thermodynamics wee

3b Thermodynamics
Specific heat capacity, cLatent heat capacity, LChange of phase, evaporationFirst law of thermodynamics,

Real Life Application of specific heat capacity, c

Thinking Question

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

Heat capacity, C
note: 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 K1

Why is C worth learning?
Demo video on Heat capacity, CWater 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 kg1 K1

Specific Heat Capacity, c

of some substances at 25 oC and atmospheric pressure
SubstanceJ/ kg oCAluminum900Bismuth123Copper386Brass380Gold126Lead128Silver233Tungsten134Zinc387Mercury140Alcohol (ethyl)2400Water4186Ice (-10 0C)2050

What does larger specific heat capacity mean?
Specific Heat with Rods and Wax.MPGHint:For the same mass, m, cAl > c steel> c lead meansQAl > Q steel> Q lead for the same

To solve problems

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 kg1 K1)
m = 0.275 kg

c = 380 J/(kg.K)

f - i = (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 kg1 K1)
m = 0.5 kg

c = 380 J/(kg.K)

f - i = (90 10)

P= 5x103 W

heater

Q = m c (f - i )(5x103)(t)= (0.5)(380)(90-10)t = 3.0 s
m = 0.5 kg

c = 380 J/(kg.K)

f - i = (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 kg1 K1; ccopper = 380 J kg1 K1)
H2O

m = 0.2 kg

c = 4200 J/(kg.K)

f - i = ( 70)

Q= ?

cup

m = 0.1 kg

c = 380 J/(kg.K)

f - i = ( 20)

Q = 0mcup.ccup ( 20) + mH20.cH20 ( 70) = 0 (0.1)(380)( 20) + (0.2)(4200)( 70)=0Solve for = 67.8 0C = 68 0C
H2O

m = 0.2 kg

c = 4200 J/(kg.K)

f - i = ( 70)

Q= ?

cup

m = 0.1 kg

c = 380 J/(kg.K)

f - i = ( 20)

Flash Simulation
heat_metal.swfThinking and reflectionThe random fluctuation of the thermometer instrument, relate back to chapter 1 called uncertainty or random errorthe assumption no heat loss to surrounding is shown in simulation but not easily explain with clarity with wordsThe 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

A

V

Heater & material under test

Connect to power supply

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

Example 4Hand write

Phase Changes

Q= mcsolid

Q= mcliquid

Q= mcgas

Change of phase-melting
Latent heat of fusion, LfHeat is supplied, QTemperature remains constant. T=constant
Q= mLf

Change of phase-boiling
Latent heat of vaporisation, LvHeat is supplied, QEnergy (Work done by system) is needed to expand against atmosphere WbyTemperature remains constant, T =constant
Q= mLv

Change of phase

Specific 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

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

Video on Real Life Application of Evaporation & Condensation
Drinking Bird.MPGClever synthesis of many Physics ideas You can read up about it by Goggling Dippy Bird http://science.howstuffworks.com/question608.htm
When water evaporates from the fuzz on the Dippy Bird's head, the head is cooled.

The temperature decrease in the head condenses the methylene chloride vapor, decreasing the vapor pressure in the head relative to the vapor pressure in the abdomen.

The greater vapor pressure in the abdomen forces fluid up through the neck and into the head.

As fluid enters the head, it makes the Dippy Bird top-heavy.

The bird tips. Liquid travels to the head. The bottom of the tube is no longer submerged in liquid.

Vapor bubbles travel through the tube and into the head. Liquid drains from the head, displaced by the bubbles.

Fluid drains back into the abdomen, making the bird bottom-heavy.

The bird tips back up.

If the bird dips into a cup of water, the fuzzy material absorbs water again and the cycle starts over.

Now lets test yourselves

Qn 1

For a given liquid at atmospheric pressure,

which process can occur at any

temperature?

Boiling

Evaporation

Melting

Solidification

Now lets test yourselves

Qn 2

Latent heat of vaporisation is the energy required to

Separate the molecules of the liquid

Force back the atmosphere to make space for the vapour

Increase the average molecular speed in the liquid phase to that in the gas phase

Separate the molecules and to force back the atmosphere

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 kg1.

No change in temperature
Specific latent heat of fusion, L fSpecific latent heat of vaporisation, L v Q = m L f Q = m L v

Example 5

summary of Specific heat capacity, Specific latent heat

Total heat gained = Total heat lost / supplied

Or

(heat) = 0

Specific heat capacity, c : Q=mc

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

A

V

Heater & material under test

Connect to power supply

Example 6Example 7Suggested to hand write, for ease of following the chain of thought

Example 6

An 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 kg1)

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 kg

Power = 240 W

Time taken = 15 min

2nd set up:

Mass = 0.15 kg

Power = 350 W

Time taken = 15 min

a) Calculate the specific latent heat of vaporisation.

