f4 chap 4 kks notes

8/15/2019 F4 Chap 4 KKS notes

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4.1 UNDERSTANDING THERMAL EQUILIBRIUM

Heat is a form of energy that flows from a hot body to a cold body. SI unit: Joule, J.

Temperature is the degree of hotness of a body. SI unit: Kelvin, K.

When two objects with different temperatures come into thermal contact, heat energy is transferred

between the two objects. Heat is transferred in both directions. When thermal equilibrium is reached,

the net rate of heat flow between the two bodies is zero.

There is no net flow of heat between two objects that are in thermal equilibrium.

Two objects in thermal equilibrium have the same temperature irrespective of shape, mass, size or type

of surface.

Comparison of two objects before and after achieving thermal equilibrium

Before After

The two objects have different temperature The two objects have equal temperature

The rate of energy transfer rate are different The rate of energy transfer rate are equal

The hotter object loses energy while the colder object

gains energy

There is not net gain or net loss of energy by

either object.

Thw hotter object colls down while the colder object

heats up

Both of the objects remain at the same

temperature

Energy is transferred between the two objects.

Applications of thermal equilibrium

Situation Explanation

Mom puts a piece of wet towel on David’s forehead

One of the ways to bring down fever is by wiping his body and forehead

using a wet towel.

Heat energy from the body will be transferred to the wet towel until

thermal equilibrium is reached.

The wet towel is then rinsed under tap water so that the heat energy from

the towel is transferred to tap water. The process is repeated until the right

amount of heat is transferred out of the body to bring fever down.


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The spoon is hot!!!

Mom: The milk is too cold for my baby!

It’s just too hot! Let’s have a cold drink

SPM 2003 SPM 2005


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Liquid in glass thermometer

The liquid used in glass thermometer should

(a) Be easily seen

(b) Expand and contract rapidly over a wide range of temperature

(c) Not stick to the glass wall of the capillary tube

The characteristic of mercury

Characteristics Explanation

Opaque Easy to take a readingExpands uniformly with heat Uniform scale

High cohesive force Does not wet the tube and does not stick to the glass wall

High boiling point Can measure high temperature

Disadvantages

o Freeze at -39OC

o Poisonous

o Expensive

The sensitivity of the thermometer can be increased by: Modification Explanation

Narrow

capillary tube

At a fixed temperature, a narrow capillary tube will produce a longer

mercury column that of a wide capillary tube as the expansion ofmercury is constant.

Thin glass wall

bulb

The heat transfer from the surrounding to the mercury is faster.

Small bulbA smaller bulb will absorb heat energy in a shorter time and responds

faster to temperature change.


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Exercise

The figure shows two metal blocks. Which the

following statement is false?

A. P and Q are in thermal contact

B. P and Q are in thermal equilibrium

C. Energy is transferred from P to Q

D. Energy is transferred from Q to P

2. When does the energy go when a cup of hot tea

cools? A. It warms the surroundings

B. It warms the water of the tea

C. It turns into heat energy and disappears.

3. Which of the following temperature corresponds to

zero on the Kelvin scale?

A. 273 OC

B. 0 OC

C. -273 OC

D. 100 OC

4. How can the sensitivity of a liquid- in –glass

thermometer be increased?

A. Using a liquid which is a better conductor of heat

B. Using a capillary tube with a narrower bore.

C. Using a longer capillary tube

D. Using a thinner-walked bulb

5. Which instrument is most suitable for measuring a

rapidly changing temperature?

A. Alcohol-in –glass thermometer

B. Thermocouple

C. Mercury-in-glass thermometerD. Platinum resistance thermometer

6. When shaking hands with Anwar, Kent Hui noticed

that Anwar’s hand was cold. However, Anwar felt tha

Kent Hui hand was warm. Why did Anwar and Kent

Hui not feel the same sensation?

A.

Both hands in contact are in thermal equilibrium.B. Heat is flowing from Kent Hui’s hand to Anwar’s

hand.

C. Heat is following from Anwar’s hand to Kent Hui

hand.

7.

The length of the mercury column at the ice point and steam point are 5.0 cm and 40.0cm respectively. When the

thermometer is immersed in the liquid P, the length of the mercury column is 23.0 cm. What is the temperature o

the liquid P?

8.

