chapter 4 - understanding thermal equilibrium

145

Click here to load reader

Upload: shimyu-tan

Post on 15-Mar-2016

300 views

Category:

Documents


17 download

DESCRIPTION

 

TRANSCRIPT

Page 1: Chapter 4 - Understanding Thermal Equilibrium

Chapter 4 HeatChapter 4 Heat

2.1 Arah Mata Angin

ITeach – Physics Form 4

4.1 Understanding Thermal Equilibrium

Page 2: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Thermal EquilibriumChapter 4 Heat

Heat

Heat is a form of energy and is measured in Joules (J).

When an object is heated, it absorbs heat energy and its temperature increases.

Example : Water absorbs heat resulting in a rise in temperature.

When an object is cooled, it releases heat energy and its temperature decreases.

Example : When we sweat, heat is released from our body and our body cools down.

Page 3: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Keseimbangan TermaBab 4 Haba

Haba

Haba ialah satu bentuk tenaga dan diukur dalam unit Joule (J).

Apabila suatu objek dipanaskan, objek itu menyerap tenaga haba dan suhu objek bertambah.

Contoh : Air menyerap haba menyebabkan suhu air bertambah.

Apabila suatu objek disejukkan, ia membebaskan tenaga haba dan menyebabkan suhu berkurang.

Contoh : Apabila kita berpeluh, haba dibebaskan daripada badan. Pembebasan haba menyebabakn badan berasa sejuk.

Page 4: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Thermal EquilibriumChapter 4 Heat

Temperature The higher the temperature of an object, the more heat energy is in the object.

Mass An object with bigger mass contains more heat energy.

Material Objects of the same mass and temperature but made of different materials contains different

amount of heat energy.

The total amount of heat contains in an object depends on

Page 5: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Keseimbangan TermaBab 4 Haba

Suhu Semakin tinggi suhu sesuatu objek, semakin banyak tenaga haba terdapat dalam objek itu.

Jisim Objek yang mempunyai jisim yang lebih besar mengandungi tenaga haba yang lebih banyak.

Bahan Objek-objek yang mempunyai jisim dan suhu yang sama tetapi berlainan bahan mengandungi tenaga

haba yang berlainan.

Jumlah haba yang terdapat dalam sesuatu objek bergantung kepada

Page 6: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Thermal EquilibriumChapter 4 Heat

Temperature

• Temperature is the degree of hotness of a body.

• A hot object will have a temperature higher than a cold object.

• Two objects having the same temperature may not contain the same amount of heat energy in them.

• Example1 litre of water at a temperature of 800C contains more heat than 0.5 litre of water of the same temperature.

Page 7: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Keseimbangan TermaBabr 4 Haba

Suhu

• Suhu ialah darjah kepanasan suatu jasad.

• Objek yang panas mempunyai suhu yang tinggi berbanding objek yang sejuk.

• Dua objek yang mempunyai suhu yang sama mungkin tidak mempunyai kendungan tenaga haba yang sama.

• Contoh1 liter air pada suhu 800C mengandungi lebih banyak haba berbanding 0.5 liter air pada suhu yang sama.

Page 8: Chapter 4 - Understanding Thermal Equilibrium

Quantity of heat absorbed

The more heat is absorbed, the bigger is the rise in temperature of an object.

Mass An object of smaller mass will experience a bigger rise in temperature than an object with bigger mass when both absorbs the same quantity of heat.

Type of material When heated, iron will experience a bigger rise in temperature than plastic.

ITeach – Physics Form 4

Understanding Thermal EquilibriumChapter 4 Heat

The rise in the temperature of an object depends on

Page 9: Chapter 4 - Understanding Thermal Equilibrium

Kuantiti haba yang diserap

Semakin banyak haba yang diserap, semakin banyak kenaikan suhu pada objek.

Jisim

Objek yang mempunyai jisim yang kecil akan mengalami kenaikan suhu yang lebih banyak berbanding objek yang mempunyai jisim yang lebih besar apabila kedua-dua objek menyerap haba dalam kuantiti yang sama.

Jenis bahan Besi akan mengalami kenaikan suhu yang lebih tinggi berbanding plastik.

ITeach – Fizik Tingkatan 4

Memahami Keseimbangan TermaBab 4 Haba

Kenaikan suhu pada objek bergantung kepada

Page 10: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Thermal EquilibriumChapter 4 Heat

Thermal Equilibrium

Body BBody A

Heat

Body A

Heat

Heat

When two objects with different temperatures touches (in thermal contact) with each other

net heat will flow from the object (body A) with higher temperature to the object with lower temperature (body B).

Body A losses heat resulting in a drop in its temperature while body B absorbs heat thereby increasing its temperature until both bodies attain the same temperature.

When this happens, the rate of heat flow from body A to body B and vice versa is the same, that is, there is no net flow of heat between the two bodies. Body B

At this stage, both bodies are said to be in thermal equilibrium.

Higher temperature Lower temperature

Page 11: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Keseimbangan TermaBab 4 Haba

Keseimbangan Terma

Jasad BJasad A

Haba

Jasad A

Haba

Haba

Apabila dua objek yang berlainan suhu bersentuhan (dalam sentuhan terma) antara satu sama lain

haba akan mengalir daripada objek (jasad A) yang bersuhu lebih tinggi kepada objek yang bersuhu lebih rendah (jasad B)

Jasad A akan kehilangan haba dan ini menyebabkan suhunya menurun manakala jasad B akan menyerap haba dan menyebabkan suhunya bertambah sehingga kedua-dua jasad mencapai suhu yang sama.

Apabila kedua-dua jasad mencapai suhu yang sama, kadar aliran haba daripada jasad A ke jasad B dalam kedua-dua arah adalah sama. Maka, tiada pemindahan haba antara dua jasad itu.

Jasad B

Pada peringkat ini, kedua-dua jasad berasa dalam keseimbangan terma.

Suhu tinggi Suhu rendah

Page 12: Chapter 4 - Understanding Thermal Equilibrium

When the temperature of mercury increases, the length of the mercury column increases.

ITeach – Physics Form 4

Understanding Thermal EquilibriumChapter 4 Heat

Thermometry – Liquid-In-Glass Thermometer

To measure temperature of an object

Use a substance that changes with temperature like mercury.

Mercury is a liquid and when temperature of mercury increases, its volume increases.

Mercury

Thin-walledGlass bulb

Celsius scale

Glass tube

Vacuum

Capillary

The mercury thermometer

Mercury is contained in a capillary tube of uniform cross sectional area.

Page 13: Chapter 4 - Understanding Thermal Equilibrium

Apabila suhu merkuri meningkat, panjang turus merkuri juga meningkat.

ITeach – Fizik Tingkatan 4

Memahami Keseimbangan TermaBab 4 Haba

Termometer – Termometer Merkuri dalam Kaca

Untuk mengukur suhu sesuatu objek.

Menggunakan bahan yang berubah dengan suhu seperti merkuri

Merkuri ialah cecair. Apabila suhu merkuri bertambah, isipadunya juga akan bertambah.

Merkuri

Bebulu kaca Berdinding nipis

Skala Celsius

Batang kaca

Vakum

Tiub kapilari

Termometer Merkuri

Merkuri terdapat di dalam tiub kapilari.

Page 14: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Thermal EquilibriumChapter 4 Heat

Why mercury is used in a liquid-in-glass thermometer?

Expand uniformly

High boiling point

Opaque – easier to read the thermometer

Not sticky

Good heat conductor

Page 15: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Keseimbangan TermaBab 4 Haba

Mengapa merkuri digunakan di dalam termometer?

Mengembang dengan seragam

Takat didih tinggi

Bahan legap – mudah untuk membaca

bacaan termometer

Tidak melekit

Konduktor haba yang baik

Page 16: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Thermal EquilibriumChapter 4 Heat

0

Calibration Procedure Of A Mercury-in-glass Thermometer

Ice Point

Steam Point

If when placed in a substance and the length of the mercury column is l when thermal equilibrium is reached,

then the temperature of the substance is C1000100

0

llll

Page 17: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Keseimbangan TermaBab 4 Haba

0

Kaedah Penentukuran Termometer Merkuri dalam Kaca

Titik ais

Titik stim

Apabila termometer diletakkan pada suatu bahan, panjang turus merkuri ialah l apabila keseimbangan terma dicapai.

Suhu bahan ialah, C1000100

0

llll

Page 18: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Thermal EquilibriumChapter 4 Heat

Thermometry – Other Types Of Thermometers

Name of thermometer Diagram

Thermocouple

Constant volume gas thermometer

Resistance thermometer

Physical quantity that changes with temperature

Electromotive force (e.m.f.) produced

Pressure of air in the flask

Resistance of the platinum wire

VCold

junction

Constantanwire

Copper wire

(0°C)

N/m

airflaskwater

Ceramicrod

PlatinumWire coil

Leads

Glasscoating

Page 19: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Keseimbangan TermaBab 4 Haba

Termometer – Lain – lain Jenis Termometer

Nama termometer Rajah

Termogandingan

Termometer isipadu gas tetap

Termometer perintang

Kuantiti fizik yang berubah dengan suhu

Daya elektromotif (e.m.f) dihasilkan

Tekanan udara di dalam kelalang

Perintang wayar platinum

VSimpang

sejuk

Wayar kekal

Wayar kuprum

(0°C)

N/m

UdaraKelalangAir

Rod seramikGegelung wayar

Platinum

Plumbum

Sadur kaca

Page 20: Chapter 4 - Understanding Thermal Equilibrium

Chapter 4 HeatChapter 4 Heat

2.1 Arah Mata Angin

ITeach – Physics Form 4

4.2 Understanding Specific Heat Capacity

Page 21: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Specific Heat CapacityChapter 4 Heat

Factors Affecting The Rise In Temperature When A Substance Is Heated

Substances (Materials)

Rise in temperature of paraffin is more than the rise in temperature of water.

1 kgwater

thermometer

1 kgparaffin

Page 22: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Muatan Haba TentuBab 4 Haba

Faktor yang Mempengaruhi Kenaikan Suhu Apabila Bahan Dipanaskan

Bahan

Kenaikan suhu pada parafin adalah lebih tinggi berbanding kenaikan suhu pada air.

