HEAT Fizik Tg4 B4 2015... · Jenis-jenis termometer Types of thermometers: (a) Termometer cecair-dalam-kaca Liquid-in-glass thermometer (b) Termometer termogandingan Thermocouple

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MODUL • Fizik TINGKATAN 4

•Menerangkankeseimbanganterma/Explain thermal equilibrium•Menerangkanbagaimanatermometercecair-dalam-kacabekerja

Explain how a liquid–in-glass thermometer works

•Definisimuatanhabatentu(c)/Define specific heat capacity (c)

•Menyatakan/State that c = Q

mθ•Menyelesaikanmasalahberkaitanmuatanhabatentu

Solve problems involving specific heat capacity•Menghuraikanaplikasimuatanhabatentu/Describe applications of specific heat capacity•Menentukanmuatanhabatentucecair/Determine the specific heat capacity of a liquid•Menentukanmuatanhabatentupepejal(blokaluminiumataublokkuprum)

Determine the specific heat capacity of a solid (aluminium block or copper block)

•MenyatakanbahawapemindahanhabasemasaperubahanfasatidakmelibatkanperubahansuhuState that transfer of heat during a change of phase does not cause a change in temperature

•Mentakrifkanhabapendamtentu(L)/Define specific latent heat (L)

•NyatakanL/State that L = Qm

•Menentukanhabapendamtentupelakuran/Determine the specific latent heat of fusion•Menentukanhabapendamtentupengewapan

Determine the specific latent heat of vaporisation•Menyelesaikanmasalahyangmelibatkanhabapendamtentu

Solve problems involving specific latent heat

•Menerangkantekanan,suhudanisipadugasdarisegikelakuanmolekulgasExplain gas pressure, temperature and volume in terms of behaviour of gas molecules

•MenentukanhubunganantaratekanandanisipadupadasuhumalarsuatugasberjisimtetapiaituPV=malarDetermine the relationship between pressure and volume at constant temperature for a fixed mass of gas i.e. PV = constant

•Menentukan hubungan antara isi padu dan suhu pada tekananmalar bagi jisim tetap gas iaituVT =malar

Determine the relationship between volume and temperature at constant pressure for a fixed mass of gas i.e. VT

= constant

•MenentukanhubunganantaratekanandansuhupadaisipadumalarbagisuatugasberjisimtetapiaituPT=pemalar

•Determine the relationship between pressure and temperature at constant volume for a fixed mass of gas i.e. PT

= constant

•Menerangkansifarmutlak/Explain absolute zero•Jelaskanskalamutlak/skalaKelvinbagisuhu

Explain the absolute / Kelvin scale of temperature•Menyelesaikanmasalahyangmelibatkantekanan,suhudanisipadusuatugasberjisimtetap

Solve problems involving pressure, temperature and volume of a fixed mass of gas

HABAHEAT4

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4.1 MEMAHAMI KESEIMBANGAN TERMA/UNDERSTANDING THERMAL EQUILIBRIUM

4.2 MEMAHAMI MUATAN HABA TENTU/UNDERSTANDING SPECIFIC HEAT CAPACITY

4.3MEMAHAMI HABA PENDAM TENTU/UNDERSTANDING SPECIFIC LATENT HEAT

4.4 MEMAHAMI HUKUM-HUKUM GAS/UNDERSTANDING THE GAS LAWS

Fizik Tg4 B4 2015(FSY4p).indd 169 10/20/15 2:15 PM

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Heat (Q)

Haba membolehkan sesuatu objek menukar suhu atau mengubah keadaan fizikal:Heat makes an object to change its temperature or change its physical state:(a) Jika keadaan fizikal tidak

berubah, suhunya akan berubah

If the physical state is unchanged, the temperature changes

(b) Jika suhu tetap, keadaan fizikal akan berubah

If the temperature is constant, the physical state changes

Pertukaran Unit Suhu:Change Unit of Temperature:θ ºC = (θ + 273) K

Contoh:Example:(i) 0ºC = (0 + 273) K = 273 K(ii) 27ºC = (27 + 273) K = 300 K(iii) 100ºC = (100 + 273) K = 373 K

Ciri-ciri fizikal (sifat termometri) termometer yang berubah dengan haba / suhuPhysical characteristics (Thermometric properties) of thermometer that change with heat/temperature:(a) Termometer berjenis

cecair-dalam-kaca – panjang / isi padu turus merkuri berubah dengan haba

Liquid-in-glass thermometer – the length / volume of mercury column changes with heat

(b) Termometer termogandingan – d.g.e

berubah terhadap haba Thermocouple

thermometer – the e.m.f/current changes with heat

(c) Termometer rintangan – rintangan berubah terhadap haba.

Resistance thermometer – the resistance changes with heat

(d) Termometer gas – tekanan gas berubah

terhadap haba. Gas thermometer – the

gas pressure changes with heat.

Haba, Q/Heat, Q:• Haba adalah suatu bentuk tenaga

Heat is a form of energy• Haba boleh memanaskan atau

menyejukkan objekHeat can heat up or cool down an object

• Untuk mengetahui suatu objek panas atau sejuk, ukur suhu objek tersebutTo know whether an object is hot or cool, measure the temperature of the object

• Haba mengalir dari kawasan panas ke kawasan sejukHeat flows from a hot region to a cool region

• Haba boleh mengubah keadaan fizikal bahanHeat can change the physical state of a substance

Contoh/Example:• Perubahan pepejal kepada cecair

Solid changes to liquid• Unit S.I. bagi haba: Joule(J)

S.I. Unit of heat: Joule(J)

Alat untuk mengukur suhu ialah termometer

The measuring instrument to measure temperature is thermometer

Jenis-jenis termometerTypes of thermometers:(a) Termometer cecair-dalam-kaca Liquid-in-glass thermometer(b) Termometer termogandingan Thermocouple thermometer(c) Termometer rintangan Resistance thermometer(d) Termometer gas Gas thermometer

Jenis termometer cecair-dalam-kacaLiquid-in-glass thermometer:(a) Termometer merkuri/Mercury thermometer(b) Termometer alkohol/Alcohol thermometer