The length of the mercury column at the steam point and ice point and are 65.0 cm and 5.0cm respectively. When

the thermometer is immersed in the liquid Q, the length of the mercury column is 27.0 cm. What is the

temperature of the liquid Q?

9.

The distance between 0OC and 100 OC is 28.0 cm. When the thermometer is put into a beaker of water, the length

of mercury column is 24.5cm above the lower fixed point. What is the temperature of the water?

10.

The distance between 0 OC and 100 OC is 25 cm. When the thermometer is put into a beaker of water, the length o

mercury column is 16cm above the lower fixed point. What is the temperature of the water? What is the length o

mercury column from the bulb at temperatures 30 OC?


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4.2 SPECIFIC HEAT CAPACITY

The heat capacity of a body is the amount of heat that must be supplied to the body to increase its

temperature by 10OC.

The heat capacity of an object depends on the

Temperature of the body

Mass of the body

Type of material

The specific heat capacity of a substance is the amount of heat that must be supplied to increase the

temperature by 1 OC for a mass of 1 kg of the substance. Unit Jkg-1 OC -1

The heat energy absorbed or given out by an object is given by Q = mc∆θ. High specific heat capacity absorb a large amount of heat with only a small temperature increase such

as plastics.

Some substances have low specific heat capacities while some have high specific heat capacities.


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Figure 1 shows three objects with different heat capacities because different amounts of heat are required

to raise their temperature by 1 OC. Note that although A and B are both made of aluminium. They have

different heat capacities because the masses are different.

Figure 2 shows the amounts of heat required to raise the temperature by 1 OC for equal masses of the

objects. Note that although B and C have equal masses, different amounts of heat are required to raise the

temperature by 1 OC because they are made of different materials. Different are said to have different

specific heat capacities

Example 1

The bulb of thermometer contains 3.4g of

mercury. What amount of heat is required to

raise the temperature of mercury from 30 OC

to 100 OC?

Solution

Mass of mercury, m = 3.4 x 10 −3 kg,

Specific heat capacity, c = 139Jkg −1 OC −1,Change of temperature, ∆θ = 100 – 30 =70OC.

Amount of heat required, Q = mc ∆θ

= 3.4 x 10 −3 kg x 139Jkg −1 OC −1 x 70 OC.

= 33J

Example 2

320 g of hot water at 80 OC is poured into an

aluminium can of mass 35g and temperature

of 20 OC. Calculate the final temperature.

Solution

Let the final temperature be T

Assuming no heat loss to the surroundings,Heat lost by water = heat gained by the

aluminium can

m1c1 ∆θ1 = m2 c2 ∆θ2

0.320 x 4200 x (80 –T) = 0.035 x 900 x (T – 20)

T = 78.63 OC.


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Exercise

1. Calculate the total heat that is observed by a

copper block of mass 500 g and which has been

heated from 31 OC to 80 OC

(Specific heat capacity of copper = 390 J kg-1 OC-1)

2. A metal of mass 2000 g is heated to 85OC and

placed in 500 g of water which is at temperature o

30OC. When a steady state is achieved, the

temperature of the mixture is 40OC. Find the

specific heat capacity of the metal. (specific heat

capacity of water is 4200 J kg-1 OC-1)

3.

An electric kettle of power 2500 W contains of

mass 3 kg and which is at a temperature of 25 OC.

What is the time required to heat the water until i

reaches its boiling limit at 100 OC.

4. 300 g of water at temperature 40OC is mixed with

900 g of water at temperature 80OC. If there is no

heat loss to the surroundings, what is the finaltemperature when thermal equilibrium is achieved

by the mixture of water?

5. An electric heater is installed in the bathroom of

Ramli’s house. It can increase the temperature of

the tap water from 15 OC to 45 OC.The heater can

provide 3 kg of hot water every minute.

Find the energy supplied to the water every

minute.

(specific heat capacity of water = 4200 J kg-1 OC-1)


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Experiment: To determine the specific heat capasity of a solid and a liquid

Aim To determine the specific heat capacity of

aluminium.

To determine the specific heat capacity of

Water.

List of

apparatus

and materials

Aluminium block, tissue paper,

polystyrene sheet, oil, immersion heater,

thermometer, power supply, triple beam

balance and stopwatch.

Polystyrene cup, water, immersion heate

thermometer, power supply, stirrer, triple

beam balance or electronic balance and

stopwatch.

