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Warm-up. Agree. Disagree. The temperature of boiling water does not increase even though energy is supplied to it continually. . Warm-up. Separating magnets that stick to each other. Allowing magnets to stick to each other. Some water molecules taken from liquid water are put in air.

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Warm up
Warm-up

Agree.

Disagree.

The temperature of boiling water doesnot increase even though energy issupplied to it continually.


Warm up1
Warm-up

Separating magnets that stick toeach other.

Allowing magnets to stick to eachother.

Some water molecules taken from liquid water are put in air.

Water molecules in air come together and form dew.

Do these requireenergy supply?


Warm up2
Warm-up

Ice molecules moves ________ (more/ less) freely than water molecules;

less

so they have ________ (more / less) KE than water molecules.

less


Introduction
Introduction

Change of state

Matter exists in 3 states:

solid,

liquid,

gas

e.g. water

vaporization (takes place at boiling point)

Fusion (at melting point)

ice

water

steam

solidification

condensation

at freezing point

at boiling point


Cooling curve

temperature

gas

liquid

solid

time

Cooling curve

When a hot liquid is cooled down,

its temperature drops.

Graph of temperature vs time?

??


Experiment 3a
Experiment 3a

Cooling curve of octadecan-1-ol

Record the temperature of the melted octadecan-1-ol as it cools down.

Video

Video


Cooling curve1
Cooling curve

  • When solid is heated, it melts to a liquid

  • On cooling, the variation of temperature with time from A to B

  • The temperature remains constant from B to C, the energy evolved is called the latent heat

  • The latent heat is used to change from liquid to solid at constant temperature, which is called the melting point


Why temperature remains constant
Why temperature remains constant?

  • From liquid to solid, energy is released to the surroundings due to the decrease in the kinetic energy of the particles

  • From solid to liquid, energy is absorbed from the surroundings due to the increase in the kinetic energy of the particles


Specific latent heat of fusion and vaporization
Specific latent heat of fusion and vaporization

  • Definition

    • Latent heat is required is proportional to the mass of the substance that undergoes state change

      i.e. Latent heat (H)  mass (m)

      H = L m

      L is a constant called the specific latent heat

    • Unit: J kg-1


Cooling curve2

temperature / C

A

B

C

D

Here is a cooling curve of octadecan-1-ol .

Cooling curve

AB drops steadily — liquid cooling (temperature falling )


Cooling curve3

temperature / C

A

B

C

D

Cooling curve

BC is flat — liquid solidifying

(temperature unchanged)


Cooling curve4

temperature / C

A

B

C

D

Cooling curve

CD drops steadily — solid cooling to room temperature (temperature falling)


Cooling curve5

temperature / C

A

B

C

D

Cooling curve

melting point:

melting point

read from the flat part BC


Cooling curve of water
Cooling curve of water

temperature

liquid-solid mixture

Simulation


Latent heat
Latent heat

The cooling curve shows:

When a substance is solidifying,

  • it loses energy continuously but...

  • its temperature remains unchanged


Latent heat1
Latent heat

During change of state:

The energy given out/absorbed is called latent heat

means‘hidden’


Latent heat2
Latent heat

Ice-water mixture stays at 0 oC until all the ice is melted.

This energy is called

latent heat of fusion of ice.

  • temperature unchanged

  • energy is absorbed

from air to change the ice to water


Latent heat3
Latent heat

Energy supplied continuously to keep water boils…

This energy is called

latent heat of vaporizationof water.

  • temperature unchanged

  • energy is absorbed

to change the water to stream


steam

condensation

vaporization

releases latent heat of vaporization

water

solidification

fusion

ice


steam

vaporization

condensation

absorbs latent heat of vaporization

water

solidification

fusion

ice


steam

condensation

vaporization

releaseslatent heat of fusion

water

solidification

fusion

ice


steam

vaporization

condensation

water

absorbs latent heat of fusion

solidification

fusion

ice


State Change

solid

liquid

gas

liquid


2 latent heat and particle motion

strongattraction

weakattraction

2 Latent heat and particle motion

molecule

Regular arrangement breaks up


2 latent heat and particle motion1

PE

2 Latent heat and particle motion

Energy has to be suppliedto oppose the

attractive forceof the particles.