MEMAHAMI KESEIMBANGAN TERMAUNDERSTANDING THERMAL EQUILIBRIUM4.1

Suhu, T/Temperature, T:

• Definisi/Definition:

Ditakrifkan sebagai darjah

kepanasan objek

Is defined as a degree of hotness

of an object

• Unit S.I.: Kelvin(K)

S.I. Unit: Kelvin(K)

• Menukar unit suhu darjah celsius (°C) kepada Kelvin (K)Change unit of temperature from degree celcius (°C) to kelvin (K)

• θ ºC = (θ + 273) K


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Penentukuran Termometer/Themometer Calibration:Termometer Merkuri/Mercury Thermometer:

Perubahan dalam panjang turus merkuri, Δl = l100 – l0The change in length of mercury column, Δl = l100 – l0

Perubahan suhu, ΔT1 = 100 ºC – 0 ºC = 100 ºCThe change in temperature, ΔT1 = 100 ºC – 0 ºC = 100 ºC

Perubahan dalam panjang pada suhu θ = lθ – l0The change in length at temperature θ = lθ – l0

Oleh itu, suhu/ Therefore, the temperature, θ = lθ – l0

l100 – l0 × 100 ºC

Merkuri digunakan dalam termometer kerana:Mercury is used in the thermometer because:

Ciri-ciriCharacteristics

Penerangan Explanation

LegapOpaque

Senang untuk mengambil bacaanEasy to take a reading

Mengembang seragam dengan habaExpands uniformly with heat

Skala seragamUniform scale

Daya lekitan yang tinggiHigh cohesive force

Tidak membasah tiub dan tidak melekat pada dinding kacaDoes not wet the tube and does not stick to the glass wall

Takat didih yang tinggiHigh boiling point

Boleh mengukur suhu yang tinggiCan measure high temperature

Kepekaan termometer boleh ditingkatkan dengan menggunakan:The sensitivity of the thermometer can be increased by using:(a) tiub kapilari yang kecil narrow capillary tube (b) dinding bebuli kaca yang

nipis thin glass wall bulb(c) saiz bebuli yang kecil small bulb(d) skala maksimum

melebihi 200 °C supaya dapat mengukur suhu yang tinggi

the maximum scale is higher than 200 °C so that it can measure higher temperature.

Termometer merkuriMercury thermometer

Air dari aisyang meleburWater from melting ice

AisIce

Takat aisIce point

100 °c0°c

θ

100

0

Kaki retortRetort stand

StimSteam

Penunu BunsenBunsen burner

Takat stimSteam point

Kasa dawaiWire gauzeXxxxxxxxxxxxxxxxxx

Hubungan antara Haba dan SuhuRelationship between Heat and Temperature:

(a) Apabila suatu objek menyerap haba, suhu meningkat .

When an object absorbs heat, the temperature increases .

(b) Apabila suatu objek membebaskan haba, suhu menurun .

When an object releases heat, the temperature decreases .


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Keseimbangan Terma:Thermal Equilibrium:

Apabila dua objek terdapat perbezaan suhu, haba dipindahkan di antara dua objek itu.When there is a temperature difference between two objects, heat is transferred between the two objects.

Objek yang panas membebaskan haba manakala objek yang sejuk menyerap haba sehingga mencapai keseimbangan terma.

The hot object releases heat whereas the cold object absorbs heat until it reaches thermal equilibrium.

Dalam keseimbangan termaAt thermal equilibrium

PanasHot

SejukCold

A B A B

Haba yang dibebaskan = Haba yang diserap (Kadar bersih pemindahan haba adalah SIFAR)Heat released = Heat absorbed (Rate of net heat transferred is ZERO)

Dua objek bersentuhan yang berlainan suhu akan mencapai keseimbangan terma apabila:Two objects in contact with each other but at different temperatures will reach thermal equilibrium when:

(a) kadar bersih haba yang dipindahkan antara dua objek ialah sifar , dan

the net rate of heat transfer between the two objects is zero , and

(b) suhu dua objek itu adalah sama .

the temperature of the two objects is the same .

Pada keseimbangan terma,At thermal equilibrium,

haba yang diserap oleh objek yang sejuk adalah sama dengan haba yang dibebaskan oleh objek yang panas.

the heat absorbed by the cool object is equal to the heat released by the hot object.

Pada keseimbangan terma,Kadar pemindahan tenaga dari A ke B

= Kadar pemindahan tenaga dari B ke A

At thermal equilibrium,Rate of transfer of energy from A to B

= Rate of transfer of energy from B to A


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Aplikasi keseimbangan termaApplications of thermal equilibrium

Untuk menurunkan suhu orang yang demam panasTo bring down the temperature of the person who has high fever

• Tualayangbasahdiletakkanpadadahiorangyangdemampanas.

The wet towel is placed on the forehead of the person who has high fever.

• Duaobjek(tualabasahdandahi) bersentuhan antara satu sama lain.

The two objects (wet towel and forehead) are in contact with each other.

• Tualabasahmempunyaitenagahabayang lebih

rendah berbanding dahi yang mempunyai tenaga

haba yang lebih tinggi .

The wet towel has lower heat energy compared with the

forehead which has higher heat energy.

• Tenagahabadialirkandaridahiketualasehingga

keseimbangan terma tercapai.

Heat energy is transferred from the forehead to the towel until

thermal equilibrium is reached.

Untuk meningkatkan suhu susu yang sejukTo bring up the temperature of cold milk

• Susuibubolehdisimpandalampetiaisdankemudiandiberi kepada bayi.

Breast milk from a mother can be kept in the refrigerator and given to the baby later.

• Sebelumitu,sebotolsusudirendamdidalamsemangkukair panas.Before that, the bottle of milk is immersed in a bowl of hot water.

• Kedua-duaobjek bersentuhan antara satu sama lain.

The two objects are in contact with each other.

• Susudidalambotolmempunyaitenagahabayang

lebih rendah berbanding mangkuk berisi air panas

yang mempunyai tenaga haba yang lebih tinggi .

The milk in the bottle has lower heat energy compared

with the bowl of hot water which has higher heat energy.

• Tenagahabadialirkandariairpanaskesususehingga


Heat energy is transferred from the hot water to the milk until

thermal equilibrium is reached.