Arrangementof

the apparatus

Procedure 1. The mass of the aluminium block, m is

determined using the triple beambalance.

2. The initial temperature of the

aluminium block, θ1, is recorded.

3. The heater is switched on and the

stopwatch is started simultaneously.

4. The heater is switched off after 10

minutes.

5.

The highest temperature, θ2, is

recorded.

1. The cup is filled with water of mass, m

(example, m = 150 g).2. The initial temperature of the water, θ1

is recorded.

3. The heater is switched on and the

stopwatch is started simultaneously.

4. The water is stirred continuously.

5. The heater is switched off after 10

minutes.

6.

The highest temperature, θ2, is

recorded.

Tabulate the

data

Power of the heater, P= ______W

Mass of aluminium block, m= ______gInitial temperature, θ1= _______°C

Final temperature, θ2= _______°C

Power of the heater, P = ________W

Mass of water, m= _______gInitial temperature, θ1=_______ °C

Final temperature, θ2= _______°C

Analysis of the

data

Calculation of specific heat capacity of

aluminium,c:

Q = mcθ

Pt = mc(θ2 – θ1)

∴ c =Pt

m(θ – θ)

Calculation of specific heat capacity of

water, c:

Q = mcθ

Pt = mc(θ2 – θ1)

∴ c =Pt

m(θ – θ)

The purpose of wrapping the aluminium blok

with wool heat loss to or absorption of heat

from the surrounding.

Oil in the holes for housing thermometer and

the immersion heater is to improve the

conduction of heat from the heater to the

thermometer throughthe aluminium block.

Specific heat capacity calculated is usually

larger than the standard value because some

heat is lost to the surroundings.


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Aim To investigate the relationship between temperature rise and mass of

water

Hypothesis When the mass of water increase the temperature rise will decrease.

Manipulated variable Mass of water

Responding variable Rise in temperature

Fixed variable Heating duration, water, weighing scale, power rating of heater usedApparatus Thermometer, water, beaker water, weighing scale

Setup

Procedure 1. 100 ml of water is placed in a 500 ml beaker.

2. A heater is placed in the water.

3. Heating process is carried out for 1minute.

4. Highest temperature achieved is recorded.

5. Step 2 to 4 is repeated for 200ml, 300ml, 400ml and 500ml of wate

Analysis

Conclusion


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Applications of Specific Heat Capacity

Water has a large specific heat capacity.

It can absorb a large amount of heat but the increase in temperature is small.

It can cool down internal combustion engines such as the car engine.

Part of a cooking pot Characteristic

Copper base

Wooden handle

Aluminium body

Claypot

Clay has a higher specific heat capacity than metals.

It is also a poor conductor of heat.

During cooking, heat is conducted slowly from the fireto the food inside the pot.

The longer cooking time brings out the taste in the

food. After the flame is switched off, the claypot is at a

higher temperature than the food inside it.

A considerable amount of heat continues to be

transferred into the food. The food remains hot for a

longer period of time.


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Sea Breeze

Land (solid) has a low specific heat capacity compared

to the sea (liquid).

In the day time, land gets hot faster than the sea.

Therefore, hot air on the land that has low density will

rise up and produces low pressure region on the land.

The cool air on the sea that has high density and highpressure will blow towards the land.

Land Breeze

Land (solid) has a low specific heat capacity compared

to the sea (liquid).

In the night time, land cools down faster than the sea.

The hot air on the sea that has low density will rise up

and produce a low pressure region.

The cool air on the land that has high density and high

pressure will blow towards the sea.

Exercise

In figure below, block A of mass 5kg at temperature 100°C is in contact with

another block B of mass 2.25kg at temperature 20°C.

Assume that there is no energy loss to the surroundings.

a) Find the final temperature of A and B if they are in thermal equilibrium.

Given the specific heat capacity of A and B are 900 Jkg-1°C-1 and 400 Jkg-1 °C-1

respectively.

b) Find the energy given by A during the process.

c) Suggest one method to reduce the energy loss to the surroundings.


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4.3 SPECIFIC LATENT HEAT

Latent Heat is the total energy absorbed or released when a substance changes its

physical state completely at a constant temperature.

Latent Heat of fusion is heat absorbed when solid changes into liquid or heat released

when liquid changes into solid at constant temperature.

Latent Heat of vaporization is heat absorbed when liquid changes into vapour or heat

released when vapour changes into liquid at constant temperature. Unit for latent heat is Joule (J).