1 kgAir

Termometer

1 kgParafin

Page 23: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Specific Heat CapacityChapter 4 Heat

Factors Affecting The Rise In Temperature When A Substance Is Heated

Mass

Temperature of water in beaker A (smaller mass) rises more than the temperature of water in beaker B.

water

beaker A

water

beaker B

Page 24: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Muatan Haba TentuBab 4 Haba

Faktor yang Mempengaruhi Kenaikan Suhu Apabila Bahan Dipanaskan

Jisim

Suhu air dalam bikar A (Jisim yang lebih kecil) meningkat lebih banyak berbanding suhu air dalam bikar B.

Air

Bikar A

Air

Bikar B

Page 25: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Specific Heat CapacityChapter 4 Heat

Factors Affecting The Rise In Temperature When A Substance Is Heated

Quantity Of Heat Absorbed

Temperature of water in beaker D rise more because more heat is absorbed by the water in beaker D.

beaker C

1 minute

beaker D

5 minutes

Page 26: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Muatan Haba TentuBab 4 Haba

Faktor yang Mempengaruhi Kenaikan Suhu Apabila Bahan Dipanaskan

Kuantiti haba yang diserap

Suhu air di dalam bikar D meningkat lebih tinggi kerana lebih banyak haba diserap oleh air dalam bikar D.

Bikar C

1 minit

Bikar D

5 minit

Page 27: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Specific Heat CapacityChapter 4 Heat

Specific Heat Capacity

This mean that

The specific heat capacity, c, of a substance is the quantity of heat that is absorbed by 1 kilogram of the substance to increase its temperature by 1°C.

The unit of specific heat capacity is J kg-1 °C-1.

The most common substance, that is, water, have a specific heat capacity of 4200 J kg-1 °C-1.

in order for the temperature of 1 kg water to increase by 1°C, the water needs to absorb 4200 J of heat energy, or

for the temperature of 1 kg of water to decrease by 1°C, the water needs to release 4200 J of heat energy.

Page 28: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Muatan Haba TentuBab 4 Haba

Muatan Haba Tentu

Ini bermaksud

Muatan haba tentu, c, suatu bahan ialah kuantiti haba yang diperlukan untuk menaikkan suhu 1 kg bahan sebanyak 1°C.

Unit bagi muatan haba tentu ialah J kg-1 °C-1.

Air mempunyai muatan haba tentu sebanyak 4200 J kg-1 °C-1.

Air perlu menyerap 4200 J tenaga haba untuk menaikkan suhu sebanyak 1°C bagi 1 kg air atau

Air perlu membebaskan 4200 J tenaga haba untuk menurunkan suhu sebanyak 1°C bagi 1 kg air.

Page 29: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Specific Heat CapacityChapter 4 Heat

The quantity of heat absorbed or released, Q, by a substance of mass m kg experiencing a change of temperature of °C, is given by the formula.

Specific Heat Capacity

When applying this formula, remember that

Q = mc

• if substance absorbs heat, then is the increase in temperature.

• if the substance releases heat, then is the drop in temperature.

Mass of water, m = 2.0 kg

Initial temperature, 1 = 30°C Final temperature , 2 = 38°C

The increase in temperature, = 2 -1 = 38 – 30 = 8°C

Hence the quantity of heat absorbed by the water , Q = mc = (2.0)(4200)(8) = 67200 J

Example:

Page 30: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Muatan Haba TentuBab 4 Haba

Kuantiti haba yang diserap atau dibebaskan, Q, oleh bahan yang mempunyai jisim m kg dan mengalami perubahan suhu °C, diberi melalui formula berikut:

Muatan Haba Tentu

Apabila menggunakan formula ini, ingat bahawa

Q = mc

• Jika bahan menyerap haba, maka ialah penambahan dalam suhu

• Jika bahan membebaskan haba, maka ialah penurunan dalam suhu.

Jisim air, m = 2.0 kg

Suhu awal, 1 = 30°C Suhu akhir , 2 = 38°C

Penambahan suhu, = 2 -1 = 38 – 30 = 8°C

Maka, kuantiti haba yang diserap oleh air, Q = mc = (2.0)(4200)(8) = 67200 J

Contoh:

Page 31: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Specific Heat CapacityChapter 4 Heat

Specific Heat Capacity Of Substances

Substance Specific heat capacity, c/JKg-1 °C-1

SolidAluminium

CopperIron

Lead

900

390470

130

Substance Specific heat capacity, c/JKg-1 °C-1

LiquidWater

ParaffinMercury

Glycerine

4 200

2 100140

2 430

Generally, the specific heat capacity of solids are lower than the specific heat capacity of liquids.

Also, the specific heat capacity of conductors are lower than the specific heat capacity of insulators.

Mercury is a metal that exists in liquid form at room temperature which explains the low specific heat capacity of mercury.

When heated, substances with small specific heat capacities will experience a large increase in temperature and vice versa.

Page 32: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Mamahami Muatan Haba TentuBab 4 Haba

Muatan Haba Tentu Bahan

Bahan Muatan haba tentu, c/JKg-1 °C-1

PepejalAluminium

KuprumFerum

Plumbum

900

390470

130

Bahan Muatan haba tentu, c/JKg-1 °C-1

CecairAir

ParafinMerkuri

Gliserin

4 200

2 100140

2 430

Muatan haba tentu bagi pepejal lebih rendah daripada muatan haba tentu bagi cecair.

Muatan haba tentu bagi konduktor adalah lebih rendah daripada muatan haba tentu bagi penebat.

Merkuri ialah logam yang wujud dalam bentuk cecair pada suhu bilik. Oleh sebab itu, muatan haba tentu merkuri adalah rendah.

Apabila bahan dipanaskan, bahan yang mempunyai muatan haba tentu yang rendah akan mengalami penambahan suhu yang lebih tinggi dan

sebaliknya.

Page 33: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Specific Heat CapacityChapter 4 Heat

Determination Of Specific Heat Capacity Of Liquid

To determine the specific heat capacity of water

Mass of empty beaker = m1 kg

Mass of beaker plus water = m2 kg Mass of water, m = (m2 –m1) kgInitial temperature of water = 1 °C Final temperature of water = 2 °C Increase in temperature , = (2 - 1) °C

Power of heater = P Watts Time heater is turned on = t seconds

Heat released by heater = Heat absorbed by water Pt = (m2 –m1) c (2 - 1)

Therefore, specific heat capacity of water , c 1212 θθmm

Pt

Assumption made during the experiment : No heat is lost to the surroundings

stirrerthermometer

beaker

cotton wool

polystyrene tile

12V a.c power supply

water

immersionheater

Page 34: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Mamahami Muatan Haba TentuBab 4 Haba

Penentuan Muatan Haba Tentu Cecair

Untuk menghitung muatan haba tentu air :

Jisim bikar kosong = m1 kg

Jisim bikar dan air = m2 kg Jisim air, m = (m2 –m1) kgSuhu awal air = 1 °C Suhu akhir air = 2 °C Kenaikan suhu , = (2 - 1) °C

Kuasa pemanas = P Watt

Tempoh masa pemanas dipasangkan = t saat

Haba yang dibebaskan oleh pemanas = Haba yang diserap oleh airPt = (m2 –m1) c (2 - 1)

Maka, muatan haba tentu air, c 1212 θθmm

Pt

Anggapan dibuat semasa eksperimen : Tiada haba dibebaskan ke persekitaran

PengacauTermometer

Bikar

Wul kapas

Jubin polistirena

12V a.u Bekalan kuasa

Air

Pemanas rendam

Page 35: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Specific Heat CapacityChapter 4 Heat

Determination Of Specific Heat Capacity Of Solid

To determine the specific heat capacity of aluminium

Mass of aluminium block = m kg

Initial temperature of aluminium = 1 °C

Final temperature of aluminium = 2 °C

Increase in temperature , = (2 - 1) °C

Power of heater = P Watts

Time heater is turned on = t seconds

Heat released by heater = Heat absorbed by aluminium Pt = m c (2 - 1)

Hence, specific heat capacity of water , c 12 θθm

Pt

Heater

Power supplyInsulation

Thermometer

Tissuepaper

Oil

Aluminiumblock

Page 36: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Muatan Haba TentuBab 4 Haba

Menentukan Muatan Haba Tentu Pepejal

Untuk menghitung muatan haba tentu aluminium

Jisim bongkah aluminium = m kg

Suhu awal aluminium = 1 °C

Suhu akhir aluminium = 2 °C

Kenaikan suhu , = (2 - 1) °C

Kuasa pemanas = P Watt

Tempoh masa pemanas dipasangkan = t saat

Haba yang dibebaskan pemanas = Haba yang diserap aluminium Pt = m c (2 - 1)

Maka, muatan haba tentu aluminium , c

12 θθmPt

Pemanas

Bekalan kuasaPenebat

Termometer

Kertas tisu

Minyak

Bongkah Aluminium

Page 37: Chapter 4 - Understanding Thermal Equilibrium

Engine cylinderDirectionof waterflow

Radiator fan

cold

air

Water absorbs heat that is produced by the engine of a car.

The heat absorbed is cooled down when the water flows through the radiator and is circulated back again into the engine block to repeat the process.

ITeach – Physics Form 4

Understanding Specific Heat CapacityChapter 4 Heat

Applications Of Specific Heat Capacity – Water as a cooling agent

Water has very high specific heat capacity.

This enables water to absorb a large quantity of heat with a small rise in temperature.

This makes water an ideal cooling agent in the cooling system of motor vehicles such as the car.

Cooling system of a car engine

Page 38: Chapter 4 - Understanding Thermal Equilibrium

Silinder enjinArah aliran air

Kipas radiator

Sejuk

Udara

Air menyerap haba yang dihasilkan oleh enjin kereta.

Haba yang terhasil diserap dan disejukkan oleh air apabila air mengalir melalui radiator dan bergerak mengelilingi bongkah enjin dan kembali semula ke radiator untuk mengulangi proses penyejukan enjin.