• Padakeseimbanganterma, kadar bersih haba

dialirkan antara tuala dan dahi ialah sifar .

At thermal equilibrium, the net rate of heat transfer

between the towel and the forehead is zero .

• Suhukedua-duaobjek sama .

The temperature of the two objects is the same .

• Kemudiantualasuamdirendamdenganairpili,tenaga haba dialirkan dari tuala kepada air sehingga keseimbangan terma dicapai lagi. Then the warm towel is immersed and rinsed with tap water. Heat energy is transferred from the towel to the water until thermal equilibrium is reached again.

• Padakeseimbanganterma, kadar bersih haba dialirkan antara mangkuk berisi air panas dan susu di

dalam botol ialah sifar .

At thermal equilibrium, the net rate of heat transfer

between the bowl of hot water and the bottle of milk is zero

.



• Pada keseimbangan terma, susu dan air akan mempunyai suhu yang lebih rendah daripada suhu awal air panas tetapi lebih tinggi daripada suhu awal susu. At thermal equilibrium, the milk and the water will have a temperature somewhere between their initial temperatures.


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Sup panas di dalam mangkukHot soup inside a bowl

• Suppanasdituangkedalamsatumangkuk./Hot soup is poured into a bowl.

• Selagimangkukmasihtidakterlalupanas,seseorangitumampu memegang mangkuk dengan tangan tanpa lapik. As the bowl is still not too hot, a person may be able to hold the bowl with his

bare hands.

• Tenagahabadialirkandarisupkemangkuksehingga


Heat energy is transferred from the hot soup to the bowl until thermal equilibrium is reached.

• Mangkuk bersentuhan dengan sup panas./The bowl is in contact with the hot soup.

• Mangkukitumempunyaitenagahabayang lebih rendah berbanding dengan sup panas yang mempunyai tenaga

haba yang yang tinggi .

The bowl has lower heat energy compared with hot soup which has higher heat energy.

• Padakeseimbanganterma, kadar bersih haba dialirkan antara sup panas dan mangkuk ialah sifar .

At thermal equilibrium, the net rate of heat transfer between the hot soup and the bowl is zero .



• Beberapaminitkemudian,mangkukdansuppanasmencapaikeseimbangantermapada suhu tertentu .

A few minutes later, the bowl and the hot soup achieve thermal equilibrium at a certain temperature .

• Padamasaitu,seseorangituperlumenggunasehelaikainuntukmemegangmangkukitu.At that time, a person needs to use a piece of cloth to hold the bowl.

1 Panjang turus merkuri pada takat ais dan pada takat stim masing-masing ialah 5.0 cm dan 40.0 cm.

Apabila termometer itu direndam dalam cecair P, panjang turusmerkuri adalah 23.0 cm.Berapakahsuhu cecair P? The lengths of the mercury column at ice point and steam point are 5.0 cm and 40.0 cm respectively. When the thermometer is immersed in liquid P, the length of the mercury column is 23.0 cm. What is the temperature of liquid P?

Suhup/Temperaturep = lθ – l0

l100 – l0 × 100 ºC

= (23.0 – 5.0) cm(40.0 – 5.0) cm

× 100 ºC

= 18.0 cm35.0 cm

× 100 ºC

= 51.43 °C

2 Panjang turus merkuri pada takat stim dan pada takat ais masing-masing ialah 65.0 cm dan 5.0 cm.

Apabila termometer itu direndam dalam cecair Q, panjang turusmerkuri adalah 27.0 cm. Berapakahsuhu cecair Q? The lengths of the mercury column at the steam point and ice point are 65.0 cm and 5.0 cm respectively. When the thermometer is immersed in liquid Q, the length of the mercury column is 27.0 cm.What is the temperature of liquid Q?

θ = (27.0 – 5.0) cm(65.0 – 5.0) cm

× 100 ºC

= 22 cm60 cm

× 100 ºC

= 36.67 ºC

Latihan/Exercises


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3 Jarak pada turus merkuri antara 0 ºC dan 100 ºC adalah 28.0 cm. Apabila termometer itu dimasukkan ke dalam bikar yang berisi air, panjang turus merkuri

adalah24.5cmdiatastitikais.Berapakahsuhuairtersebut?The distance between 0 ºC and 100 ºC on the mercury column is 28.0 cm. When the thermometer is put into a beaker of water, the length of the mercury column is 24.5 cm above the ice point. What is the temperature of the water?

θ = 24.5 cm28.0 cm

× 100 ºC = 87.5 ºC

4 Jarak pada turus merkuri antara 0 ºC dan 100 ºC adalah 25 cm. Apabila termometer diletakkan dalam bikar yang berisi air, panjang turus merkuri adalah 16 cm di atas titik ais.

The distance between 0 ºC and 100 °C on the mercury coloumn is 25 cm. When the thermometer is put into a beaker of water, the length of mercury column is 16 cm above the ice point.

(a) Berapakahsuhuairtersebut?What is the temperature of the water?

(b) Berapakahpanjangturusmerkuridaribebulikaca pada suhu 30 ºC? What is the length of the mercury column from the bulb at 30 ºC?

(a) θ = 16 cm25 cm

× 100 ºC

= 64 ºC

(b) 30 °C = lθ – l0

25 cm × 100 ºC

lθ – l0 = 30 °C × 25 cm

100 °C

= 7.5 cm

1 Muatan haba tentu air ialah 4 200 J kg-1 ºC-1. Specific heat capacity of water is 4 200 J kg-1 ºC-1.2 Ini bermakna untuk 1 kg air meningkat

suhu sebanyak 1 ºC, air perlu 4 200 J haba.

This means that for 1 kg of water, to increase its temperature by 1 ºC, the water needs 4 200 J of heat.

Penerangan menggunakan teori kinetik jirim:Explanation using kinetic theory of matter:• Apabilabahanmenyeraphaba,getaran

atom/molekul sangat kuat. When a substance absorbs heat, the vibrations of atoms/molecules are strong.• Atom/molekulbergerakdenganlaju

yang lebih tinggi. Atoms/molecules move with a higher

speed.• Tenagakinetikatom/molekulmeningkat. The kinetic energy of the atoms/

molecules increases.• Suhubahantersebutmeningkat. The temperature of the substance

increases.• Keadaanfizikalbahantidakberubah. The physical state of matter is

unchanged.