Heat supplied = heat gained by liquid + heat loss

P t = m lv + h

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

h = P t m lv

Example 7 (b)

= 8.3 %

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

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

Percentage of heat loss =

=

First law of thermodynamics
Conservation of energyTo change the internal energy of system
Forces to do work

By means of heat transfer

Combination of both

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.

Increase in internal energy of system

Heat supplied to system

Work done by system

(+)

(+)

(+)

Volume will increase.

Temperature will increase.


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.

to

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

Increase in internal energy of system

Work done by system

Heat supplied to 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.

Mechanical Equivalent of Heat.MPG
Video showing Q is a form of energy.

Nice explanation
Remember one form


Sign convention:
DescriptorsMeaning SignsHeat enters systemHeat supplied to system+ QHeat leaves systemHeat given out by system Q


Sign convention:
DescriptorsMeaning SignsVolume of system decreasesWork done on system- WbyVolume of system increasesWork done by system+ Wby


Sign convention:
DescriptorsMeaning SignsTemperature increasesIncrease in internal energy+ UTemperature decreasesDecrease in internal energy U

Sign conventions
conventionsOpposite conventionUIncrease in internal energyDecrease in internal energyQHeat supplied to the systemHeat loss by the systemWbyWork done by the systemWork done on the system

Internal energy of ideal gas
Recall internal energy does not possess internal potential energyPE=0 by definition for gasis the sum of the molecular KE of particlesKE temperature

Internal energy of ideal gas

Average translational KE per molecule =

Total KE for N molecules in the gas =

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 =

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

Work done-cycle

p

V

By gas

On gas

A

B

C

D

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 region

WD ABCDA = WD by gas = area of enclosed region

WD 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 tableProcessWhat it meansIso-baricConstant pressureIso-choric /Iso-volumetricConstant volumeIso-thermalConstant temperatureAdiabaticNo 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

P=constant

V=constant

Thermodynamic processes

There are 4 special processes to consider:

Isothermal processes
ProcessWhat it meansIso-baricConstant pressureIso-choric /Iso-volumetricConstant volumeIso-thermalConstant temperatureAdiabaticNo heat enters or leaves the system
p

V

T2

T1

Thermodynamic Processes

There are 4 special thermodynamic processes:
ProcessWhat it meansImplicationsIso-baricConstant pressureP = constantIso-choric/ Iso-volumetricConstant volumeV = constantV = 0 m3W = P V = 0 JIso-thermalConstant temperatureT = constantU = 0 JAdiabaticNo heat enters or leaves systemQ = 0 JU = W


ProcessWhat it meansImplicationsIso-baricConstant pressureP = constantIso-choric/ Iso-volumetricConstant volumeV = constantV = 0 m3Wby = P V = 0 JIso-thermalConstant temperatureT = constantU = 0 JAdiabaticNo heat enters or leaves systemQ = 0 JU = -Wby


ProcessWhat it meansImplicationsIso-baricConstant pressureP = constantIso-choric/ Iso-volumetricConstant volumeV = constantV = 0 m3W = P V = 0 JIso-thermalConstant temperatureT = constantU = 0 JAdiabaticNo heat enters or leaves systemQ = 0 JU = -Wby

Cool video on Q=0
Adiabatic Expansion.MPG

pV diagram

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

p / Pa

V / m3

V = volume of container

p = Pressure exerted on the walls by gas molecules

pV diagram

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

p / Pa

V / m3

U = Q - Wby

Wby = P V

= area under pV diagram

Heat flow in

U= (3/2).nRT

So

U = (3/2)nRT

pV diagram

Iso-volumetric process

Constant volume

p / Pa

V / m3

U = Q - Wby

pV diagram

Area = 0 J

Wby = 0 J

U = Q

V remains unchanged when P changes.

60

Iso-baric process

Constant pressure

p / Pa

V / m3

U = Q - Wby

pV diagram

Area = 100 x 50 = 5000 J

Wby = + 5000 J or 5000 J?

V increases

Work done BY gas

+ Wby

Wby = + 5000 J

P remains unchanged when V changes.

100

V = 50

Iso-baric process

Constant 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 process

Constant Temperature

p / Pa

V / m3

U = Q - Wby

pV diagram

Area = W

No need to calculate W!

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

T = 80 oc

T remains unchanged along the curve.