ITeach – Fizik Tingkatan 4

Memahami Muatan Haba TentuBab 4 Haba

Aplikasi Muatan Haba Tentu – Air Sebagai Agen Penyejuk

Air mempunyai muatan haba tentu yang tinggi

Ini membolehkan air untuk menyerap haba pada kuantiti yang banyak dengan kenaikan suhu yang sedikit.

Ini menjadikan air sebagai agen penyejuk yang baik dalam sistem penyejuk kenderaan seperti kereta.

Sistem penyejuk enjin

kereta

Page 39: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Specific Heat CapacityChapter 4 Heat

Pots And Pans

The body of pots and pans is made of metal so that when heated, its temperature will rise quickly to enable the food to be cooked in a short period of time.

The handle is made of insulator which will experience only a small rise in temperature so that the pots and pans can be handled without the handler’s hand being burnt.

Made of metalMade of insulator

Page 40: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Muatan Haba TentuBab 4 Haba

Periuk dan Kuali

Bahagian badan periuk dan kuali diperbuat daripada logam supaya apabila ia dipanaskan, suhu akan bertambah dengan cepat untuk membolehkan makanan dapat dimasak dalam masa singkat.

Pemegang periuk dan kuali diperbuat daripada bahan penebat. Ini kerana pertambahan suhu pada bahan penebat adalah kecil. Jadi, hanya sedikit haba daripada pemegang dipindah kepada orang yang memegang periuk atau kuali.

Diperbuat daripada logam

Diperbuat daripada bahan penebat

Page 41: Chapter 4 - Understanding Thermal Equilibrium

Warm land

ITeach – Physics Form 4

Understanding Specific Heat CapacityChapter 4 Heat

SEA BREEZE

Warm air above the land rises

Cool air from the sea moves towards the land to replace the rising air.

Applications Of Specific Heat Capacity - Sea Breeze

During the day, land gets heated up faster than the sea water.

Hot air from the surface of the land rises creating a region of low pressure.

Cool air from the sea then flows towards the land creating sea breeze.

Page 42: Chapter 4 - Understanding Thermal Equilibrium

Permukaan darat lebih panas

ITeach – Fizik Tingkatan 4

Memahami Muatan Haba TentuBab 4 Haba

BAYU LAUT

Udara panas di atas permukaan darat naik

Udara sejuk dari laut bergerak ke arah darat menggantikan udara panas.

Aplikasi Muatan Haba Tentu - Bayu Laut

Pada waktu siang, permukaan darat menjadi panas lebih cepat daripada air laut.

Udara panas daripada permukaan darat naik dan membentuk kawasantekanan rendah.

Udara sejuk dari laut mengalir ke arah darat membentuk bayu laut.

Page 43: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Specific Heat CapacityChapter 4 Heat

Applications Of Specific Heat Capacity – Land Breeze

The specific heat capacity of land is higher than that of sea water.

Land cools faster than the sea at night.

Hot air above the surface of the sea rises.

WARM SEA

Warm air above the sea rises

LAND BREEZECool air fro

m the land

moves towards the sea

to replace the rising air.

The cool air from the land flows towards the sea creating land breeze.

Page 44: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Muatan Haba TentuBab 4 Haba

Aplikasi Muatan Haba Tentu – Bayu Darat

Muatan haba tentu di darat lebih tinggi daripada muatan haba tentu air laut.

Pada waktu malam, darat menjadi sejuk lebih cepat daripada laut.

Udara panas di atas permukaan laut naik.

Air laut yang panas

Udara panas di atas laut naik

Bayu DaratUdara sejuk daripada

darat bergerak ke arah

laut menggantikan

udara panas

Udara sejuk daripada darat bergerak ke arah laut menghasilkan bayu darat.

Page 45: Chapter 4 - Understanding Thermal Equilibrium

Chapter 4 HeatChapter 4 Heat

2.1 Arah Mata Angin

ITeach – Physics Form 4

4.3 Understanding Specific Latent Heat

Page 46: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Specific Latent HeatChapter 4 Heat

The Heating Curve

• At B, the solid begins to melt.

• Heat that is absorbed by the solid is used to overcome the force of attraction between the molecules in the solid.

• The heat that is absorbed does not increase the kinetic energy of the molecules, hence when melting occurs, the temperature of the substance remains constant.

• At D, the liquid begins to vapourize.

• The heat absorbed is used to overcome the force of attraction between the molecules in the liquid and to overcome the atmospheric pressure as the liquid changes into gas.

• The kinetic energy of the molecules does not increase, hence when vapourization occurs, the temperature remains constant.

AB C

D E

FLiquidand gas Gas

LiquidSolidand liquid

Solid

Temperature (°C)

Time (s)

Boiling point

Melting point

0

Roomtemperature

Page 47: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Haba Pendam TentuBab 4 Haba

Lengkung Pemanasan

• Pada B, pepejal mula melebur.

• Haba yang diserap oleh pepejal digunakan untuk mengatasi daya tarikan antara molekul-molekul pepejal.

• Semasa peleburan, suhu bahan adalah malar kerana haba yang diserap tidak menambahkan tenaga kinetik molekul.

• Pada D, cecair mula mengewap.

• Haba yang diserap digunakan untuk mengatasi daya tarikan antara molekul-molekul cecair dan mengatasi tekanan atmosfera semasa cecair bertukar kepada gas.

• Semasa pengewapan berlaku, suhu bahan adalah malar kerana tenaga kinetik molekul tidak bertambah.

AB C

D E

FCecair dan gas Gas

CecairPepejaldan cecair

Pepejal

Suhu (°C)

Masa (s)

Takat didih

Takat lebur

0

Suhu bilik

Page 48: Chapter 4 - Understanding Thermal Equilibrium

Example

ITeach – Physics Form 4

Understanding Specific Latent HeatChapter 4 Heat

SolidLiquid

(latent heat absorbed)

Melting

Specific Latent Heat Of Fusion

The quantity of heat absorbed or released at constant temperature when a substance changes state is known as latent heat.

Latent heat of fusion is the heat absorbed by a melting solid

Specific latent heat of fusion is the quantity of heat that is needed to change 1 kg of a substance in its solid form into liquid at its melting point (no change in temperature).

• The specific latent heat of fusion of ice is 334000 J kg-1.

• This means that 334000 J of heat is needed to be absorbed by 1 kg of ice to completely melt at its melting point of 0°C.

Change of state form solid to liquid.

Page 49: Chapter 4 - Understanding Thermal Equilibrium

Contoh

ITeach – Fizik Tingkatan 4

Memahami Haba Pendam Tentu PelakuranBab 4 Haba

PepejalCecair

(haba pendam diserap)

Peleburan

Haba Pendam Tentu Pelakuran

Kuantiti haba yang diserap atau dibebaskan pada suhu malar apabila bahan berubah keadaan dipanggil haba pendam.

Haba pendam pelakuran ialah haba yang diserap oleh pepejal yang sedang melebur.

Haba pendam tentu pelakuran ialah kuantiti haba yang diperlukan untuk menukarkan 1 kg bahan daripada keadaan pepejal kepada cecair pada takat lebur (tiada perubahan suhu).

• Haba pendam tentu pelakuran ialah 334000 J kg-1. • Ini bererti 334000 J haba diperlukan bagi 1 kg ais untuk melebur

sepenuhnya menjadi air iaitu berubah keadaan daripada pepejal kepada cecair pada takat lebur 0°C.

Perubahan keadaan daripada pepejal kepada cecair.

Page 50: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Specific Latent HeatChapter 4 Heat

Determination Of The Specific Latent Heat Of Fusion Of Ice Electrical heaterSwitched off

Ice

Beaker A

Apparatus A

With the electrical heater switched off (Apparatus A) : Time taken = t seconds

Mass of water collected in beaker = mA kg

With the electrical heater switched on (Apparatus B) : Power of heater = P watt

Time heater is switched on = t seconds rheostat

power supply(transformer)

beaker B

ice

Apparatus B

Mass of water collected in beaker = mB kg Mass of ice melts due to the heater , m = (mB – mA) kg Hence specific latent heat of vapourization of ice ,

AB mmP

mP

ttlice

Page 51: Chapter 4 - Understanding Thermal Equilibrium

ITeach – FizikTingkatan 4

Memahami Muatan Haba Tentu PelakuranBab 4 Haba

Menentukan Muatan Haba Pendam Tentu Pelakuran AisPemanas elektrikSuis ditutup

Ais

Bikar A

Radas A

Suis pemanas elektrik ditutup (Radas A) :

Masa yang diambil = t saat Jisim air terkumpul dalam bikar = mA kg

Suis pemanas elektrik dipasang (Radas B) : Kuasa pemanas = P watt

Tempoh masa pemanas dipasang = t saat

Reostat

Bekalan kuasa(transformer)

Bikar B

Ais

Radas B

Jisim air terkumpul dalam bikar = mB kg Jisim ais cair disebabkan pemanas, m = (mB – mA) kg Haba pendam tentu pelakuran ais,

AB mmP

mP

ttlais

Page 52: Chapter 4 - Understanding Thermal Equilibrium

Example Specific latent heat of vapourization of water is 2260000 J kg-1. This mean that in order to change 1 kg of water to 1 kg of steam, the boiling water needs to absorb 2260000 J of heat.

ITeach – Physics Form 4

Understanding Specific Latent HeatChapter 4 Heat

Specific Latent Heat Of Vapourization

GasLiquid(boiling)

latent heat absorbed

Latent heat of vapourization is the heat absorbed by a boiling liquid.

The specific latent heat of vapourization is the quantity of heat that is required to change 1 kg of a liquid at its boiling point into gas or vapour without any change in temperature.

Page 53: Chapter 4 - Understanding Thermal Equilibrium

ContohHaba pendam tentu pengewapan ialah 2260000 J kg-1. Ini bermaksud 2260000 J haba diperlukan untuk menukarkan air kepada gas atau wap pada takat didih dengan suhu malar.

ITeach – Fizik Tingkatan 4

Memahami Haba Pendam TentuBab 4 Haba

Haba Pendam Tentu Pengewapan

GasCecair(Pendidihan)

haba pendam diserap

Haba pendam pengewapan ialah haba yang diserap oleh cecair yang mendidih.