Definisi/Definition:Muatan haba tentu ialah kuantiti haba (Q) yang diperlukan oleh sesuatu bahan berjisim 1 kg untuk menaikkan suhunya sebanyak 1 ºC.The specific heat capacity is the quantity of heat (Q) required by a substance of mass 1 kg to increase its temperature by 1 ºC.

Persamaan/Equation:

Muatan haba tentu, c = Haba (Q)

Jisim (m) × Perubahan suhu (θ)

Specific heat capacity, c = Heat (Q)

Mass (m) × Temperature change (θ)

c = Q

mθ ⇒ Q = mcθ

Unit S.I. bagi c ialah J kg–1 ºC–1

S.I. unit of c is J kg–1 ºC–1

MEMAHAMI MUATAN HABA TENTUUNDERSTANDING SPECIFIC HEAT CAPACITY4.2


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Hubungan antara muatan haba tentu dan perubahan suhu:Relationship between specific heat capacity and temperature change:

• Muatan haba tentu adalah berkadar songsang dengan perubahan suhu.

Specific heat capacity is inversely proportional to the change in temperature.

• Bahandenganmuatanhabatentuyangtinggiakanmempunyaipeningkatansuhuyang kecil .

A material with high specific heat capacity will have a small increase in temperature.

• Ini bermakna bahawa bahan lambat menjadi panas/sejuk.

This means that the material is slow to get hot/cool.

• Contoh: Muatan haba tentu air ialah 4 200 J kg-1 ºC-1

Example: The specific heat capacity of water is 4 200 J kg-1 ºC-1

•Penerangan: Untuk 1 kg air, ia menyerap 4 200 J haba tetapi suhunya meningkat sebanyak 1 ºC.Explanation: For 1 kg of water, it absorbs 4 200 J of heat but the temperature increases by 1 ºC.

• Air menyerap kuantiti haba yang besar tetapi peningkatan suhu adalah kecil .

Water absorbs a big quantity of heat but the temperature increased is small .

• Air bertindak sebagai agen penyejuk bagi enjin kereta.Water acts as cooling agent in a car engine.

• Bahanyangmempunyaimuatanhabatentuyangrendahakanmengalamipeningkatansuhuyang

besar .

Materials with low specific heat capacity will have a large increase in temperature.

• Ini bermaksud bahan mudah menjadi panas/sejuk.

This means that the material easily gets hot/cool.

• Contoh: Muatan haba tentu kuprum ialah 390 J kg-1 ºC-1

Example: Specific heat capacity of copper is 390 J kg-1 ºC-1

•Penerangan: Untuk 1 kg kuprum, ia menyerap 390 J haba tetapi suhu meningkat sebanyak 1 ºC.Explanation: For 1 kg of copper, it absorbs 390 J of heat but the temperature increases by 1 ºC.

• Kuprum menyerap kuantiti haba yang kecil tetapi peningkatan suhu adalah tinggi.

Copper absorbs a small quantity of heat but the temperature increase is high.

• Kuprum digunakan sebagai elemen pemanas.Copper is used as a heating element.


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(1) Air sebagai penyejuk dalam enjin kereta Water as coolant in a car engine

Air mempunyai muatan haba tentu yang besar .

Water has a large specific heat capacity.

Ia boleh menyerap jumlah tenaga yang besar tetapi suhu yang meningkat adalah kecil.

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

Ia boleh menyejukkan pembakaran dalam enjin seperti enjin kereta.It can cool down internal combustion engines such as the car engine.

Kipasradiator

Radiator fan

Aliran air memasuki enjinWater enters the engine

Air dipam melaluipam airWater is circulated bythe water pump

Udara masuk melaluikipas radiatorAir drawn in by theradiator fan

Haba hilang dari sirippenyejuk ke persekitaranHeat is lost from the cooling �ns to the surroundings

Haba yang dihasilkan dari pembakaran dalam silinder dipindahkan ke airHeat generated from the combustion in the cyclinders is transferred to the water

Aplikasi muatan haba tentu:Applications of specific heat capacity:

1 Air sebagai penyejuk dalam enjin keretaWater as coolant in a car engine

2 BayulautSea Breeze

3 BayudaratLand Breeze

4 Periuk memasakCooking pot

5 Periuk tanah liatClay pot


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(2) Bayu laut/Sea breeze

Darat (pepejal) mempunyai muatan haba tentu yang rendah berbanding dengan laut (cecair).Land (solid) has a low specific heat capacity compared to the sea (liquid).

Pada waktu siang, darat menjadi panas lebih cepat berbanding laut.

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

Maka udara panas di darat yang mempunyai ketumpatan rendah akan bergerak naik ke atas membentuk

kawasan daratan yang bertekanan rendah .

Therefore, hot air on the land that has low density will rise up and produces low pressure region on the land.

Udara sejuk di laut yang mempunyai ketumpatan tinggi dan bertekanan tinggi akan bertiup ke darat.

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

(3) Bayu darat/Land breeze

Darat (pepejal) mempunyai muatan haba tentu yang rendah berbanding dengan laut (cecair).Land(solid) has a low specific heat capacity compared to the sea (liquid).

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

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

Udara panas di laut yang mempunyai ketumpatan rendah akan bergerak naik ke atas dan menghasilkan

kawasan bertekanan rendah .

The hot air on the sea that has low density will rise up and produce a low pressure region.

Udara sejuk di darat yang mempunyai ketumpatan tinggi dan bertekanan tinggi akan bertiup ke laut.

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

Udara panasHot air

Udara sejukCool air

Waktu siangDay time

Laut/SeaDarat/Land

Udara panasHot air

Waktu malamNight time

Udara sejukCool air

Laut/SeaDarat/Land


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(4) Periuk memasak Cooking pot

Tapak/Base

• Diperbuatdaripada keluli .

Made of steel .

• Muatanhabatentu rendah . Ia cepat panas dan boleh memasak makanan dengan cepat. Low specific heat capacity. It becomes hot very

quickly and enables quick cooking of the food in the pot.