Iso-thermal process

Constant 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

U also remains constant

[ Wby = negative number if process is decrease in volume

Wby = postive number if process is increase in volume]

Higher T?

T = 90 oc

50

60

70

Finding Wby from pV diagram

Important points:

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

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

p / Pa

V / m3

Example 8Example 9Example 10Example 11Example 12Example 13Example 14

Example 8

C

200

500

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

A

B

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

= 2.20 x 105 J

Magnitude of Wby:

Wrong answer! Can you spot the mistake?

Draw this!

Example 8

C

200

500

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

A

B

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

3

1

20

40

A B

B C

C A

2

+0.04

0

-0.06

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

+

+

+

+

pV diagram

Iso-thermal curve

1st law of thermodynamics

U = Q - Wby

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

3

1

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 J

Therefore net work done by system = 0.04 + 0 + ( 0.06) = 0.02 J

(area enclosed by loop)

Example 13

An 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

A

B

D

C

Initial state

0.3

0.1

100

200

A B

B C

C D

D A

50

25

140

+ 20

0

40

0

1. Draw table

2. Known values

3. Find W from pV diagram

U = Q - Wby

Example 13


V/10-3 m3

p/KPa

A

B

D

C

Initial state

0.3

0.1

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 Q

Q may be found using mc

Summary

For cycle, total U = 0 J

Example 13: Won explanation


V/10-3 m3

p/KPa

A

B

D

C

Initial state

0.3

0.1

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

Example 14 (a)

Mass of gas = 1 gInitial pressure = 1.01 x 105 Pa

Molar mass = 18Initial temperature = 27 oc

V/m3

p/ x 105 Pa

A

B

C

V1

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 temperature

P2

Initial state?

V2

27 oc

127 oc

Example 14 (b)



V/m3

p/ x 105 Pa

A

B

C

V1

1.01

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

P2

Find V1:

V1 = 1.37 x 10-3 m3

V2

Example 14 (b)




Find V2:

V/m3

p/ x 105 Pa

A

B

C

V1

1.01

P2

V1 = 1.37 x 10-3 m3

V2 = 1.83 x 10-3 m3

V2

Example 14 (b)




Therefore work done by gas

= + (area under graph)

V/m3

p/ x 105 Pa

A

B

C

V1

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

V2

Example 14 (c)



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

V/m3

p/ x 105 Pa

A

B

C

V1

1.01

P2

= + 69.3 J

Assuming negligible potential energy, U = KE = (3/2) NK T = (3/2) nR T

V2

27 oc

127 oc

Example 14 (d)



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

V/m3

p/ x 105 Pa

A

B

C

V1

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

V2

Example 14 (e)



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

V/m3

p/ x 105 Pa

A

B

C

V1

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?

V2

Example 14 (f)



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

V/m3

p/ x 105 Pa

A

B

C

V1

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

V2

(

)

q

D

=

Q

C

on

by

W

W

-

=

nRT

2

3

U

gas

ideal

=

by

in

W

Q

U

-

=

D

NkT

2

3

U

gas

ideal

=

kT

2

3

NkT

2

3

nRT

2

3

dv

p

q

D

=

mc

Q

i

f

q

q

q

-

=

D

(

)

(

)

(

)

J

3

10

99

.

8

14

100

380

275

.

0

,

=

-

=

D

=

q

mc

Q

copper

by

gained

Heat

(

)

q

D

=

m

Q

c

mL

Q

=

on

in

W

Q

U

+

=

D

=

0

gain

Q

t

P

m

=

v

l

)

)(

2000

(

)

10

2

)(

5

.

1

(

6

t

=

%

Pt

h

100

%

)

)(

(

.

100

60

15

240

10

80

1

4

molecules

p

k

)

E

E

(

U

+

=

by

W

Q

U

-

=

D

by

W

Q

U

-

=

D

by

W

Q

U

-

=

D

by

W

Q

U

-

=

D

by

W

Q

U

-

=

D

3

3

1

1

5

10

37

1

273

27

31

8

18

1

10

01

1

m

.

V

)

)(

.

(

V

)

.

(

nRT

pV

Use

-

=

+

=

=

3

3

2

2

3

2

2

1

1

10

83

1

273

127

273

27

10

37

1

m

.

V

V

.

T

V

T

V

-

-

=

+

=

+

=

)

)(

.

(

T

nR

U

27

127

31

8

18

1

2

3

2

3

-

=

D

=

D

thermodynamics wee

Documents

heat q

heat loss

clatent heat capacity

j kg1 k1specific heat

cwater balloon heat

meltinglatent heat of

c worth

j kg1 k1m