Haba pendam tentu pengewapan ialah kuantiti haba yang diperlukan untuk menukarkan 1 kg cecair kepada gas atau wap pada takat didih tanpa sebarang perubahan suhu.

Page 54: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Specific Latent HeatChapter 4 Heat

Determination Of The Specific Latent Heat Of Vapourization Of Water

Electronic balance

Water

Immersion heaterBeaker

To power supply

Power of heater = P watt

Mass of water when water starts to boil = m1 kg Mass of water after boiling for t seconds = m2 kg Mass of water that changed into steam(vapour), m = (m1 – m2) kg

hence, the specific latent heat of vapourization of water ,

21 mmP

mP

ttl ionvapourizat

Page 55: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Haba Pendam Tentu PengewapanBab 4 Haba

Menentukan Haba Pendam Tentu Pengewapan Air

Penimbang elektronik

Air

Pemanas rendamBikar

Ke bekalan kuasa

Kuasa pemanas = P watt

Jisim air semasa mula mendidih= m1 kg Jisim air selepas pendidihan selama t saat = m2 kg Jisim air yang berubah kepada stim/wap, m = (m1 – m2) kg

Haba pendam tentu pengewapan air,

21 mmP

mP

ttlpengewapan

Page 56: Chapter 4 - Understanding Thermal Equilibrium

Autoclave

ITeach – Physics Form 4

Understanding Specific Latent HeatChapter 4 Heat

Applications Of Specific Latent Heat

Keeping fish fresh with ice When ice melts, a large quantity of latent heat is absorbed by the fish to maintain the freshness of the fish.

Steaming food • When steam condenses, it releases a large amount

of latent heat which is absorbed by the food.

• This will cook the food faster.

• Autoclave is used to sterilise medical equipments in the hospital.

• The large amount of heat released when steam in the autoclave condensed kills germs and bacteria on the medical equipments.

Page 57: Chapter 4 - Understanding Thermal Equilibrium

Autoklaf

ITeach – Fizik Tingkatan 4

Memahami Haba Pendam TentuBab 4 Haba

Aplikasi Haba Pendam Tentu

Menyimpan ikan dengan ais

Apabila ais cair, ikan menyerap haba pendam tentu ais dalam kuantiti yang banyak untuk mengekalkan kesegaran ikan.

Makanan berstim • Apabila stim terkondensasi, ia membebaskan haba

pendam tentu yang banyak yang kemudiannya diserap oleh makanan.

• Ini membolehkan makanan dimasak dengan cepat.

• Autoklaf digunakan untuk mensteril peralatan perubatan di hospital.

• Haba dalam amaun yang banyak dibebaskan apabila stim di dalam autoklaf terkondensasi. Haba yang dibebaskan membunuh bekteria dan mikroorganisma pada alatan perubatan.

Page 58: Chapter 4 - Understanding Thermal Equilibrium

Chapter 4 HeatChapter 4 Heat

2.1 Arah Mata Angin

ITeach – Physics Form 4

4.4 Understanding The Gas Laws

Page 59: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding The Gas LawsChapter 4 Heat

The Kinetic Theory of Gas

Gases can be studies from the motion of gas molecules.

The Kinetic Theory Of Gas is used to study the motion of gas molecules.

The Kinetic Theory of Gas is based on below assumptions:• Gases consist of molecules , • The molecules are in constant, random motion and frequently collide with each

other and with the walls of any container. • The motion of gas molecules obeyed The Newton’s Motion Law,• The collisions between the molecules and the walls of the container are

elastic.• The volume occupied by the molecules is very small and can be neglected.• The tima of impact during collisions can be neglected compared to the time

between the collisions.

Page 60: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Hukum GasBab 4 Haba

Teori Kinetik Gas

Gas boleh dikaji melalui pergerakan molekul-molekul gas.

Teori Kinetik Gas digunakan untuk mengkaji pergerakan molekul-molekul gas.

Teori Kinetik Gas adalah berdasarkan anggapan-anggapan yang berikut:• Gas terdiri daripada molekul-molekul, • molekul-molekul gas bergerak secara rawak pada semua arah dengan kelajuan

yang tinggi, • pergerakan molekul-molekul gas mematuhi Hukum Gerakan Newton,• pelanggaran antara molekul-molekul gas dan juga antara molekul gas dengan

dinding bekas adalah kenyal.• isipadu yang dipenuhi oleh molekul gas adalah sangat kecil dan boleh

diabaikan.• masa hentaman ketika pelanggaran boleh diabaikan berbanding masa di

antara pelanggaran.

Page 61: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding The Gas LawsChapter 4 Heat

Boyle’s Law

For a fixed mass of gas at a fixed temperature, the pressure of the gas is inversely proportional to its volume.

Robert Boyle

Page 62: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Tingkatan 4

Hukum Boyle

Menyatakan bahawa tekanan P, suatu jisim tetap gas adalah berkadar songsang dengan isi padunya, V, pada suhu malar.

Robert Boyle Jisim

Tekanan Suhu

Memahami Hukum GasBab 4 Haba

Page 63: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding The Gas LawsChapter 4 Heat

Explaining Boyle’s Law Using Kinetic Theory of Gas

When the volume of a gas is decreased, The density (number of gas molecules per unit volume) increases

The surface area of the container decreases

The rate of collision between gas molecules and the walls of the container increases

The rate of change of momentum (force) of the gas molecules exert on the walls of the container increases,

The force per unit surface area (pressure) of the gas exerts on the wall increases

Page 64: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Hukum GasBab 4 Haba

Menerangkan Hukum Boyle Menggunakan Teori Kinetik Gas

Apabila isipadu gas berkurang,, Ketumpatan (bilangan molekul gas per isipadu) bertambah.

Luas permukaan bekas berkurang.

Kadar pelanggaran antara gas molekul dengan dinding bekas bertambah.

Kadar perubahan momentum (daya) yang dikenakan molekul gas pada dinding bekas bertambah.

Daya per unit luas permukaan (tekanan) yang dikenakan oleh molekul gas pada dinding bekas bertambah.

Page 65: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding The Gas LawsChapter 4 Heat

Boyle’s Law

Mathematically,

Where k = a constant, and when the temperature is constant.

Hence P1V1 = P2V2

Whereby P1 = initial pressureP2 = final pressure

V1 = initial volumeV2 = final volume

0 V

P

0

PV

P 0 V

PVP

01V

The relationship between pressure and volume of a gas at a fixed temperature is shown as in

VkP

V1P

Page 66: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Hukum GasBab 4 Haba

Hukum Boyle

Di mana k = a pemalar, dan apabila suhu malar.

Maka P1V1 = P2V2

Di mana P1 = Tekanan awalP2 = Tekanan akhir

V1 = Isipadu awalV2 = Isipadu akhir

0 V

P

0

PV

P 0 V

PVP

01V

Hubungan antara tekanan dan isipadu gas pada suhu malar ditunjukkan seperti di bawah:

VkP

V1P

Page 67: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding The Gas LawsChapter 4 Heat

Boyle’s Law Demonstration

Apparatus and materials• Boyle's Law apparatus • Foot pump and adaptor

The apparatus has been specially designed to give quick and clear readings.

A sample of dry air is trapped in a tall, wide glass tube by a piston of oil.

The volume is found from the length of the air column, which should be clearly visible.

The pressure is read from a Bourdon gauge connected to the air over the oil reservoir.

The foot pump is attached to the oil reservoir and is used to change the pressure

Page 68: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Hukum GasBab 4 Haba

Demostrasi Hukum Boyle

Bahan dan radas• Radas Hukum Boyle • Penyesuai dan Pam kaki

Radas direka khas untuk memberi bacaan yang jelas dan mudah.

Satu sampel udara kering diperangkap tiub kaca tinggi dan lebar oleh omboh minyak.

Isipadu dihitung daripada panjang turus udara yang boleh dilihat dengan jelas.

Bacaan tekanan dibaca daripada tolok Bourdon yang bersambung dengan udara dalam tekungan minyak.Pam kaki disambungkan kepada takungan minyak dan digunakan untuk mengubah tekanan.

Page 69: Chapter 4 - Understanding Thermal Equilibrium

• For example : temperature = 300 K is correct but 3000K is wrong

ITeach – Physics Form 4

Understanding The Gas LawsChapter 4 Heat

• When the velocity of a gas decreases, its kinetic energy decreases, hence its temperature drops.

• -273oC is the lowest possible temperature that can be achieved.

• At this temperature, the volume of a gas becomes negligible and its volume becomes zero.

• -273oC on the Celsius temperature scale is also known as the absolute zero, or 0 Kelvin ( 0 K ) on the absolute temperature scale.

• Note : Temperatures measures in the absolute temperature scale does not have the word “degree” before the unit Kelvin.

Absolute Zero And The Kelvin Temperature Scale

Page 70: Chapter 4 - Understanding Thermal Equilibrium

• Contoh : suhu = 300 K adalah betul tetapi 3000K adalah salah.

ITeach – Fizik Tingkatan 4

Memahami Hukum GasBab 4 Haba

• Apabila halaju gas berkurang, tenaga kinetik gas juga berkurang. Maka, suhu gas menurun.

• -273oC adalah suhu paling rendah yang boleh dicapai.

• Pada suhu ini, isipadu gas menjadi sangat rendah dan isipadunya menjadi sifar.

• -273oC pada skala suhu Celsius dipanggil sifar mutlak, atau 0 Kelvin ( 0 K ) pada skala suhu sifar mutlak.

• Nota : Suhu yang diukur dalam skala suhu sifar mutlak tiada perkataan “darjah” sebelum unit Kelvin.

Suhu Sifar Mutlak dan Skala Suhu Kelvin

Page 71: Chapter 4 - Understanding Thermal Equilibrium

(100 + 273) = 373 K

ITeach – Physics Form 4

Understanding The Gas LawsChapter 4 Heat

The Relationship Between The Kelvin Temperature Scale And The Celsius Temperature

where T is the temperature on the Kelvin scale and is the temperature on the Celsius scale.