Temperature-time Heating Graph

At QR and ST, the physical states change but the temperature is constant:

The heat absorbed is used to break the bonds between the atoms/molecules.

The physical state changes.

The kinetic energy of the atoms/molecules remains unchanged.

The temperature is constant.

At PQ, RS and TU, the temperature increases but the physical states are unchanged:

The heat absorbed is used to increase the kinetic energy of the atoms/molecules.

The temperature increases.

The bonds between atoms/molecules do not break.

The physical state is unchanged.


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15/30

Experiment: To determine the specific latent of fusion of ice and specific latent vaporisation

of water.

Aim To determine the specific latent heat of

fusion of ice.

To determine the specific latent heat o

vaporisation of water.

Variables Manipulated variable: Heat supplied, Q

Responding variable: Mass of ice that

melts, m

Fixed variable: Period of heating, t

Manipulated variable: Heat supplied, Q

Responding variable: Mass of water

which has vaporised, m

Fixed variable: Period of heating, t

Arrangement

of

the

apparatus

Procedure 1. Apparatus B is used as a control.

2.

The heater is set up andconnected as shown in the

diagram.

3. The circuit is closed and the

stopwatch is started

simultaneously.

4. Water which drips from the filter

funnels is collected in beakers A

and B.

5. The heater and the stopwatch are

switched off simultaneously after

t = 3 minutes

6. The mass of the water in both

beakers is measured.

1. A heater with power, P is

connected to the power supply.2. Pour water into a beaker until it i

almost full.

3. The circuit is closed.

4. When the water starts to boil, the

reading on the lever balance, m1,

is recorded and the stopwatch is

started simultaneously.

5.

After t = 4 minutes, the stopwatc

is stopped and final mass of the

water, m2 is recorded.

Tabulate the

data

Power of the heater, P =_____ W

Time of heating, t = _____s

Mass of water in beaker A, mA =____ g

Mass of water in beaker B, mB = ____g

Mass of water from the ice melted by

the heater, m = mA – mB =_____ g

Power of the heater, P =_____ W

Time of heating, t = _____s

Mass of water in beaker A, m1 =____ g

Mass of water in beaker B, m2 = ____g

Mass of water that has vaporised,

m = m1 – m2 =_____ gAnalysis of

the

data

Calculation of specific latent heat of

fusion of ice, Lf :

Q = mLf

Pt = mLf

Lf =

The unit of Lf is J kg –1

Calculation of specific latent heat of

vaporisation of water, LV:

Q = mLv

Pt = mLv

Lv =

The unit of Lv is J kg –1


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Example 1

Example2

Example 3

Example 4


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Example 5

Example 6


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4.4 THE GAS LAWS

The 4 physical quantities involved are:

I.

Mass (m) – always kept constant by

using a closed container

II.

Pressure (P)III.

Volume (V) can

change

IV.

Absolute temperature (T)

Explanation Using Kinetic Theory of Matter

Quantity Explanation

Mass (m) Amount of matter is in a closed container, the amount of matter is constant.

The mass is constant.

Volume (V) Volume of gas is equal to the volume of container.

Pressure (P) Rate of collision between the gas molecules and the walls of the container.

When the rate of collisions between gas molecules and the walls of the

container increases, the gas pressure also increases.

Temperature

(T)

Kinetic energy of the gas molecules. When the gas moves at a higher speed,

the kinetic energy increases; so the gas temperature also increases.

GasLaw

Boyle’s

Law

PressureLaw

Charles’

Law


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Boyle’s Law: (m&T -constant: P

) Charles’ Law: (m P -constant: V T) Pressure Law: (m&V -constant:P

Boyle's Law states that for a fixed

mass of gas, the gas pressure is

inversely proportional to the volume

of the gas if the temperature is kept

constant.

Charles' Law states that for a fixed

mass of gas, the volume of the gas is

directly proportional to the absolute

temperature of the gas if the pressure

is kept constant.

Pressure Law states that for a fixed

mass of gas, the pressure of gas is

directly proportional to the absolut

temperature of the gas if the volum

of the gas is kept constant. When the volume of the container

decreases, the volume of the gas

also decreases. (V2 P1

When the gas is heated,

temperature increases. (T2 > T1)

The volume of the gas also increases

to keep the rate of collision between

the gas molecules and the walls of

the container is constant.

