signs

Volume of system decreases

Work done on system

- Wby

Volume of system increases

Work done by system

+ Wby

Sign convention:byWQU

onby WW

First law of Thermodynamics

Sign convention:

Descriptors Meaning Signs

Temperature increases

Increase in internal energy

+ ΔU

Temperature decreases

Decrease in internal energy

– ΔU

byWQU

onby WW

Sign conventions

conventions Opposite convention

∆U Increase in internal energy

Decrease in internal energy

Q Heat supplied to the system

Heat loss by the system

Wby Work done by the system

Work done on the system

byin WQU

onby WW

Internal energy of ideal gas

• Recall internal energy • does not possess internal potential energy• PE=0 by definition for gas• is the sum of the molecular KE of particles• KE temperature

nRT2

3 U gas ideal NkT

2

3 U gas ideal

Internal energy of ideal gas

nRT2

3 U gas ideal NkT

2

3 U gas ideal

Average translational KE per molecule = kT2

3

Total KE for N molecules in the gas = NkT2

3

nRT2

3

Work done by system

V

A

B

p

Work done by gas in compression is a negative number

p

V

A

B

Work done by gas in expansion is a positive number

Work done by system (gas)

V

p

Wby = area under the p-v graph =

dvp

Wby= p ∆V = p (Vf – Vi) = negative number

Work done-cycle

p

V

By gas

On gas

A B

CD

Process

A B : by gas

B C : zero

C D : on gas

D A : zero

Overall : work done by gas

WD in a cycle = area of enclosed regionWD ABCDA = WD by gas = area of enclosed regionWD ADCBA = - WD by gas = - area of enclosed region

Video on Fire Syringe

• After watching the video, think about how this video is related to the table

Process What it means

Iso-baric Constant pressure

Iso-choric /Iso-volumetric Constant volume

Iso-thermal Constant temperature

AdiabaticNo heat enters or leaves the system

Notice isothermal lines.

• http://www.lon-capa.org/~mmp/applist/pvt/pvt.htm

• http://mysite.verizon.net/pmrenault/thermo.html

T=200K

T=150K

• http://mysite.verizon.net/pmrenault/thermo.html

P=constant

• http://mysite.verizon.net/pmrenault/thermo.html

V=constant

Thermodynamic processesThere are 4 special processes to consider:

Process What it means

Iso-baric Constant pressure

Iso-choric /Iso-volumetric Constant volume

Iso-thermal Constant temperature

AdiabaticNo heat enters or leaves the system

p

V

T2T1

Isothermal processes

Thermodynamic ProcessesThere are 4 special thermodynamic processes:

Process What it means Implications

Iso-baric Constant pressure P = constant

Iso-choric/ Iso-volumetric

Constant volume V = constant

∆V = 0 m3

W = P ∆V = 0 J

Iso-thermal Constant temperature

T = constant

∆U = 0 J

Adiabatic No heat enters or leaves system

Q = 0 J

∆U = W

Thermodynamic ProcessesThere are 4 special thermodynamic processes:

Process What it means Implications

Iso-baric Constant pressure P = constant

Iso-choric/ Iso-volumetric

Constant volume V = constant

∆V = 0 m3

W = P ∆V = 0 J

Iso-thermal Constant temperature

T = constant

∆U = 0 J

Adiabatic No heat enters or leaves system

Q = 0 J

∆U = W

Thermodynamic ProcessesThere are 4 special thermodynamic processes:

Process What it means Implications

Iso-baric Constant pressure P = constant

Iso-choric/

Iso-volumetric

Constant volume V = constant

∆V = 0 m3

Wby = P ∆V = 0 J

Iso-thermal Constant temperature

T = constant

∆U = 0 J

Adiabatic No heat enters or leaves system

Q = 0 J

∆U = -Wby

Thermodynamic ProcessesThere are 4 special thermodynamic processes:

Process What it means Implications

Iso-baric Constant pressure P = constant

Iso-choric/

Iso-volumetric

Constant volume V = constant

∆V = 0 m3

W = P ∆V = 0 J

Iso-thermal Constant temperature

T = constant

∆U = 0 J

Adiabatic No heat enters or leaves system

Q = 0 J

∆U = -Wby

Cool video on Q=0

• Adiabatic Expansion.MPG

pV diagramIndicator diagram or pV diagram is simply a Pressure vs Volume graph

p / Pa

V / m3

V = volume of containerp = Pressure exerted on the walls by gas molecules

pV diagramIndicator diagram or pV diagram is simply a Pressure vs Volume graph

p / Pa

V / m3

∆U = Q - Wby

Heat flow in

U= (3/2).nRT

So

ΔU = (3/2)nRΔT pV diagram

Wby = P ∆V= area under pV diagram

Iso-volumetric processConstant volume

p / Pa

V / m3

V remains unchanged when P changes.