Temperature

Celsius KelvinT = ( + 273) K

Situations

(0 – 273) = -273C 0 K (zero Kelvin) absolute zero

0 C (0 + 273) = 273 Kfreezing point of water

30 C (30 + 273) = 303 Kroom temperature

100 C boiling point of water

T = ( + 273) K

Page 72: Chapter 4 - Understanding Thermal Equilibrium

(100 + 273) = 373 K

ITeach – Fizik Tingkatan 4

Memahami Hukum GasBab 4 Haba

Hubungan antara Skala Suhu Kelvin dan Skala Suhu Celcius

Dimana T adalah suhu pada skala Kelvin dan adalah suhu pada skala Celcius.

Suhu

Celsius KelvinT = ( + 273) K

Keadaan

(0 – 273) = -273C 0 K (sifar Kelvin) Sifar mutlak

0 C (0 + 273) = 273 KTakat beku air

30 C (30 + 273) = 303 KSuhu bilik

100 C Takat didih air

T = ( + 273) K

Page 73: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding The Gas LawsChapter 4 Heat

Charles’ Law

For a fixed mass of gas at constant pressure, the volume of the gas is directly proportional to its absolute temperature.

Jacques Charles

Page 74: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Hukum GasBab 4 Haba

Hukum Charles

Bagi suatu jisim gas yang tetap pada tekanan malar, isipadu gas berkadar langsung dengan suhu mutlaknya.

Jacques Charles

Page 75: Chapter 4 - Understanding Thermal Equilibrium

the force exerted by the gas

molecules on the walls of the container

to maintain the same pressure in the container, the volume of the gas must increase.

ITeach – Physics Form 4

Understanding The Gas LawsChapter 4 Heat

Explaining Charles’ Law Using Kinetic Theory Of Gas

gas temperature velocity kinetic energy

frequency of collision between

the gas molecules with the walls of the

container

Page 76: Chapter 4 - Understanding Thermal Equilibrium

Daya yang dikenakan oleh

molekul gas pada dinding bekas

Isipadu gas mesti ditambah untuk mengekalkan tekanan yang sama pada bekas.

ITeach – Fizik Tingkatan 4

Memahami Hukum GasBab 4 Haba

Menerangkan Hukum Charles Menggunakan Teori Kinetik Gas

Suhu gas Halaju Tenaga kinetik

Kekerapan pelanggaran

antara molekul gas dan dinding

bekas

Page 77: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding The Gas LawsChapter 4 Heat

Charles’ Law

The magnitude of constant depends on pressure, mass and nature of a gas.

Charles' law is useful for calculating the volume of a gas at any required temperature if the volume at some other temperature is known by using the

following equation.

Whereby

V1 = initial volume

V2 = final volume

T1 = initial temperature in Kelvin.

T2 = final temperature in Kelvin.

2

2

1

1

TV

TV

TV TconstantV constantTV

Page 78: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Hukum GasBab 4 Haba

Hukum Charles

Magnitud bagi pemalar bergantung kepada tekanan, jisim dan keadaan gas.

Huku Charles digunakan untuk menghitung isipadu gas pada suhu yang dikehendaki jika isipadu pada suhu yang berlainan diketahui melalui persamaan yang berikut.

Dimana

V1 = Isipadu awal

V2 = Isipadu akhir

T1 = Suhu awal dalam Kelvin.

T2 = Suhu akhir dalam Kelvin.

2

2

1

1

TV

TV

TV TPemalarV PemalarTV

Page 79: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding The Gas LawsChapter 4 Heat

Graphical representations of Charles’ Law

If the temperature on the temperature axis is measured in units of Kelvin, then

If the unit of temperature on the temperature axis is measured in the unit of °C, then

V

T0 V

TV

xx

xx

0 T

V

°C (°C)

Page 80: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Hukum GasBab 4 Haba

Graf-graf Mengenai Hukum Charles

Jika suhu pada paksi suhu diukur dalam unit Kelvin, maka

Jika unit suhu pada paksi suhu diukur dalam unit °C, maka

V

T0 V

TV

xx

xx

0 T

V

°C (°C)

Page 81: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding The Gas LawsChapter 4 Heat

Charles’ Law Experiment

As temperature increases, the air column increases in length, showing that the volume of the trapped air has increased.

The data obtained from the experiment is represented graphically as shown. x x x x

thermometer

rulersulphuric acid column

trapped air

bunsen burner

water bath

retort stand

x x x x

0

Length of air column (cm)

Temperature (°C)-273

Page 82: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Hukum GasBab 4 Haba

Eksperimen Hukum Charles

Apabila suhu bertambah, panjang turus udara bertambah. Ini menunjukkan isipadu udara yang terperangkap telah bertambah.

Data yang diperolehi daripada eksperimen di atas ditunjukkan dalam bentuk graf seperti di sebelah: x x x x

Termometer

PembarisTurus asid sulfurik

Udara terperangkap

Penunu Bunsen

Rendaman air

Kaki retort

x x x x

0

Panjang turus udara (cm)

Suhu (°C)-273

Page 83: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding The Gas LawsChapter 4 Heat

Pressure Law

For a fixed mass of gas at constant volume, the pressure of the gas is directly proportional to its absolute temperature.

According to the Kinetic Theory of Gas

• When a gas is heated at constant volume, its molecules gain kinetic energy.

• The gas thus moves with higher speed resulting in an increase in the frequency of collision between the gas molecules and the walls of the container.

• This results in an increase in the force and hence pressure in the container.

Page 84: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Hukum GasBab 4 Haba

Hukum Tekanan

Bagi jisim gas yang tetap, tekanan gas itu berkadar langsung dengan suhu mutlaknya.

Mengikut Teori Kinetik Gas:

• Apabila gas dipanaskan pada isipadu malar, molekul-molekul gas mendapat tenaga kinetik.

• Molekul gas akan bergerak pada kelajuan yang tinggi. Ini menyebabkan kekerapan pelanggaran antara molekul gas dengan dinding bekas bertambah.

• Ini mengakibatkan daya yang dikenakan molekul gas pada dinding bekas bertambah. Maka, tekanan di dalam bekas juga bertambah.

Page 85: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding The Gas LawsChapter 4 Heat

Pressure Laws

Mathematically,

Where k = a constant, and when the volume is constant.

Hence to Pressure Law

P1 = initial pressureP2 = final pressure

T1 = initial temperature in KelvinT2 = final temperature in Kelvin

P

T0 P

TP

0

kTPTP

2

2

1

1

TP

TP

The graphs illustrate the Pressure law

Page 86: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Hukum GasBab 4 Haba

Hukum Tekanan

Di mana k = a pemalar, dan apabila isipadu malar.

Maka, hukum tekanan

P1 = Tekanan awalP2 = Tekanan akhir

T1 = Suhu awal dalam KelvinT2 = Suhu akhir dalam Kelvin

P

T0 P

TP

0

kTPTP

2

2

1

1

TP

TP

Graf menjelaskan Hukum Tekanan

Page 87: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding The Gas LawsChapter 4 Heat

Pressure Law Experiment

The water is heated to increase the temperature of the trapped air in the round base flask.

Relort stand

GasRound base flask (250 ml

Water

Bunsen burner

Bourdon gauge

Block of wood

Asbestos sheet

Rubber tube

Thermometer

A sample data obtained

The pressure of the gas in the flask is read from the Bourbon Gauge.

Page 88: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Hukum GasBab 4 Haba

Eksperimen Hukum Tekanan

Air dipanaskan untuk meningkatkan suhu udara terperangkap di dalam kelalang dasar bulat.

Kaki retort

GasKelalang dasar bulat (250 ml)

Air

Penunu Bunsen

Tolok Bourdon

Bongkah kayu

Kepingan simen

Tiub getah

Termometer

Sampel data diperolehi

Tekanan gas di dalam kelalang di baca menggunakan Tolok Bourbon.

Page 89: Chapter 4 - Understanding Thermal Equilibrium

The End

i - Teach

Page 90: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Thermal EquilibriumChapter 4 Heat

Heat

Heat is a form of energy and is measured in Joules (J).

When an object is heated, it absorbs heat energy and its temperature increases.

Example : Water absorbs heat resulting in a rise in temperature.

When an object is cooled, it releases heat energy and its temperature decreases.

Example : When we sweat, heat is released from our body and our body cools down.

Page 91: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Keseimbangan TermaBab 4 Haba

Haba

Haba ialah satu bentuk tenaga dan diukur dalam unit Joule (J).

Apabila suatu objek dipanaskan, objek itu menyerap tenaga haba dan suhu objek bertambah.

Contoh : Air menyerap haba menyebabkan suhu air bertambah.

Apabila suatu objek disejukkan, ia membebaskan tenaga haba dan menyebabkan suhu berkurang.

Contoh : Apabila kita berpeluh, haba dibebaskan daripada badan. Pembebasan haba menyebabakn badan berasa sejuk.

Page 92: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Thermal EquilibriumChapter 4 Heat

Temperature The higher the temperature of an object, the more heat energy is in the object.

Mass An object with bigger mass contains more heat energy.

Material Objects of the same mass and temperature but made of different materials contains different

amount of heat energy.

The total amount of heat contains in an object depends on

Page 93: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Keseimbangan TermaBab 4 Haba

Suhu Semakin tinggi suhu sesuatu objek, semakin banyak tenaga haba terdapat dalam objek itu.

Jisim Objek yang mempunyai jisim yang lebih besar mengandungi tenaga haba yang lebih banyak.

Bahan Objek-objek yang mempunyai jisim dan suhu yang sama tetapi berlainan bahan mengandungi tenaga

haba yang berlainan.

Jumlah haba yang terdapat dalam sesuatu objek bergantung kepada

Page 94: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Thermal EquilibriumChapter 4 Heat

Temperature

• Temperature is the degree of hotness of a body.

• A hot object will have a temperature higher than a cold object.

• Two objects having the same temperature may not contain the same amount of heat energy in them.

• Example1 litre of water at a temperature of 800C contains more heat than 0.5 litre of water of the same temperature.

Page 95: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Keseimbangan TermaBabr 4 Haba

Suhu

• Suhu ialah darjah kepanasan suatu jasad.

• Objek yang panas mempunyai suhu yang tinggi berbanding objek yang sejuk.