PE related to the forces of attraction between the particles

solid  liquidor liquid  gas

average potential energy 


2 latent heat and particle motion2
2 Latent heat and particle motion

The transfer of energy does not change the KE.

Temperature does not change.

latent heat

= change in PE during change of state

Simulation


Specific latent heat
Specific latent heat

Specific = for 1 kg of a substance

e.g.

energy E

without temperature change

1 kg solid X

1 kg liquid X

E= latent heat for 1 kg of X

= specific latent heat of X


Specific latent heat1
Specific latent heat

Energy transferred to change the state of 1 kg of the substance without a changein temperature.

symbol: l

or E = ml

unit: J kg-1


Specific latent heat of fusion of ice l f
Specific latent heat of fusion of ice (lf )

lf = energy needed to change 1 kg ofice to water (without temperature change)

Find (1) mass melted m

and (2) energy transferred E

 lf = E/m


Experiment 3b
Experiment 3b

Measuring the specific latent heat of fusion of ice

Video

Simulation


Experiment 3b1
Experiment 3b

control apparatus

experimental apparatus

Ice also melts at roomtemperature, so a control is needed.


Experiment 3b

Measuring the specific latent heat of fusion of ice

For ice, lf = 3.34  105 J kg-1

ice (0 C)

water (0 C)


B specific latent heat of vaporization of water l v
b Specific latent heat of vaporization of water (lv)

lv = energy needed to change 1 kg of water to steam (without change of temperature)

Find (1) mass boiled away m

and (2) energy transferred E

 lv = E/m


Experiment 3c
Experiment 3c

Video

Simulation


Experiment 3c1
Experiment 3c

For water, lv = 2.26  106 J kg-1

steam (0 C)

water (0 C)


Summary change of state

condensation

energy out 2260 kJ

energy out 334 kJ

solidification

Summary: change of state

vaporization

energy in 2260 kJ

steam

(1 kg)

water

(1 kg)

energy in 334 kJ

ice

(1 kg)

fusion


Summary from ice to steam

ice

(ice and water) melting

boiling

(water and stream)

(334)

(2260)

water

Summary: from ice to steam

stream

100

0

energy / kJ

(420)

Energy involved in heating 1 kg of water


Consider a cup of water
Consider a cup of water...

Consider a cup of water (mass m) being heated from0 °C until it starts to boil at 100°C.

Since E = mcT and E = mlv

mcT = mlv

 lv= c T = 4200×100 = 420 kJ kg-1

Is the student correct? (Yes/No)


When vapour condenses
When vapour condenses...

When vapour condenses, is the surrounding air warmedor cooled?

(warmed/cooled)


Jimmy melts
Jimmy melts...

Temperature / oC

Z

X

Y

time / s

0 0.5 1.2 2.3 3.7

Jimmy melts three materials X, Yand Zof equal mass at the same time and place.


...

Temperature / °C

Z

X

Y

time / s

0 0.5 1.2 2.3 3.7

Which material(s) has/have the greatest melting point?( X / Y / Z )


Temperature / °C

Z

X

Y

time / s

0 0.5 1.2 2.3 3.7

Which material(s) has/have the largest value of specificlatent heat of fusion? ( X / Y / Z )


Temperature / °C

Z

X

Y

time / s

0 0.5 1.2 2.3 3.7

Which material(s) release(s) largest amount of energy(per kg) when they freeze? ( X / Y / Z )


Fusion and boiling
Fusion and Boiling

  • Boiling of water and fusion of ice

    • when water boils, the temperature is found to remained at 100oC. The energy supplied is only used to change from water to steam without any change in temperature

    • When the steam condenses, the temperature remains at 100oC and energy is given out


Fusion and boiling1
Fusion and Boiling

  • Fusion of ice

    • when ice melts, the temperature is found to remained at 0oC. The energy supplied is only used to change from ice to water without any change in temperature

    • When the water freezes, the temperature remains at 0oC and energy is given out


Latent heat of fusion and vaporization
Latent Heat of Fusion and Vaporization

  • Energies used in fusion and vaporization are called the latent heat of fusion and vaporization

  • The constant temperatures are called the melting point and boiling point


Example 1
Example 1

Finding specific latent heat of fusion of ice

Result of melting experiment

Mass of water

in experimental cup: m1 = 0.050 kg

in control cup: m2 = 0.014 kg

Joulemeter reading

initial: j1 = 15 000 J

final: j2 = 29 200 J


Example 11
Example 1

Finding specific latent heat of fusion of ice

(a) Find the specific latent heat of fusion of ice.