• Ketumpatan tinggi , tapak yang berat menyebabkan ia lebih stabil dan tidak jatuh dengan senang. High density, the heavy base ensures that the pot is

stable and will not topple over easily.

Pemegang/Handle

• Pemegangdiperbuatdaripadabahansintetik.Handle is made of synthetic material.

• Muatanhabatentu tinggi , ia tidak akan menjadi panas apabila haba diserap. High specific heat capacity. It will not become too hot when

heat is absorbed.

• Konduktorhaba lemah , sangat sedikit haba dipindahkan kepada tangan seseorang yang mengangkat periuk itu. Poor conductor of heat, very little heat from the

pot is transferred to the hand of the person holding the pot.

Bekas/Body

• Bekasaluminium. Aluminium body.

• Muatan haba tentu rendah , maka ia cepat menjadi panas. Low specific heat capacity, so it becomes hot quickly.

• Ketumpatan rendah , maka ianya sangat ringan. Low density so it is very light.

• Tidak bertindak balas dengan makanan di dalamnya.

• Doesnot react with the food in the pot.

(5) Periuk tanah liat/Clay pot

• Tanahliatmempunyaimuatanhabatentu

lebih tinggi berbanding besi.

Clay has a larger specific heat capacity than metal.

• Semasamemasak, haba dialirkan

perlahan dari api ke makanan di dalam periuk.

During cooking, heat is conducted

slowly from the fire to the food inside the

pot.

• Masamemasakmakanan lama .

A longer cooking time is needed to cook

the food.

• Selepasapidipadam,periuktanahliatmempunyai suhu yang lebih tinggi berbanding makanan di dalamnya.

After the flame is switched off, the clay pot is at

higher temperature than the food inside it.

• Sejumlahbesar haba masih terus dialirkan kepada makanan itu.

A considerable amount of heat continues

to be transferred into the food.

• Selepasperiuktanahliatitudialihkandaripadaapi, makanan di dalam periuk terus mendidih dalam masa beberapa minit. The food inside continues to boil for a few minutes after the clay pot has been removed from the fire.

BekasBody

TapakBase

PemegangHandle


4

UNIT



Tujuan Aim

Untuk menentukan muatan haba tentu aluminium.

To determine the specific heat capacity of aluminium.

Untuk menentukan muatan haba tentu air.

To determine the specific heat capacity of water.

Senarai radas dan bahanList of apparatus and materials

Blok aluminium, kertas tisu, helaian polisterina,minyak, pemanas rendam, termometer, bekalan kuasa, neraca tuas dan jam randikAluminium block, tissue paper, polystyrene sheet, oil, immersion heater, thermometer, power supply, lever balance and stopwatch.

Cawan polisterina, air, pemanas perendam, termometer, bekalan kuasa, pengacau, neraca tuas atau penimbang elektronik dan jam randik.Polystyrene cup, water, immersion heater, thermometer, power supply, stirrer, lever balance or electronic balance and stopwatch.

Susunan radasArrangement of the apparatus

TermometerThermometer

Blok aluminiumAluminium block

Lubang diisi denganminyak untukmeningkatkan konduksiHole �lled withoil to increasethermal conductivity

PemanasHeater

Bekalan kuasaPower supply

Blok kayuWooden block

AirWater


PemanasHeater


Kertas tisuTissue paper

Jam randikStopwatch

SilinderpenyukatMeasuringcyclinder

39

12

6

Prosedur Procedure

1. Jisim blok aluminium, m, ditentukan

dengan menggunakan neraca tuas.

The mass of the aluminium block, m is

determined using the lever balance.

2. Suhu awal blok aluminium, θ1 direkodkan.

The initial temperature of the aluminium block, θ1,

is recorded.

3. Pemanas dihidupkan dan pada masa yang sama

jam randik dimulakan. / The heater is switched

on and the stopwatch is started simultaneously.

4. Pemanas ditutup selepas 10 minit.

The heater is switched off after 10 minutes.

5. Suhu tertinggi, θ2, direkodkan.

The highest temperature, θ2, is recorded.

1. Cawan diisikan dengan air, berjisim m.

(contoh, m = 150 g)./The cup is filled with

water of mass, m (example, m = 150 g).

2. Suhu awal air, θ1, direkodkan.

The initial temperature of the water, θ1, is recorded.

3. Pemanas dihidupkan dan pada masa yang sama

jam randik dimulakan. / The heater is switched

on and the stopwatch is started simultaneously.

4. Air dikacau berterusan.

The water is stirred continuously.

5. Pemanas ditutup selepas 10 minit.

The heater is switched off after 10 minutes.

6. Suhu tertinggi, θ2, direkodkan.

The highest temperature, θ2, is recorded.

Penjadualan dataTabulate the data

Kuasa pemanas, P/Power of the heater, P

= WJisim blok aluminium, mMass of aluminium block, m

= gSuhu awal, θ1 /Initial temperature, θ1

= °CSuhu akhir, θ2 /Final temperature, θ2

= °C

Kuasa pemanas, P/Power of the heater, P

= WJisim air, mMass of water, m

= gSuhu awal, θ1 /Initial temperature, θ1

= °CSuhu akhir, θ2 /Final temperature, θ2

= °C

Untuk menentukan muatan haba tentu pepejal dan cecairTo determine the specific heat capacity of a solid and a liquid

EksperimenExperiment


4

UNIT



Analisis dataAnalysis of the data

Pengiraan muatan haba tentu aluminium, c:Calculation of specific heat capacity of aluminium, c: Q = mcθ Pt = mc(θ2 – θ1)

∴ c = Pt

m(θ2 – θ1)

Pengiraan muatan haba tentu air, c:Calculation of specific heat capacity of water, c: Q = mcθ Pt = mc(θ2 – θ1)

∴ c = Pt

m(θ2 – θ1)

1 Sebuah blok logam berjisim 2 kg. Hitungkan jumlah haba yang mesti dipindahkan kepada logam

untuk meningkatkan suhu dari 30 ºC kepada 70 ºC. (Muatan haba tentu logam = 500 J kg-1 ºC-1)

A metal block has a mass of 2 kg. Calculate the amount of heat that must be transferred to the metal to raise its temperature from 30 ºC to 70 ºC (Specific heat capacity of the metal = 500 J kg-1 ºC-1)

Penyelesaian/Solution: Q = mcθ = (2 kg) × (500 J kg–1°C–1) × (70 – 30)°C = (2 × 500 × 40) J = 40 000 J

2 8.4 × 105 J tenaga haba meningkatkan suhu 4 kg air dari 40 ºC hingga 90 ºC. Berapakahmuatan habatentu air tersebut?8.4 × 105 J of heat energy raises the temperature of4 kg of water from 40 ºC to 90 ºC. What is the specific heat capacity of the water?