• Dua objek yang mempunyai suhu yang sama mungkin tidak mempunyai kendungan tenaga haba yang sama.

• Contoh1 liter air pada suhu 800C mengandungi lebih banyak haba berbanding 0.5 liter air pada suhu yang sama.

Page 96: Chapter 4 - Understanding Thermal Equilibrium

Quantity of heat absorbed

The more heat is absorbed, the bigger is the rise in temperature of an object.

Mass An object of smaller mass will experience a bigger rise in temperature than an object with bigger mass when both absorbs the same quantity of heat.

Type of material When heated, iron will experience a bigger rise in temperature than plastic.

ITeach – Physics Form 4

Understanding Thermal EquilibriumChapter 4 Heat

The rise in the temperature of an object depends on

Page 97: Chapter 4 - Understanding Thermal Equilibrium

Kuantiti haba yang diserap

Semakin banyak haba yang diserap, semakin banyak kenaikan suhu pada objek.

Jisim

Objek yang mempunyai jisim yang kecil akan mengalami kenaikan suhu yang lebih banyak berbanding objek yang mempunyai jisim yang lebih besar apabila kedua-dua objek menyerap haba dalam kuantiti yang sama.

Jenis bahan Besi akan mengalami kenaikan suhu yang lebih tinggi berbanding plastik.

ITeach – Fizik Tingkatan 4

Memahami Keseimbangan TermaBab 4 Haba

Kenaikan suhu pada objek bergantung kepada

Page 98: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Thermal EquilibriumChapter 4 Heat

Thermal Equilibrium

Body BBody A

Heat

Body A

Heat

Heat

When two objects with different temperatures touches (in thermal contact) with each other

net heat will flow from the object (body A) with higher temperature to the object with lower temperature (body B).

Body A losses heat resulting in a drop in its temperature while body B absorbs heat thereby increasing its temperature until both bodies attain the same temperature.

When this happens, the rate of heat flow from body A to body B and vice versa is the same, that is, there is no net flow of heat between the two bodies. Body B

At this stage, both bodies are said to be in thermal equilibrium.

Higher temperature Lower temperature

Page 99: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Keseimbangan TermaBab 4 Haba

Keseimbangan Terma

Jasad BJasad A

Haba

Jasad A

Haba

Haba

Apabila dua objek yang berlainan suhu bersentuhan (dalam sentuhan terma) antara satu sama lain

haba akan mengalir daripada objek (jasad A) yang bersuhu lebih tinggi kepada objek yang bersuhu lebih rendah (jasad B)

Jasad A akan kehilangan haba dan ini menyebabkan suhunya menurun manakala jasad B akan menyerap haba dan menyebabkan suhunya bertambah sehingga kedua-dua jasad mencapai suhu yang sama.

Apabila kedua-dua jasad mencapai suhu yang sama, kadar aliran haba daripada jasad A ke jasad B dalam kedua-dua arah adalah sama. Maka, tiada pemindahan haba antara dua jasad itu.

Jasad B

Pada peringkat ini, kedua-dua jasad berasa dalam keseimbangan terma.

Suhu tinggi Suhu rendah

Page 100: Chapter 4 - Understanding Thermal Equilibrium

When the temperature of mercury increases, the length of the mercury column increases.

ITeach – Physics Form 4

Understanding Thermal EquilibriumChapter 4 Heat

Thermometry – Liquid-In-Glass Thermometer

To measure temperature of an object

Use a substance that changes with temperature like mercury.

Mercury is a liquid and when temperature of mercury increases, its volume increases.

Mercury

Thin-walledGlass bulb

Celsius scale

Glass tube

Vacuum

Capillary

The mercury thermometer

Mercury is contained in a capillary tube of uniform cross sectional area.

Page 101: Chapter 4 - Understanding Thermal Equilibrium

Apabila suhu merkuri meningkat, panjang turus merkuri juga meningkat.

ITeach – Fizik Tingkatan 4

Memahami Keseimbangan TermaBab 4 Haba

Termometer – Termometer Merkuri dalam Kaca

Untuk mengukur suhu sesuatu objek.

Menggunakan bahan yang berubah dengan suhu seperti merkuri

Merkuri ialah cecair. Apabila suhu merkuri bertambah, isipadunya juga akan bertambah.

Merkuri

Bebulu kaca Berdinding nipis

Skala Celsius

Batang kaca

Vakum

Tiub kapilari

Termometer Merkuri

Merkuri terdapat di dalam tiub kapilari.

Page 102: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Thermal EquilibriumChapter 4 Heat

Why mercury is used in a liquid-in-glass thermometer?

Expand uniformly

High boiling point

Opaque – easier to read the thermometer

Not sticky

Good heat conductor

Page 103: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Keseimbangan TermaBab 4 Haba

Mengapa merkuri digunakan di dalam termometer?

Mengembang dengan seragam

Takat didih tinggi

Bahan legap – mudah untuk membaca

bacaan termometer

Tidak melekit

Konduktor haba yang baik

Page 104: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Thermal EquilibriumChapter 4 Heat

0

Calibration Procedure Of A Mercury-in-glass Thermometer

Ice Point

Steam Point

If when placed in a substance and the length of the mercury column is l when thermal equilibrium is reached,

then the temperature of the substance is C1000100

0

llll

Page 105: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Keseimbangan TermaBab 4 Haba

0

Kaedah Penentukuran Termometer Merkuri dalam Kaca

Titik ais

Titik stim

Apabila termometer diletakkan pada suatu bahan, panjang turus merkuri ialah l apabila keseimbangan terma dicapai.

Suhu bahan ialah, C1000100

0

llll

Page 106: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Thermal EquilibriumChapter 4 Heat

Thermometry – Other Types Of Thermometers

Name of thermometer Diagram

Thermocouple

Constant volume gas thermometer

Resistance thermometer

Physical quantity that changes with temperature

Electromotive force (e.m.f.) produced

Pressure of air in the flask

Resistance of the platinum wire

VCold

junction

Constantanwire

Copper wire

(0°C)

N/m

airflaskwater

Ceramicrod

PlatinumWire coil

Leads

Glasscoating

Page 107: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Keseimbangan TermaBab 4 Haba

Termometer – Lain – lain Jenis Termometer

Nama termometer Rajah

Termogandingan

Termometer isipadu gas tetap

Termometer perintang

Kuantiti fizik yang berubah dengan suhu

Daya elektromotif (e.m.f) dihasilkan

Tekanan udara di dalam kelalang

Perintang wayar platinum

VSimpang

sejuk

Wayar kekal

Wayar kuprum

(0°C)

N/m

UdaraKelalangAir

Rod seramikGegelung wayar

Platinum

Plumbum

Sadur kaca

Page 108: Chapter 4 - Understanding Thermal Equilibrium

Chapter 4 HeatChapter 4 Heat

2.1 Arah Mata Angin

ITeach – Physics Form 4

4.2 Understanding Specific Heat Capacity

Page 109: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Specific Heat CapacityChapter 4 Heat

Factors Affecting The Rise In Temperature When A Substance Is Heated

Substances (Materials)

Rise in temperature of paraffin is more than the rise in temperature of water.

1 kgwater

thermometer

1 kgparaffin

Page 110: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Muatan Haba TentuBab 4 Haba

Faktor yang Mempengaruhi Kenaikan Suhu Apabila Bahan Dipanaskan

Bahan

Kenaikan suhu pada parafin adalah lebih tinggi berbanding kenaikan suhu pada air.

1 kgAir

Termometer

1 kgParafin

Page 111: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Specific Heat CapacityChapter 4 Heat

Factors Affecting The Rise In Temperature When A Substance Is Heated

Mass

Temperature of water in beaker A (smaller mass) rises more than the temperature of water in beaker B.

water

beaker A

water

beaker B

Page 112: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Muatan Haba TentuBab 4 Haba

Faktor yang Mempengaruhi Kenaikan Suhu Apabila Bahan Dipanaskan

Jisim

Suhu air dalam bikar A (Jisim yang lebih kecil) meningkat lebih banyak berbanding suhu air dalam bikar B.

Air

Bikar A

Air

Bikar B

Page 113: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Specific Heat CapacityChapter 4 Heat

Factors Affecting The Rise In Temperature When A Substance Is Heated

Quantity Of Heat Absorbed

Temperature of water in beaker D rise more because more heat is absorbed by the water in beaker D.

beaker C

1 minute

beaker D

5 minutes

Page 114: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Muatan Haba TentuBab 4 Haba

Faktor yang Mempengaruhi Kenaikan Suhu Apabila Bahan Dipanaskan

Kuantiti haba yang diserap

Suhu air di dalam bikar D meningkat lebih tinggi kerana lebih banyak haba diserap oleh air dalam bikar D.

Bikar C

1 minit

Bikar D

5 minit

Page 115: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Specific Heat CapacityChapter 4 Heat

Specific Heat Capacity

This mean that

The specific heat capacity, c, of a substance is the quantity of heat that is absorbed by 1 kilogram of the substance to increase its temperature by 1°C.

The unit of specific heat capacity is J kg-1 °C-1.

The most common substance, that is, water, have a specific heat capacity of 4200 J kg-1 °C-1.

in order for the temperature of 1 kg water to increase by 1°C, the water needs to absorb 4200 J of heat energy, or

for the temperature of 1 kg of water to decrease by 1°C, the water needs to release 4200 J of heat energy.

Page 116: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Muatan Haba TentuBab 4 Haba

Muatan Haba Tentu

Ini bermaksud

Muatan haba tentu, c, suatu bahan ialah kuantiti haba yang diperlukan untuk menaikkan suhu 1 kg bahan sebanyak 1°C.

Unit bagi muatan haba tentu ialah J kg-1 °C-1.

Air mempunyai muatan haba tentu sebanyak 4200 J kg-1 °C-1.

Air perlu menyerap 4200 J tenaga haba untuk menaikkan suhu sebanyak 1°C bagi 1 kg air atau

Air perlu membebaskan 4200 J tenaga haba untuk menurunkan suhu sebanyak 1°C bagi 1 kg air.

Page 117: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Specific Heat CapacityChapter 4 Heat

The quantity of heat absorbed or released, Q, by a substance of mass m kg experiencing a change of temperature of °C, is given by the formula.