The gas pressure is constant.

T2> T1 and V2 > V1

When the gas is heated, its

temperature

increases. (T2 > T1)

The volume of the container is ke

constant, so the volume of the ga

is constant.

The rate of collision between the

gas molecules and the walls of th

container increases.

The gas pressure increases.

T2> T1 and P2 > P1

Relationship:

Gas pressure (P) is inversely

proportional to the gas volume (V)，

P ∝

V

Relationship:

Volume of gas (V) is directly

proportional to the absolute

temperature of gas (T)；V ∝ T

Relationship:

Pressure of gas (P) is directly

proportional to the absolute

temperature of gas (T). P ∝ T

Equation: P = k (

V) ; k is a constant

k = P V = constant ∴ P1 V1 = P2 V2

Equation: V = k (T) ; k is a constant

k =V

T= constant ∴

V

T=

V

T

Equation: P = k (T) ; k is a constant

k =P

T = constant ∴

P

T=

P

T


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Boyle’s Law Charles’ Law Pressure Law

P is inversely

proportional

to V

P is constant

P is dire

proporti

l to T

P is directly

proportional

to

V is directly

proportional

to T

P varie

linear

with θ

P is constant

V varies

linearly with

θ

P is

consta


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Boyle’s Law

1. An air bubble of capacity 15 cm3 at a pressure

of 320 cm Hg is released from the sea.

Determine the volume of the bubble when it

reaches the surface of the water, if the

atmospheric pressure is 76 cm Hg. Assume

that the temperature of the water is

constant.Solution

Initial volume, V 1 = 15 cm3

Initial pressure, P1 = 320 cm Hg

Final volume, V 2 = ?

Final pressure, P2 = 76 cm Hg

P1V 1 = P2V 2

V 2 = P1V 1 /P2

= (320)(15)/(76)

= 63 cm

3

2.

A rod of tube with one end closed has a

strip of mercury 3 cm long. When the tube

is held vertically, the length of air trapped

is 8 cm as in figure (a) above. Atmosphericpressure is taken as 76 cm Hg.

(a) Calculate the length, x of air which is

trapped when the tube is placed

horizontally as in figure (b) above.

(b) Calculate the length, y of air which is

trapped when the tube is inverted as

shown in figure (c) above.

Solution

(a) P1V1 = P2V2

(76 + 3)8 = 76 (x)

x = 632/76

= 8.3 cm

(b) P1V1 = P2V2

(76 + 3) 8 = (76 – 3) y

y = 632/73

= 8.7 cm

3. Figure 1 shows trapped air inside a thin

glass tube in three different positions.

Determine the value of x and y.

[Atmospheric pressure = 76 cm Hg]

Solution

Position (a) P1 = (76 + 4) cm Hg

= 80 cm Hg

Position (b)P2 = (76 – 4) cm Hg

= 72 cm Hg

Position (c)P3 = 76 cm Hg

Compare (a) and (b)

P1V 1 = P2V 2

80 (12) =(72)

= 13.3 cm

Compare (a) and (c)

P1V 1 = P2V 2

80(12) = (76)

= 12.6 cm

Pressure Law

1. An iron cylinder containing gas has a

pressure of 360 kPa when it is kept in a

store at temperature 27OC. What is the

pressure of the gas when the cylinder is

moved outdoors where the temperature is

40 OC.

Solution

P1 = 360 kPa

T1 = (27 + 273) = 300 K

P2 = Final pressure

T2 = (40 + 273) = 313 K

Using the pressure law,

P1/T1 = P2/T2

360/300 = P2/313

P2 = (360 x 313)/300

P2 = 375.6 kPa


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2. Before a journey from Alor Setar to Ipoh,

the air in a car tyre has a pressure of 200

kPa and a temperature of 27OC. After the

journey, the air pressure in the tyre is 240

kPa. What is the temperature of the air in

the tyre after the journey? [Assume the

volume of a tyre is constant]

Solution

P1/T1 = P2/T2

200/(27 + 273) = 240/T2

T2 = (300 x 240)/200

T2 = 360 K

T2 = 360 – 273

= 87 OC

3. The pressure of nitrogen gas in a light bulb

is 60 kPa at 20OC. Calculate the

temperature of the gas when the Pressure

inside the bulb rises to 90 kPa after the

bulb is lighted up.