Penyelesaian/Solution:

c = Q

mθ

= (8.4 × 105) J

4 kg × (90 – 40)°C

= 4 200 J kg-1 ºC-1

3 0.2 kg air pada suhu 100 ºC dicampur dengan 0.25 kg air pada suhu 10 ºC. Berapakah suhu tertinggiyang dicapai oleh campuran tersebut?0.2 kg of water at 100 ºC is mixed with 0.25 kg of water at 10 ºC. What is the maximum temperature reached by the mixture? Penyelesaian/Solution: Haba yang dibebaskan = Haba yang diserapHeat released = Heat absorbed

(0.2 kg)(ca)(100 – θ)°C = (0.25 kg)(ca)(θ – 10)°C 20 – 0.2 θ = 0.25θ – 2.5 20 + 2.5 = 0.25θ + 0.2θ 22.5 = 0.45 θ

θ = 22.50.45

= 50 ºC

di mana/where ca = muatan haba tentu air specific heat capacity of water

4 Sebuah 2 kW, 240 V pemanas elektrik digunakan untuk memanaskan 3 kg air. Jika kuasa dibekalkan

selama 8 minit, berapakah peningkatan suhu air tersebut? (Muatan haba tentu air adalah 4 200 J kg-1 ºC-1)

A 2 kW, 240 V electric heater is used to heat up 3 kgof water. If the power is supplied for 8 minutes, whatis the increase in temperature of the water? (The specific heat capacity of water is 4 200 J kg-1 ºC-1)

Penyelesaian/Solution: Pt = mcθ (2 000 W)(8 × 60 s) = 3 kg × 4 200 J kg–1°C–1 × θ

θ = (2 × 103 × 8 × 60) J

3 × 4 200 J °C–1

= 76.2 ºC

5 Sebuah pemanas rendam 1.2 kW digunakan untuk meningkatkan 0.2 kg air dalam bekas kuprum berjisim 0.05 kg. Kirakan masa yang diambil agar suhu air dan bekas dinaikkan sehingga 20 °C.

(Muatan haba tentu air, ca = 4 200 J kg-1 °C-1). (Muatan haba tentu kuprum, ck = 400 J kg-1 °C-1)

A 1.2 kW immersion heater is used to raise the temperature of 0.2 kg water in a copper container of mass 0.05 kg. Calculate the time taken so that the temperature of the water and the container is increased by 20 °C. (The specific heat capacity of water, ca = 4 200 J kg-1 °C-1)(The specific heat capacity of copper, ck = 400 J kg-1 °C-1)

Penyelesaian/Solution: Pt = mcθ (1.2 × 103 W) × t = macaθ + mkckθ (1 200 W) × t = θ (maca + mkck) = 20°C [(0.2 kg × 4 200 J kg–1 °C–1

+ (0.05 kg × 400 J kg °C–1)]

t = 17 200 J

1 200 J s–1

= 14.33 s

KBAT

KBAT

KBAT

Latihan/Exercises


4

UNIT



Graf Pemanasan Suhu-masa / Temperature-time Heating Graph

Haba boleh menyebabkan objek:Heat can make an object:(i) berubah suhu, θ atau change in

temperature,θ or(ii) berubah keadaan

fizikal change its physical

state

Graf pemanasan (Haba diserap)Heating graph (Heat is absorbed):

PerubahanChanges

MalarConstant

PersamaanEquation

Suhu meningkat, θTemperatureincrease, θ

Keadaan fizikalPhysical state

Q = mcθ

Keadaan fizikalPhysical state

SuhuTemperature

Q = mL

Pada QR dan ST, keadaan fizikal berubah tetapi suhu adalah tetap:At QR and ST, the physical states change but the temperature is constant:

• Habadiserapdigunakanuntukmemecahkan ikatan antara atom/molekul.

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

• Keadaanfizikal berubah . / The physical state changes .

• Tenagakinetikatom/molekul tidak berubah .

The kinetic energy of the atoms/molecules remains unchanged .

• Suhu tetap . / The temperature is constant .

Pada PQ, RS dan TU, suhu menaik tetapi keadaan fizikal tidak berubah:At PQ, RS and TU, the temperature increases but the physical states are unchanged:

• Habayangdiserapdigunakanuntuk meningkatkan tenaga kinetik atom/molekul.

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

• Suhu meningkat .

The temperature increases .

• Ikatanantaraatom/molekul tidak terputus .

The bonds between atoms/molecules do not break .

• Keadaanfizikal tidak berubah .

The physical state is unchanged .

Suhu/°CTemperature/°C

Haba/Heat, Q4 = mL

Masa/sTime/sHaba/Heat, Q2 = mL

P

Q0

ST

U

R

Pepejal + CecairSolid + Liquid

PepejalSolid

Cecair + GasLiquid + Gas

CecairLiquid

GasGas

1Haba/Heat, Q1 = mc 1θ

θ


θ


θ

MEMAHAMI HABA PENDAM TENTU (L)UNDERSTANDING SPECIFIC LATENT HEAT (L)4.3


4

UNIT



Jenis Haba Pendam Tentu, L/Types of Specific Latent Heat, L

Haba Pendam Tentu Pengewapan (Lv)Specific Latent Heat of Vaporisation (Lv)

Definisi/DefinitionHaba (Q) yang diperlukan untuk menukarkan cecair berjisim 1 kg kepada wap pada takat didih.Heat (Q) required to change 1 kg of liquid to vapour at its boiling point.