Specific Heat Capacity

When applying this formula, remember that

Q = mc

• if substance absorbs heat, then is the increase in temperature.

• if the substance releases heat, then is the drop in temperature.

Mass of water, m = 2.0 kg

Initial temperature, 1 = 30°C Final temperature , 2 = 38°C

The increase in temperature, = 2 -1 = 38 – 30 = 8°C

Hence the quantity of heat absorbed by the water , Q = mc = (2.0)(4200)(8) = 67200 J

Example:

Page 118: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Muatan Haba TentuBab 4 Haba

Kuantiti haba yang diserap atau dibebaskan, Q, oleh bahan yang mempunyai jisim m kg dan mengalami perubahan suhu °C, diberi melalui formula berikut:

Muatan Haba Tentu

Apabila menggunakan formula ini, ingat bahawa

Q = mc

• Jika bahan menyerap haba, maka ialah penambahan dalam suhu

• Jika bahan membebaskan haba, maka ialah penurunan dalam suhu.

Jisim air, m = 2.0 kg

Suhu awal, 1 = 30°C Suhu akhir , 2 = 38°C

Penambahan suhu, = 2 -1 = 38 – 30 = 8°C

Maka, kuantiti haba yang diserap oleh air, Q = mc = (2.0)(4200)(8) = 67200 J

Contoh:

Page 119: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Specific Heat CapacityChapter 4 Heat

Specific Heat Capacity Of Substances

Substance Specific heat capacity, c/JKg-1 °C-1

SolidAluminium

CopperIron

Lead

900

390470

130

Substance Specific heat capacity, c/JKg-1 °C-1

LiquidWater

ParaffinMercury

Glycerine

4 200

2 100140

2 430

Generally, the specific heat capacity of solids are lower than the specific heat capacity of liquids.

Also, the specific heat capacity of conductors are lower than the specific heat capacity of insulators.

Mercury is a metal that exists in liquid form at room temperature which explains the low specific heat capacity of mercury.

When heated, substances with small specific heat capacities will experience a large increase in temperature and vice versa.

Page 120: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Mamahami Muatan Haba TentuBab 4 Haba

Muatan Haba Tentu Bahan

Bahan Muatan haba tentu, c/JKg-1 °C-1

PepejalAluminium

KuprumFerum

Plumbum

900

390470

130

Bahan Muatan haba tentu, c/JKg-1 °C-1

CecairAir

ParafinMerkuri

Gliserin

4 200

2 100140

2 430

Muatan haba tentu bagi pepejal lebih rendah daripada muatan haba tentu bagi cecair.

Muatan haba tentu bagi konduktor adalah lebih rendah daripada muatan haba tentu bagi penebat.

Merkuri ialah logam yang wujud dalam bentuk cecair pada suhu bilik. Oleh sebab itu, muatan haba tentu merkuri adalah rendah.

Apabila bahan dipanaskan, bahan yang mempunyai muatan haba tentu yang rendah akan mengalami penambahan suhu yang lebih tinggi dan

sebaliknya.

Page 121: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Specific Heat CapacityChapter 4 Heat

Determination Of Specific Heat Capacity Of Liquid

To determine the specific heat capacity of water

Mass of empty beaker = m1 kg

Mass of beaker plus water = m2 kg Mass of water, m = (m2 –m1) kgInitial temperature of water = 1 °C Final temperature of water = 2 °C Increase in temperature , = (2 - 1) °C

Power of heater = P Watts Time heater is turned on = t seconds

Heat released by heater = Heat absorbed by water Pt = (m2 –m1) c (2 - 1)

Therefore, specific heat capacity of water , c 1212 θθmm

Pt

Assumption made during the experiment : No heat is lost to the surroundings

stirrerthermometer

beaker

cotton wool

polystyrene tile

12V a.c power supply

water

immersionheater

Page 122: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Mamahami Muatan Haba TentuBab 4 Haba

Penentuan Muatan Haba Tentu Cecair

Untuk menghitung muatan haba tentu air :

Jisim bikar kosong = m1 kg

Jisim bikar dan air = m2 kg Jisim air, m = (m2 –m1) kgSuhu awal air = 1 °C Suhu akhir air = 2 °C Kenaikan suhu , = (2 - 1) °C

Kuasa pemanas = P Watt

Tempoh masa pemanas dipasangkan = t saat

Haba yang dibebaskan oleh pemanas = Haba yang diserap oleh airPt = (m2 –m1) c (2 - 1)

Maka, muatan haba tentu air, c 1212 θθmm

Pt

Anggapan dibuat semasa eksperimen : Tiada haba dibebaskan ke persekitaran

PengacauTermometer

Bikar

Wul kapas

Jubin polistirena

12V a.u Bekalan kuasa

Air

Pemanas rendam

Page 123: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Specific Heat CapacityChapter 4 Heat

Determination Of Specific Heat Capacity Of Solid

To determine the specific heat capacity of aluminium

Mass of aluminium block = m kg

Initial temperature of aluminium = 1 °C

Final temperature of aluminium = 2 °C

Increase in temperature , = (2 - 1) °C

Power of heater = P Watts

Time heater is turned on = t seconds

Heat released by heater = Heat absorbed by aluminium Pt = m c (2 - 1)

Hence, specific heat capacity of water , c 12 θθm

Pt

Heater

Power supplyInsulation

Thermometer

Tissuepaper

Oil

Aluminiumblock

Page 124: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Muatan Haba TentuBab 4 Haba

Menentukan Muatan Haba Tentu Pepejal

Untuk menghitung muatan haba tentu aluminium

Jisim bongkah aluminium = m kg

Suhu awal aluminium = 1 °C

Suhu akhir aluminium = 2 °C

Kenaikan suhu , = (2 - 1) °C

Kuasa pemanas = P Watt

Tempoh masa pemanas dipasangkan = t saat

Haba yang dibebaskan pemanas = Haba yang diserap aluminium Pt = m c (2 - 1)

Maka, muatan haba tentu aluminium , c

12 θθmPt

Pemanas

Bekalan kuasaPenebat

Termometer

Kertas tisu

Minyak

Bongkah Aluminium

Page 125: Chapter 4 - Understanding Thermal Equilibrium

Engine cylinderDirectionof waterflow

Radiator fan

cold

air

Water absorbs heat that is produced by the engine of a car.

The heat absorbed is cooled down when the water flows through the radiator and is circulated back again into the engine block to repeat the process.

ITeach – Physics Form 4

Understanding Specific Heat CapacityChapter 4 Heat

Applications Of Specific Heat Capacity – Water as a cooling agent

Water has very high specific heat capacity.

This enables water to absorb a large quantity of heat with a small rise in temperature.

This makes water an ideal cooling agent in the cooling system of motor vehicles such as the car.

Cooling system of a car engine

Page 126: Chapter 4 - Understanding Thermal Equilibrium

Silinder enjinArah aliran air

Kipas radiator

Sejuk

Udara

Air menyerap haba yang dihasilkan oleh enjin kereta.

Haba yang terhasil diserap dan disejukkan oleh air apabila air mengalir melalui radiator dan bergerak mengelilingi bongkah enjin dan kembali semula ke radiator untuk mengulangi proses penyejukan enjin.

ITeach – Fizik Tingkatan 4

Memahami Muatan Haba TentuBab 4 Haba

Aplikasi Muatan Haba Tentu – Air Sebagai Agen Penyejuk

Air mempunyai muatan haba tentu yang tinggi

Ini membolehkan air untuk menyerap haba pada kuantiti yang banyak dengan kenaikan suhu yang sedikit.

Ini menjadikan air sebagai agen penyejuk yang baik dalam sistem penyejuk kenderaan seperti kereta.

Sistem penyejuk enjin

kereta

Page 127: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Specific Heat CapacityChapter 4 Heat

Pots And Pans

The body of pots and pans is made of metal so that when heated, its temperature will rise quickly to enable the food to be cooked in a short period of time.

The handle is made of insulator which will experience only a small rise in temperature so that the pots and pans can be handled without the handler’s hand being burnt.

Made of metalMade of insulator

Page 128: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Muatan Haba TentuBab 4 Haba

Periuk dan Kuali

Bahagian badan periuk dan kuali diperbuat daripada logam supaya apabila ia dipanaskan, suhu akan bertambah dengan cepat untuk membolehkan makanan dapat dimasak dalam masa singkat.

Pemegang periuk dan kuali diperbuat daripada bahan penebat. Ini kerana pertambahan suhu pada bahan penebat adalah kecil. Jadi, hanya sedikit haba daripada pemegang dipindah kepada orang yang memegang periuk atau kuali.

Diperbuat daripada logam

Diperbuat daripada bahan penebat

Page 129: Chapter 4 - Understanding Thermal Equilibrium

Warm land

ITeach – Physics Form 4

Understanding Specific Heat CapacityChapter 4 Heat

SEA BREEZE

Warm air above the land rises

Cool air from the sea moves towards the land to replace the rising air.

Applications Of Specific Heat Capacity - Sea Breeze

During the day, land gets heated up faster than the sea water.

Hot air from the surface of the land rises creating a region of low pressure.

Cool air from the sea then flows towards the land creating sea breeze.

Page 130: Chapter 4 - Understanding Thermal Equilibrium

Permukaan darat lebih panas

ITeach – Fizik Tingkatan 4

Memahami Muatan Haba TentuBab 4 Haba

BAYU LAUT

Udara panas di atas permukaan darat naik

Udara sejuk dari laut bergerak ke arah darat menggantikan udara panas.

Aplikasi Muatan Haba Tentu - Bayu Laut

Pada waktu siang, permukaan darat menjadi panas lebih cepat daripada air laut.

Udara panas daripada permukaan darat naik dan membentuk kawasantekanan rendah.

Udara sejuk dari laut mengalir ke arah darat membentuk bayu laut.

Page 131: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Specific Heat CapacityChapter 4 Heat

Applications Of Specific Heat Capacity – Land Breeze

The specific heat capacity of land is higher than that of sea water.

Land cools faster than the sea at night.

Hot air above the surface of the sea rises.