Solution

P1/T1 = P2/T2

60/293 = 90/T2

T2 = (90 x 293)/60

T2 = 439.5 K

Final T = 439.5 – 273 = 166.5 OC

4. A closed flask contains gas at a

temperature of 95 OC and pressure of 152

kPa. If the temperature is reduced to -17OC, what is the new pressure of the gas?

Solution

P1/T1 = P2/T2

P2 = (152)(256)/368

P2 = 106 kPa

5.

A motorcycle tyre is found to possess a

pressure of 3.2 atmospheric when it was

tested in a garage at a temperature of 27OC. By assuming that the volume of the air

in the tyre is fixed, what is the

surrounding temperature if the tyre is

fixed, what is the surrounding temperature

if the tyre is found to possess a pressure of

2.8 atmospheric?

Solution

P1/T1 = P2/T2

3.2/300 = 3.5/T2

T2 = 328 K The surrounding temp.

= (328 – 273)

= 55OC

Charle’s Law

1. A syringe in a refrigerator contains 4.5 ml

of air at -3OC. When the syringe was taken

out and placed in a room where the

temperature was 27

O

C, the air in itexpands. Calculate the final volume of the

air in the syringe.

Solution

V1/T1 = V2/T2

4.5/270 = V2/300

V2 = (4.5 x 300)/270

V2 = 5.0 ml

2.

The figure above shows a glass tube

containing some trapped air inside it. At 27OC, the horizontal column of trapped air is

9 cm. If the atmospheric pressure does not

vary, how much is the horizontal column ofair at a temperature of 45 OC?

Solution

V1/T1 = V2/T2

9A/300 = xA/318

V2 = 9.54 cm3


23/30

Experiment (i) To investigate the relationship between the pressure and volume of a gas

(ii) To investigate the relationship between the volume and temperature of a gas

Aim To investigate the relationship

between the pressure and volume for

a fixed mass of gas at constant

temperature.

To investigate the relationship

between the volume and temperature

for a fixed mass of gas at constant

pressure.

Variables Manipulated: Pressure of air, P

Responding: Volume of air, V

Fixed: Mass and temperature of air

Manipulated:Temperature of air, T

Responding: Volume of air, V

Fixed: Mass and pressure of air

Apparatus

and

materials

Bicycle pump, capillary tube, rubber

tube and Bourdon gauge.

Capillary tube, sulphuric acid, half

metre rule, retort stand, beaker,

tripod stand, rubber tube,

thermometer and Bunsen burner.

Arrangement

of

the apparatus

Procedure 1.

The apparatus is set up as shownin the diagram above.

2. The initial pressure and volume of

the air in the capillary tube are

recorded.

3. Push the pump until the pressure

of air is 100 kPa.

4. Record the volume of the air in

the capillary tube.

5.

Repeat the experiment withdifferent pressures of air that is,

P = 120 kPa, 140 kPa, 160 kPa and

180 kPa.

6. The values of pressure, P, volume,

V, and

are tabulated.

7. Plot a graph P against V and a

graph of P against

.

1.

The apparatus is set up as shownin the diagram above.

2. The water in the beaker is heated

slowly and stirred uniformly and

gently. When the reading of the

thermometer is 30 °C, the length

of the trapped air, l cm is

recorded.

3. Repeat the experiment with

different temperatures of waterthat is θ = 40 °C, 50° C, 60 °C and

70 °C.

4. The values of temperature,

absolute temperature, T, and

length of air trapped, l, are

tabulated.

5. Plot a graph l against T (in Kelvin).


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Tabulate the

data

Pressure of

air, P/kPa

Volume of air/cm3

/ cm3 V1 V2 Vavg

100

120

140

160180

Temperature

of air, θ/°C

Absolute

temperature,

T/K

Length of

trapped air,

l/cm

l1 l2 lavg

30

40

5060

70

Analysis of the

data


25/30

Aim To investigate the relationship between the pressure and temperatur

for a fixed mass of gas at constant volume.

Manipulated variable Temperature of air, T

Responding variable Pressure of air, P

Fixed variable Mass and volume of air

Apparatus Round-bottomed flask, thermometer, retort stand, Bourdon gauge,

stirrer, ice, rubber tube, tripod stand, wire gauze and Bunsen burner.Setup

Procedure 1.

The apparatus is set up as shown in the diagram above.

2.

The round-bottomed flask is immersed in a beaker of water

containing ice.