∆U = Q - Wby

pV diagram

Area = 0 J

Wby = 0 J

∆U = Q

60

Iso-baric processConstant pressure

p / Pa

V / m3

P remains unchanged when V changes.

∆U = Q - Wby

pV diagram

100

∆V = 50

Area = 100 x 50 = 5000 J

Wby = + 5000 J or – 5000 J?

→ V increases→ Work done BY gas

→ + Wby

Wby = + 5000 J

Iso-baric processConstant pressure

p / Pa

V / m3

P remains unchanged when V changes.

∆U = Q -Wby

pV diagram

100

∆V = 50

Area = 100 x 50 = 5000 J

Wby = + 5000 J or – 5000 J?

→ V decreases→ Work done ON gas

→ Wby negative number

Wby = - 5000 J

Iso-thermal processConstant Temperature

p / Pa

V / m3

∆U = Q - Wby

pV diagram

Area = W

No need to calculate W!

T = 80 oc

T remains unchanged along the curve.

However, calculation of W in Iso-thermal process is not in A level syllabus!

Iso-thermal processConstant Temperature

p / Pa

V / m3

T remains unchanged along the curve.

∆U = Q - Wby

pV diagram

T remains constant

0= Q – Wby

∆U = 0 J

T = 80 oc

T = 90 oc

U also remains constant

[ Wby = negative number if process is decrease in volume

Wby = postive number if process is increase in volume]

50

60

70 Higher T?

Finding Wby from pV diagram

Important points:

1. Find the area under graph = magnitude of W.

p / Pa

V / m3

2. If volume decreases (compression), work is done ON gas → Wby is negative number

3. If volume increases (expansion), work is done BY gas → Wby is positive number

4. Work done ON gas = – Work done BY gas

• Example 8

• Example 9

• Example 10

• Example 11

• Example 12

• Example 13

• Example 14

C200

500

Example 8

Find the work done by the gas in going from the situation represented by point A, through point B, to point C.

300 500 800 V/cm3

p/KPa

ABArea = (200+500)(200/2) + (500)(300)

= 2.20 x 105 J

Magnitude of Wby:

Wrong answer! Can you spot the mistake?

Draw this!

C200

500

Example 8

Find the work done on the gas in going from the situation represented by point A, through point B, to point C.

Wby gas = + 220 J

= 220 J

300 500 800 V/cm3

p/KPa

AB

Since volume decreases (compression) → Wby is a positive number

Area = (2 x 105 + 5 x 105)(200 x 10-6)/2 + (5 x 105)(300 x 10-6)

Magnitude of W:

Example 12 – (a)A thermodynamic system is taken from initial state A to state B and back again to A via state C.

V/cm3

p/KPa

A B

C

Initial state

31

20

40

∆U = Q - Wby

A → B

B → C

C → A – 2

+0.04

0

-0.06

a) Fill up the table with ‘+’, ‘–’ or ‘0’.

+

+

–

+

+

pV diagramIso-thermal curve1st law of thermodynamics

Example 12 – (b)A thermodynamic system is taken from initial state A to state B and back again to A via state C.

V/cm3

p/KPa

A B

C

Initial state

31

20

40

2

b) Calculate net work done by the system for one complete cycle.

WDA → B = + (20 x 103)(2 x 10-6) = 0.04 J

WDB → C = 0 J

WDC → A = – [(20 + 40) x 103)](2 x 10-6)/2 = – 0.06 JTherefore net work done by system = 0.04 + 0 + (– 0.06)

= – 0.02 J

(area enclosed by loop)

Example 13An ideal gas undergoes a cycle of changes A → B → C → D as shown in the figure below. Complete the table.

V/10-3 m3

p/KPa

AB

DC

Initial state

0.30.1

100

200

∆U = Q - Wby

A → B

B → C

C → D

D → A

– 50

25

140

+ 20

0

– 40

0

1. Draw table2. Known

values3. Find W from

pV diagram

Example 13An ideal gas undergoes a cycle of changes A → B → C → D as shown in the figure below. Complete the table.