WARM SEA

Warm air above the sea rises

LAND BREEZECool air fro

m the land

moves towards the sea

to replace the rising air.

The cool air from the land flows towards the sea creating land breeze.

Page 132: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Muatan Haba TentuBab 4 Haba

Aplikasi Muatan Haba Tentu – Bayu Darat

Muatan haba tentu di darat lebih tinggi daripada muatan haba tentu air laut.

Pada waktu malam, darat menjadi sejuk lebih cepat daripada laut.

Udara panas di atas permukaan laut naik.

Air laut yang panas

Udara panas di atas laut naik

Bayu DaratUdara sejuk daripada

darat bergerak ke arah

laut menggantikan

udara panas

Udara sejuk daripada darat bergerak ke arah laut menghasilkan bayu darat.

Page 133: Chapter 4 - Understanding Thermal Equilibrium

Chapter 4 HeatChapter 4 Heat

2.1 Arah Mata Angin

ITeach – Physics Form 4

4.3 Understanding Specific Latent Heat

Page 134: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Specific Latent HeatChapter 4 Heat

The Heating Curve

• At B, the solid begins to melt.

• Heat that is absorbed by the solid is used to overcome the force of attraction between the molecules in the solid.

• The heat that is absorbed does not increase the kinetic energy of the molecules, hence when melting occurs, the temperature of the substance remains constant.

• At D, the liquid begins to vapourize.

• The heat absorbed is used to overcome the force of attraction between the molecules in the liquid and to overcome the atmospheric pressure as the liquid changes into gas.

• The kinetic energy of the molecules does not increase, hence when vapourization occurs, the temperature remains constant.

AB C

D E

FLiquidand gas Gas

LiquidSolidand liquid

Solid

Temperature (°C)

Time (s)

Boiling point

Melting point

0

Roomtemperature

Page 135: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Haba Pendam TentuBab 4 Haba

Lengkung Pemanasan

• Pada B, pepejal mula melebur.

• Haba yang diserap oleh pepejal digunakan untuk mengatasi daya tarikan antara molekul-molekul pepejal.

• Semasa peleburan, suhu bahan adalah malar kerana haba yang diserap tidak menambahkan tenaga kinetik molekul.

• Pada D, cecair mula mengewap.

• Haba yang diserap digunakan untuk mengatasi daya tarikan antara molekul-molekul cecair dan mengatasi tekanan atmosfera semasa cecair bertukar kepada gas.

• Semasa pengewapan berlaku, suhu bahan adalah malar kerana tenaga kinetik molekul tidak bertambah.

AB C

D E

FCecair dan gas Gas

CecairPepejaldan cecair

Pepejal

Suhu (°C)

Masa (s)

Takat didih

Takat lebur

0

Suhu bilik

Page 136: Chapter 4 - Understanding Thermal Equilibrium

Example

ITeach – Physics Form 4

Understanding Specific Latent HeatChapter 4 Heat

SolidLiquid

(latent heat absorbed)

Melting

Specific Latent Heat Of Fusion

The quantity of heat absorbed or released at constant temperature when a substance changes state is known as latent heat.

Latent heat of fusion is the heat absorbed by a melting solid

Specific latent heat of fusion is the quantity of heat that is needed to change 1 kg of a substance in its solid form into liquid at its melting point (no change in temperature).

• The specific latent heat of fusion of ice is 334000 J kg-1.

• This means that 334000 J of heat is needed to be absorbed by 1 kg of ice to completely melt at its melting point of 0°C.

Change of state form solid to liquid.

Page 137: Chapter 4 - Understanding Thermal Equilibrium

Contoh

ITeach – Fizik Tingkatan 4

Memahami Haba Pendam Tentu PelakuranBab 4 Haba

PepejalCecair

(haba pendam diserap)

Peleburan

Haba Pendam Tentu Pelakuran

Kuantiti haba yang diserap atau dibebaskan pada suhu malar apabila bahan berubah keadaan dipanggil haba pendam.

Haba pendam pelakuran ialah haba yang diserap oleh pepejal yang sedang melebur.

Haba pendam tentu pelakuran ialah kuantiti haba yang diperlukan untuk menukarkan 1 kg bahan daripada keadaan pepejal kepada cecair pada takat lebur (tiada perubahan suhu).

• Haba pendam tentu pelakuran ialah 334000 J kg-1. • Ini bererti 334000 J haba diperlukan bagi 1 kg ais untuk melebur

sepenuhnya menjadi air iaitu berubah keadaan daripada pepejal kepada cecair pada takat lebur 0°C.

Perubahan keadaan daripada pepejal kepada cecair.

Page 138: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Specific Latent HeatChapter 4 Heat

Determination Of The Specific Latent Heat Of Fusion Of Ice Electrical heaterSwitched off

Ice

Beaker A

Apparatus A

With the electrical heater switched off (Apparatus A) : Time taken = t seconds

Mass of water collected in beaker = mA kg

With the electrical heater switched on (Apparatus B) : Power of heater = P watt

Time heater is switched on = t seconds rheostat

power supply(transformer)

beaker B

ice

Apparatus B

Mass of water collected in beaker = mB kg Mass of ice melts due to the heater , m = (mB – mA) kg Hence specific latent heat of vapourization of ice ,

AB mmP

mP

ttlice

Page 139: Chapter 4 - Understanding Thermal Equilibrium

ITeach – FizikTingkatan 4

Memahami Muatan Haba Tentu PelakuranBab 4 Haba

Menentukan Muatan Haba Pendam Tentu Pelakuran AisPemanas elektrikSuis ditutup

Ais

Bikar A

Radas A

Suis pemanas elektrik ditutup (Radas A) :

Masa yang diambil = t saat Jisim air terkumpul dalam bikar = mA kg

Suis pemanas elektrik dipasang (Radas B) : Kuasa pemanas = P watt

Tempoh masa pemanas dipasang = t saat

Reostat

Bekalan kuasa(transformer)

Bikar B

Ais

Radas B

Jisim air terkumpul dalam bikar = mB kg Jisim ais cair disebabkan pemanas, m = (mB – mA) kg Haba pendam tentu pelakuran ais,

AB mmP

mP

ttlais

Page 140: Chapter 4 - Understanding Thermal Equilibrium

Example Specific latent heat of vapourization of water is 2260000 J kg-1. This mean that in order to change 1 kg of water to 1 kg of steam, the boiling water needs to absorb 2260000 J of heat.

ITeach – Physics Form 4

Understanding Specific Latent HeatChapter 4 Heat

Specific Latent Heat Of Vapourization

GasLiquid(boiling)

latent heat absorbed

Latent heat of vapourization is the heat absorbed by a boiling liquid.

The specific latent heat of vapourization is the quantity of heat that is required to change 1 kg of a liquid at its boiling point into gas or vapour without any change in temperature.

Page 141: Chapter 4 - Understanding Thermal Equilibrium

ContohHaba pendam tentu pengewapan ialah 2260000 J kg-1. Ini bermaksud 2260000 J haba diperlukan untuk menukarkan air kepada gas atau wap pada takat didih dengan suhu malar.

ITeach – Fizik Tingkatan 4

Memahami Haba Pendam TentuBab 4 Haba

Haba Pendam Tentu Pengewapan

GasCecair(Pendidihan)

haba pendam diserap

Haba pendam pengewapan ialah haba yang diserap oleh cecair yang mendidih.

Haba pendam tentu pengewapan ialah kuantiti haba yang diperlukan untuk menukarkan 1 kg cecair kepada gas atau wap pada takat didih tanpa sebarang perubahan suhu.

Page 142: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Physics Form 4

Understanding Specific Latent HeatChapter 4 Heat

Determination Of The Specific Latent Heat Of Vapourization Of Water

Electronic balance

Water

Immersion heaterBeaker

To power supply

Power of heater = P watt

Mass of water when water starts to boil = m1 kg Mass of water after boiling for t seconds = m2 kg Mass of water that changed into steam(vapour), m = (m1 – m2) kg

hence, the specific latent heat of vapourization of water ,

21 mmP

mP

ttl ionvapourizat

Page 143: Chapter 4 - Understanding Thermal Equilibrium

ITeach – Fizik Tingkatan 4

Memahami Haba Pendam Tentu PengewapanBab 4 Haba

Menentukan Haba Pendam Tentu Pengewapan Air

Penimbang elektronik

Air

Pemanas rendamBikar

Ke bekalan kuasa

Kuasa pemanas = P watt

Jisim air semasa mula mendidih= m1 kg Jisim air selepas pendidihan selama t saat = m2 kg Jisim air yang berubah kepada stim/wap, m = (m1 – m2) kg

Haba pendam tentu pengewapan air,

21 mmP

mP

ttlpengewapan

Page 144: Chapter 4 - Understanding Thermal Equilibrium

Autoclave

ITeach – Physics Form 4

Understanding Specific Latent HeatChapter 4 Heat

Applications Of Specific Latent Heat

Keeping fish fresh with ice When ice melts, a large quantity of latent heat is absorbed by the fish to maintain the freshness of the fish.

Steaming food • When steam condenses, it releases a large amount

of latent heat which is absorbed by the food.

• This will cook the food faster.

• Autoclave is used to sterilise medical equipments in the hospital.

• The large amount of heat released when steam in the autoclave condensed kills germs and bacteria on the medical equipments.

Page 145: Chapter 4 - Understanding Thermal Equilibrium

Autoklaf

ITeach – Fizik Tingkatan 4

Memahami Haba Pendam TentuBab 4 Haba

Aplikasi Haba Pendam Tentu

Menyimpan ikan dengan ais

Apabila ais cair, ikan menyerap haba pendam tentu ais dalam kuantiti yang banyak untuk mengekalkan kesegaran ikan.

Makanan berstim • Apabila stim terkondensasi, ia membebaskan haba

pendam tentu yang banyak yang kemudiannya diserap oleh makanan.

• Ini membolehkan makanan dimasak dengan cepat.

• Autoklaf digunakan untuk mensteril peralatan perubatan di hospital.

• Haba dalam amaun yang banyak dibebaskan apabila stim di dalam autoklaf terkondensasi. Haba yang dibebaskan membunuh bekteria dan mikroorganisma pada alatan perubatan.