3.

The mixture of water and ice is stirred so that the air in the flas

has the same temperature as the water.

4.

When the reading of the thermometer is 30 °C, record the

reading of the pressure, P on Bourdon gauge.5.

Repeat the experiment with different temperatures of water,

that is θ = 40 °C, 50 °C, 60 °C and 70 °C.

6.

The values of temperature, θ, absolute temperature, T and

pressure of the air trapped, P are tabulated.

7.

Plot a graph P against T (in Kelvin).

Analysis Temperature of air, θ/°C Absolute temperature, T/K Air pressure,P/kPa30

40

5060

70

Conclusion


26/30

Exercise

1.

A 5kg iron sphere of temperature 500C is put in

contact with a 1kg copper sphere of temperature

273K and they are put inside an insulated box.

Which of the following statements is correct when

they reach thermal equilibrium?

A.

A iron sphere will have a temperature of 273K

B.

The copper sphere will have a temperature of

500

C.C.

Both the sphere have the same temperature.

D.

The temperature of the iron sphere will be

lower than 500C

2.

In the process to transfer heat from one object to

another object, which of the following processes

does not involve a transfer to material?

A.

Convection

B.

Vaporisation

C.

Radiation

D.

Evaporation

3.

When we use a microwave oven to heat up some

food in a lunch box, we should open the lid slightly.

Which of the following explanations is correct?

A.

To allow microwave to go inside the lunch box

B.

To allow the water vapors to go out, otherwise

the box will explode

C.

To allow microwave to reflect more times

inside the lunch box

D.

To allow microwave to penetrate deeper into

the lunch box.

4.

Water is generally used to put out fire. Which of

the following explanation is not correct?

A.

Water has a high specific heat capacity

B.

Steam can cut off the supply of oxygen

C.

Water is easily available

D.

Water can react with some material

5.

Given that the heat capacity of a certain sample is

5000 J0C-1. Which of the following is correct?

A.

The mass of this sample is 1kg.

B.

The energy needed to increase thetemperature of 1 kg of this sample is 5000 J.

C.

The energy needed to increase the

temperature of 0.5kg of this sample is 2500J.

D.

The temperature of this sample will increase

10C when 5 000 J energy is absorbed by this

sample.

6.

Which of the following statement is correct?

A.

The total mass of the object is kept constan

when fusion occurs.

B.

The internal energy of the object is increased

when condensation occurs

C.

Energy is absorbed when condensation occurs

D.

Energy is absorbed when vaporization occurs.

7.

Water molecules change their states between theliquid and gaseous states

A.

only when water vapour is saturated

B.

at all times because evaporation and

condensation occur any time

C.

only when the vapour molecules produce a

pressure as the same as the atmospheric

pressure

D.

only when the water is boiling

8.

Based on the kinetic theory of gas which one of the

following does not explain the behaviour of gas

molecules in a container?

A.

Gas molecules move randomly

B.

Gas molecules collide elastically with the walls

of the container

C.

Gas molecules move faster as temperature

increases

D.

Gas molecules collide inelastically with each

other

9.

A cylinder which contains gas is compressed at

constant temperature of the gas increase because

A.

the average speed of gas molecules increasesB.

the number of gas molecules increases

C.

the average distance between the ga

molecules increases

D.

the rate of collision between the gas molecule

and the walls increases

10.

A plastic bag is filled with air. It is immersed in the

boiling water as shown in diagram below.

Which of the following statements is false?

A.

The volume of the plastic bag increases.

B.

The pressure of air molecules increases

C.

The air molecules in the bag move faster

D.

The repulsive force of boiling water slows

down the movement of air molecule


27/30

1. Research student wishes to carry out an investigation on the temperature change of the substance in

the temperature range -500C to 500C. The instrument used to measure the temperature is a liquid in

glass thermometer.

Table 1

(i) State the principle used in a liquid- in –glass thermometer.(1m)

.........................................................................................................................................................................