Results:

m1 = 0.050 kg

m2 = 0.014 kg

j1 = 15 000 J

j2 = 29 200 J

lf =E /m

= (j2 – j1) / (m2 – m1)

= (29 200 – 15 000) / (0.050 – 0.014)

= 3.94  105 J kg–1


Example 12

Finding specific latent heat of fusion of ice

Example 1

(b) Account for any difference of the valueobtained from the standardvalue, 3.34 105 J kg–1.

Experimental value = 3.94  105 J kg–1

There is a rather large error of 18%.

The possiblesources of error are:

  • Difficulty in keeping the water dripping down the two funnels at the same rate.

  • Energy lost to the surroundings.


Example 2
Example 2

How much energy is required to melt 0.5 kg of ice at 0 °Ctemperature raised to 80°C?

Total energy required

=latent heat(ice at 0 °C → water at 0 °C)

+ energy (water: 0 °C → 80 °C)

= mlf+ mcT

= 0.53.34105 + 0.5420080

= 3.35105 J


Example 3
Example 3

Result of boiling experiment

Mass of water boiled away = 0.10 kg

KW h meter calibration = 600 turns/kW h

Number of rotations counted = 41

(a) Find the specific latent heat of vaporization of water.


Example 31
Example 3

kW h meter calibration =600 turns/kW h

1 kW h = 1 kW  1h = 3.6  106 J

Energy supplied per revolution of the disc

= 3.6  106/600 = 6000 J

Energy supplied to boil 0.10 kg of water

= 600041= 246 000 J

Number of rotations = 41

Specific latent heat of vaporization of water

lv=E /m = 246 000/0.10 = 2.46  106 J kg–1


Example 32
Example 3

(b) Account for any difference of the value obtained from the standard value, 2.26  106 J kg–1.

Experimental value = 2.26  106 J kg–1

There is an error of about 9%.

1 Steam condensing on the heater drips back into the cup.

2 Some water ‘bubbles’ out of the cup.

3 Energy lost to the surroundings.


Example 4
Example 4

How much energy is required to change 0.5 kg of water at 0°C to stream at 100 °C .

m = 0.5 kg

T = 100 °C– 0°C = 100 °C


Example 41
Example 4

m = 0.5 kg, T = 100 °C – 0°C = 100 °C

The total energy required

=energy(water: 0°C → 100 °C)

+latent heat (water at 100 °C → steam at 100 °C)

=mcT+mlv

=0.5 × 4200 × 100+0.5 × 2.26 × 106

=2.1 × 105+11.3 × 105

=1.34 × 106 J


Example 5
Example 5

An expresso coffee machine injects 0.025 kg of steam at 100 °C into a cup of cold coffee of mass 0.15 kg at 20 °C.

specific heat capacity

of the coffee = 5800 J kg–1°C

Find the final temeperature of the expresso coffee.


Example 51
Example 5

0.025 kg of steam at 100 oC

0.15 kg of coffee at 20 oC

Let Tbe the final temperature of the coffee.

Assuming no energy loss to the surroundings,

energy loss by steam

energy gained by coffee

=

0.0252.26 106

+ 0.0254200 (100 – T)

=

0.15 5800(T– 20)

Solving the equation,

The temperature T of the coffee is 86.6°C


Effect of pressure on melting point of water
Effect of pressure on melting point of water

  • melting point of ice at normal pressure is 0oC

  • it becomes lower when the external pressure is increased (i.e. below 0oC )

  • vice versa


Effect of pressure on boiling point of water
Effect of pressure on boiling point of water

  • boiling point of ice at normal pressure is 100oC

  • it becomes higher when the external pressure is increased (i.e. above 100oC )

  • vice versa


Effect of impurities on melting and boiling point of water
Effect of impurities on melting and boiling point of water

  • Impurities such as salt is added to ice to lower the melting point, causing the ice to melt

  • Addition of impurities into water raises the boiling point of water above 100oC


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