Air / Water Stim/Wap / Steam Persamaan/Equation: Haba pendam tentu, Lv =

Haba (Q)Jisim (m)

Specific latent heat, Lv = Heat (Q)

Mass (m)

Lv = Qm

⇒ Q = mLv

Unit S.I. bagi Lv ialah J kg-1

S.I. unit of Lv is J kg-1

Haba Pendam Tentu Pelakuran (Lf)Specific Latent Heat of Fusion (Lf)

Definisi/DefinitionHaba (Q) yang diperlukan untuk mengubah pepejal berjisim 1 kg kepada cecair pada takat lebur.Heat (Q) required to change 1 kg of solid to liquid at its melting point.

Persamaan/Equation:

Haba pendam tentu, Lf = Haba (Q)Jisim (m)

Specific latent heat, Lf = Heat (Q)

Mass (m)

Lf = Q

m ⇒ Q = mLf

Unit S.I. bagi Lf ialah J kg-1

S.I. unit of Lf is J kg-1

Penerangan menggunakan Teori Kinetik jirim:Explanation using Kinetic Theory of matter:• Apabilabahanmenyeraphaba, getaran atom/molekul adalah kuat. When a substance absorbs heat, the

vibrations of atoms/molecules are strong.

• Ikatanantaraatom/molekulmenjadilemah.

The bonds between atoms/molecules become weak.

• Atom/molekuldengantenagayangtinggi boleh memecahkan ikatan dan mengubah kepada keadaan fizikal yang baru.

Atoms/molecules with very high energy can break the bonds and change to a new physical state.

• Keadaanfizikalbahanberubah. The physical state of matter changes.• Tenagakinetikatom/molekultidak

berubah. The kinetic energy of the atoms/

molecules is unchanged.• Suhutetap/tidakberubah. The temperature is constant.

Pelakuran – proses di mana pepejal berubah kepada cecair apabila haba diserapFusion – A process where solid changes to liquid when heat is absorbed.

Pengewapan – proses di manacecair berubah kepada wapapabila haba diserap Vaporisation - A process where liquid changes to gas when heat is absorbed.

Haba Pendam Tentu / Specific Latent HeatDefinisi/Definition:Haba pendam tentu (Q) ialah kuantiti haba yang diperlukan untuk mengubah bahan berjisim 1 kg dari satu keadaan fizikal kepada keadaan fizikal yang lain pada suhu yang tetap.Specific latent heat (Q) is the quantity of heat required to change 1 kg of a substance from one physical state to another physical state at constant temperature.

Persamaan/Equation:

Haba pendam tentu, L = Haba (Q)Jisim (m)

/ Specific latent heat, L = Heat (Q)Mass (m)

L = Qm

⇒ Q = mL

Unit S.I. bagi L ialah J kg–1

S.I. unit of L is J kg–1

Perubahan keadaan fizikal (Proses pemanasan):Change of physical state (Heating process):

Jenis Haba Pendam Tentu:Types of Specific Latent Heat:(i) Haba pendam tentu pelakuran (Lf) Specific latent heat of fusion (Lf)(ii) Haba pendam tentu pengewapan (Lv) Specific latent heat of vaporisation (Lv)

CecairLiquid

Pepejalsolid

GasGas

PengewapanVaporisation

PelakuranFusion

AisIce

AirWater


4

UNIT



Aplikasi haba pendam tentuApplication of specific latent heat

• Airmempunyaihabapendamtentupengewapan besar .

Water has a large specific latent heat of vaporisation.

• Olehitu,iamemerlukantenagahabayang banyak untuk meruap.

So it needs more heat to vaporise.

• Jikastimdariperiuksupyangmendidih mengkondensasi menjadi

air di atas lengan, kuantiti haba pendam yang besar dibebaskan.

If the steam from a pot of boiling soup condenses to water on the skin of

our arm, a very large amount of latent heat of vaporisation is released

.

• Iabolehmemberikesanmelecuryangbahaya. It can give us a serious scalding.

• Minumanbolehdisejukkandenganmenambahbeberapaketulanais.Drinks can be cooled by adding in several cubes of ice.

• Apabilaaismencair, haba pendam pelakuran diserap dari minuman.

When the ice is melting, the latent heat of fusion is

absorbed from the drinks.

• Kuantiti haba diperlukan untuk menukarkan beberapa ketulan ais

kepada fasa cecair .

The amount of heat required to change the several cubes of ice to

liquid phase.

• Tanpaperubahan suhu ais.

Without change in temperature of ice.

• Suhuminuman menurun .

The temperature of the drinks is lowered .


4

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• Apabilakitaterlibatdalamaktiviti-aktivitiberat,peluhmenyejukkanbadankita.

When we are engaged in strenuous activities, sweating cools our bodies.

• Peluh menyejat dan haba dari badan dibebaskan

sebagai haba pendam pengewapan .

The sweat evaporates and the body heat is removed as the

latent heat of vaporisation .

• Menyebabkansuhubadankita menurun .

Thus our body temperature decreases .

• Kuantiti haba yang besar diperlukan untuk menukar air kepada stim.

A large quantity of heat is required to change water into steam.

• Menggunakanprinsipkeabadiantenagaiaitukuantiti

haba yang besar akan dibebaskan apabila stim

mengkondensasi menjadi air.

Using of principle of conservation of energy that a large quantity of

heat will be released when steam condenses to water.

• Makanansepertikek,telur,ikan,paudanlain-lain

menerima kuantiti tenaga yang besar apabila

haba pendam pengewapan stim dibebaskan daripada stim yang terkondensasi.

Food such as cakes, eggs, fish, buns and others receive a large amount of

energy when the latent heat of vaporisation of steam released from condensing steam.


4

UNIT



TujuanAim

Untuk menentukan haba pendam tentu

pelakuran ais.

To determine the specific latent

heat of fusion of ice.

Untuk menentukan haba pendam tentu

pengewapan air.

To determine the specific latent

heat of vaporisation of water.