(ii) Briefly explain the principle stated in (a)(i) (3m)

…………………………………………………………………………………….………………………………………………………………

………….……………………………………………………………………………………………………………………………..………………………………….……………………………………………………………………………………………………………………………

(iii) Table 1 shows the characteristic of 4 types of thermometer: A, B C and D. On the basis of the

information given in Table 1, explain the characteristics of, and suggest a suitable thermometer

for the experiment.(5 m)

…………………………………………………………..………………………….……………………………………………………………

…………….…………………………………………..…...……………………………………………………………………………………

…….……………………….…………………………..…………………………………………………………………………………………

……………….………………………….………………..………………………………………………………………………………………

………………………………….…………………………..….…………………………………………………………………………………

…………………………………………………………………….…………………………………………………..……………….…………(iv) The length of the mercury column in uncalibrated thermometer is 6.0cm and 18.5 cm at 00C

and 1000C respectively. When the thermometer is placed in a liquid, the length of the mercury

column is 14.0cm

Calculate the temperature of the liquid.

(v) State two thermometric properties which can be used to calibrate a thermometer. (6m)

…………………………………………………………………………………….………………………………………………………………

………….……………………………………………………………………………………………………………………………………….

Thermometer A B C D

Liquid Mercury Mercury Alcohol Alcohol

Freezing point of liquid (0C) -39 -39 -112 -112

Boiling point of liquid (

0

C) 360 360 360 360Diameter of capillary tube Large Small Large Small

Cross section


28/30

2. A metal block P of mass 500 g is heated is boiling water at a temperature of 100°C. Block P is then

transferred into the water at a temperature of 30°C in a polystyrene cup. The mass of water in the

polystyrene cup is 250 g. After 2 minutes, the water temperature rises to 42°C.

Assuming that the heat absorbed by the polystyrene cup and heat loss to the surroundings are

negligible. (Specific heat capacity of water 4 200 j kg-1 C-1)

Calculate

(i) the quantity of heat gained by water the polystyrene cup

(ii) the rate of heat supplied to the water

(iii) the specific heat capacity of the metal block P

Figure 2


29/30

3. A student performs an experiment to investigate the energy change in a system. He prepares a

cardboard tube 50.0 cm long closed by a stopper at one end. Lead shot of mass 500 g is placed in the

tube and the other end of the tube is also closed by a stopper. The height of the lead shot in the tube is

5.0 cm as shown in Figure 3.1. The student then holds both ends of the tube and inverts it 100 times

(Figure 3.2).

a) State the energy change each time the tube is inverted.

…………………………………………………………………………………….………………………………………………………………

………….…………………………………………………………………………………………………………………………………………

b) What is the average distance taken by the lead shot each time the tube is inverted?

…………………………………………………………………………………….………………………………………………………………

………….…………………………………………………………………………………………………………………………………………

c) Calculate the time taken by the lead shot to fall from the top to the bottom of the tube.

d) After inverting the tube 100 times, the temperature of the lead shot is found to have increased

by 30C.

i. Calculate the work done on the lead shot.

ii. Calculate the specific heat capacity of lead.

iii. State the assumption used in your calculation in (d)ii

…………………………………………………………………………………….……………………………………………………………

…………….……………………………………………………………………………………………………………………………………

Figure 3.1 Figure 3.2


30/30

4.

Heat produced in an engine block of car needs to be transferred out promptly to prevent overheating. This is

done by circulating a suitable cooling liquid through the engine block.

a) What is meant by ‘specific heat capacity of water is 4200 Jkg-1oC-1 ?

…………………………………………………………………………………….………………………………………………………………

………….…………………………………………………………………………………………………………………………………………

b) Based on the table above,

(i) Explain the suitable characteristics of the cooling liquid to extract heat out of an engine

block.…………………………………………………………..………………………….……………………………………………………………

…………….…………………………………………..…...……………………………………………………………………………………

…….……………………….…………………………..…………………………………………………………………………………………

……………….………………………….………………..………………………………………………………………………………………

………………………………….…………………………..….…………………………………………………………………………………

…………………………………………………………………….…………………………………………………..……………….…………

(ii) Decide which liquid is the most suitable and give reasons for your choice.

………………………………….…………………………..….…………………………………………………………………………………

…………………………………………………………………….…………………………………………………..……………….…………

c) Total energy released by an engine in 1 hour = 9.0 x 107 J

Energy breakdown : mechanical 40% and heat 60%

Mass of cooling liquid circulating in 1 hour = 150 kg

Temperature of water entering the engine = 30oC

Temperature of water exiting the engine = 60oC

Based on the information above,

(i) Calculate the power of the engine

(ii) Calculate the amount of heat produced by the engine in one hour.

f4 chap 4 kks notes

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