V/10-3 m3

p/KPa

AB

DC

Initial state

0.30.1

100

200

∆U = Q - Wby

A → B

B → C

C → D

D → A

– 50 – 70

25

140

– 20

+ 25

100

0

+40

0– 75– 75

4. Find ∆U 3. Find W from pV diagram

5. Finally, find QQ may be found

using mc∆θ

Summary

For cycle, total ∆U = 0 J

Example 13: Won explanationAn ideal gas undergoes a cycle of changes A → B → C → D as shown in the figure below. Complete the table.

V/10-3 m3

p/KPa

AB

DC

Initial state

0.30.1

100

200

∆U = Q - Wby

A → B

B → C

C → D

D → A

– 50 – 70

25

140

– 20

+ 25

100

0

+40

0– 75– 75

Summary

+ 20

0

– 40

0

Won

onby WW onin WQU

Example 14 – (a)Mass of gas = 1 g Initial pressure = 1.01 x 105 PaMolar mass = 18 Initial temperature = 27 oc

V/m3

p/ x 105 Pa

A

BC

V2V1

1.01

a) Illustrates these changes on a pV diagram.

A: the gas is heated at constant pressure to 127 oc

B: compressed isothermally to its original volume

C: the gas is cooled at constant volume to its original temperature27 oc

127 oc

P2

Initial state?

Example 14 – (b)Mass of gas = 1 g Initial pressure = 1.01 x 105 PaMolar mass = 18 Initial temperature = 27 oc

V/m3

p/ x 105 Pa

A

BC

V2V1

1.01

b) How much work does the gas do in pushing the piston during stage A?

P2

Find V1:

331

15

10371

2732731818

110011

m.V

))(.(V).(

nRTpVUse

V1 = 1.37 x 10-3 m3

Example 14 – (b)Mass of gas = 1 g Initial pressure = 1.01 x 105 PaMolar mass = 18 Initial temperature = 27 oc

b) How much work does the gas do in pushing the piston during stage A?

Find V2:

332

23

2

2

1

1

10831

27312727327

10371

m.V

V.

T

V

T

V

V/m3

p/ x 105 Pa

A

BC

V2V1

1.01

P2

V1 = 1.37 x 10-3 m3

V2 = 1.83 x 10-3 m3

Example 14 – (b)Mass of gas = 1 g Initial pressure = 1.01 x 105 PaMolar mass = 18 Initial temperature = 27 oc

b) How much work does the gas do in pushing the piston during stage A?

Therefore work done by gas

= + (area under graph)

V/m3

p/ x 105 Pa

A

BC

V2V1

1.01

P2

V1 = 1.37 x 10-3 m3

= (1.01 x 105)[(1.83 – 1.37) x 10-3]

= 46.5 J

V2 = 1.83 x 10-3 m3

Example 14 – (c)Mass of gas = 1 g Initial pressure = 1.01 x 105 PaMolar mass = 18 Initial temperature = 27 oc

c) What is the change in the internal energy in stage A?

V/m3

p/ x 105 Pa

A

BC

V2V1

1.01

P2

= + 69.3 J

))(.(

TnRU

2712731818

1

2

32

3

Assuming negligible potential energy,

∆U = ∆KE = (3/2) NK ∆T = (3/2) nR ∆T

27 oc127 oc

Example 14 – (d)Mass of gas = 1 g Initial pressure = 1.01 x 105 PaMolar mass = 18 Initial temperature = 27 oc

d) What is the heat input to the cylinder in stage A?

V/m3

p/ x 105 Pa

A

BC

V2V1

1.01

P2

∆U = Q – Wby

69.3 = Q – 46.5

115.8 J = Q

Note: Wby gas = + 46.5 J → Won gas = – 46.5 J

Example 14 – (e)Mass of gas = 1 g Initial pressure = 1.01 x 105 PaMolar mass = 18 Initial temperature = 27 oc

e) What is the change in internal energy of the gas in stage B?

V/m3

p/ x 105 Pa

A

BC

V2V1

1.01

P2

Since ∆T = 0,

therefore ∆U = 0 J,

since ∆U = (3/2) nR ∆T

Is there any change in temperature in stage B?

Example 14 – (f)Mass of gas = 1 g Initial pressure = 1.01 x 105 PaMolar mass = 18 Initial temperature = 27 oc

f) How much heat must be extracted from the gas in stage C?

V/m3

p/ x 105 Pa

A

BC

V2V1

1.01

P2

Therefore ∆U for stage C = – 69.3 J

For a cycle, total ∆U = 0 J

∆U = + 69.3 J

∆U = 0 J

∆U = ? J

∆U = Q - Wby

– 69.3 = Q - 0– 69.3 J =Q = Qin

Q to be released = +69.3 J