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Warm-up

Agree.

Disagree.

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

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-up

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

less

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

less

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

gas

liquid

solid

time

Cooling curveWhen a hot liquid is cooled down,

its temperature drops.

Graph of temperature vs time?

??

Experiment 3a

Cooling curve of octadecan-1-ol

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

Video

Video

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?

- 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

- 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

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

A

B

C

D

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

Cooling curveAB drops steadily — liquid cooling (temperature falling )

A

B

C

D

Cooling curveCD drops steadily — solid cooling to room temperature (temperature falling)

Latent heat

The cooling curve shows:

When a substance is solidifying,

- it loses energy continuously but...

- its temperature remains unchanged

Latent heat

During change of state:

The energy given out/absorbed is called latent heat

means‘hidden’

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 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

condensation

vaporization

releases latent heat of vaporization

water

solidification

fusion

ice

strongattraction

weakattraction

2 Latent heat and particle motionmolecule

Regular arrangement breaks up

PE

2 Latent heat and particle motionEnergy 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 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 = 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 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 (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

control apparatus

experimental apparatus

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

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 (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

energy out 2260 kJ

energy out 334 kJ

solidification

Summary: change of statevaporization

energy in 2260 kJ

steam

(1 kg)

water

(1 kg)

energy in 334 kJ

ice

(1 kg)

fusion

(ice and water) melting

boiling

(water and stream)

(334)

(2260)

water

Summary: from ice to steamstream

100

0

energy / kJ

(420)

Energy involved in heating 1 kg 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 = mcT and E = mlv

mcT = mlv

lv= c T = 4200×100 = 420 kJ kg-1

Is the student correct? (Yes/No)

When vapour condenses...

When vapour condenses, is the surrounding air warmedor cooled?

(warmed/cooled)

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 )

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 )

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

- 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 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

- 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

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 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

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

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+ mcT

= 0.53.34105 + 0.5420080

= 3.35105 J

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 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

= 600041= 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 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

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 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)

=mcT+mlv

=0.5 × 4200 × 100+0.5 × 2.26 × 106

=2.1 × 105+11.3 × 105

=1.34 × 106 J

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 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

=

0.0252.26 106

+ 0.0254200 (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

- 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

- 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

- 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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