Pemboleh ubahVariables

Pemboleh ubah dimanipulasi:Manipulated variable:Haba yang dibekalkan, Q

Heat supplied, Q

Pemboleh ubah bergerak balasResponding variable:Jisim ais yang cair, m

Mass of ice that melts, m

Pemboleh ubah dimalarkanFixed variable:Tempoh pemanasan, t

Period of heating, t

Pemboleh ubah dimanipulasi:Manipulated variable:Haba yang dibekalkan, Q

Heat supplied, Q

Pemboleh ubah bergerak balasResponding variable:Jisim air yang telah diwapkan, m

Mass of water which has vaporised, m

Pemboleh ubah dimalarkanFixed variable:Tempoh pemanasan, t

Period of heating, t

Senarai radas dan bahanList of apparatus and materials Bikar A

Beaker A Bikar BBeaker B

Pemanas elektrikElectrical heater


AisIce

AisIce


Neraca tuasLever balance

PemanasHeater

Menentukan haba pendam tentu pelakuran ais dan haba pendam tentu pengewapan aisTo determine the specific latent of fusion of ice and specific latent vaporisation of water



4

UNIT



ProsedurProcedure

1. Radas B digunakan sebagai kawalan.

Apparatus B is used as a control.

2. Pemanas disediakan dan disambung seperti

ditunjukkan dalam rajah.

The heater is set up and connected as shown in

the diagram.

3. Litar ditutup dan jam randik dimulakan

serentak.

The circuit is closed and the stopwatch is started

simultaneously.

4. Air yang turun dari penapis corong

dikumpul dalam bikar A dan B.

Water which drips from the filter funnels is

collected in beakers A and B.

5. Pemanas dan jam randik diberhentikan

serentak selepas t = 3 minit

The heater and the stopwatch are switched off

simultaneously after t = 3 minutes

6. Jisim air dalam kedua-dua bikar diukur.

The mass of the water in both beakers is

measured.

1. Pemanas dengan kuasa, P disambungkan

kepada bekalan kuasa.

A heater with power, P is connected to the power

supply.

2. Tuang air ke dalam bikar sehingga hampir

penuh./Pour water into a beaker until it is

almost full.

3. Litar ditutup.

The circuit is closed.

4. Apabila air mula mendidih, bacaan neraca

tuas, m1, direkodkan dan jam randik juga

dimulakan serentak.

When the water starts to boil, the reading

on the lever balance, m1, is recorded and

the stopwatch is started simultaneously.

5. Selepas t = 4 minit, jam randik

diberhentikan dan jisim terakhir air, m2

direkodkan.

After t = 4 minutes, the stopwatch is stopped

and final mass of the water, m2 is recorded.


Kuasa pemanas, P = WPower of the heater, P = W

Masa pemanasan, t = sTime of heating, t = s

Jisim air dalam bikar A, mA = gMass of water in beaker A, mA = g

Jisim air dalam bikar B, mB = gMass of water in beaker B, mB = g

Jisim air dari ais yang mencair oleh pemanas, m = mA– mB = gMass of water from the ice melted by the heater, m = mA – mB = g

Kuasa pemanas, P = WPower of the heater, P = W

Masa pemanasan, t = sTime of heating, t = s

Jisim air dalam bikar, m1 = ………….. gInitial mass of water, m1 = g

Jisim terakhir air, m2 = gFinal mass of water, m2 = g

Jisim air yang telah diwapkan, m = m1 – m2 = gMass of water that has vaporised, m = m1 – m2 = g

Data dianalisisAnalysis the data

Kirakan haba pendam tentu pelakuran ais, Lf:Calculation of specific latent heat of fusion of ice, Lf :

Q = mLf

Pt = mLf

Lf = Ptm

Unit, Lf ialah J kg–1

The unit of Lf is J kg–1

Kirakan haba pendam tentu pengewapan air, LV:Calculation of specific latent heat of vaporisation of water, Lv: Q = mLV

Pt = mLV

LV = Ptm

Unit, Lv ialah J kg–1

The unit of Lv is J kg–1


4

UNIT



1 Berapakah kuantiti haba yang diperlukan untukmencairkan 2.0 kg ais pada 0 ºC? (Haba pendam tentu pelakuran ais = 3.34 × 105 J kg-1) What is the quantity of heat required to melt 2.0 kg ice at 0 ºC? (The specific latent heat of fusion of ice = 3.34 × 105 J kg-1)

Penyelesaian/Solution: Q = mLf

= (2.0 kg) × (3.34 × 105 J kg–1) = 6.68 × 105 J

2 Berapakah tenaga harus dialihkan dari 4.0 kg airpada 20 ºC untuk menghasilkan ais pada 0 ºC?How much energy has to be removed from 4.0 kg of water at 20 ºC to produce a block of ice at 0 ºC?

[Muatan haba tentu air = 4.2 × 103 J kg-1 ºC-1, haba pendam tentu pelakuran ais = 3.34 × 105 J kg-1][Specific heat capacity of water = 4.2 × 103 J kg-1 ºC-1,specific latent heat of fusion of ice = 3.34 × 105 J kg-1]

Air/Water20 °C

mcθ

Air/Water0 °C

mLf

Ais/Ice0 °C

Jumlah tenaga haba/Total heat energy= mcθ + mLf

= [(4.0 kg) × (4.2 × 103 J kg–1°C–1) × (20 – 0)°C] + [(4 kg) × (3.34 × 105 J kg–1)]= 3.36 × 105 J + 13.36 × 105 J= 16.72 × 105 J = 1.672 × 106 J

3 Sebuah elektrik pemanas 800 W digunakan untuk memanaskan air. Berapakah masa yangdiperlukan untuk mengurangkan air sebanyak 4 kg

selepas air mencapai takat didih? [Haba pendam tentu pengewapan air = 2.26 × 106 J kg-1]

A 800 W electric heater is used to boil water. What is the time required to reduce the mass of water by 4 kg after the water has reached its boiling point?[Specific latent heat of vaporisation of water = 2.26 × 106 J kg-1]


Tenaga elektrik yang dibekalkanElectric energy provided

= Tenaga haba yang diterimaHeat absorbed

Pt = mLv

(800 W) t = 4 kg × 2.26 × 106 J kg–1

t = (4 × 2.26 × 106) J

800 J s–1

= 1.13 × 104 s

4 Kirakan haba yang diperlukan untuk menukar 4 kg ais pada –15 ºC kepada stim pada 100 ºC.Calculate the heat required to convert 4 kg of ice at –15 ºC to steam at 100 ºC.

[Muatan haba tentu ais/Specific heat capacity of ice = 2.1 × 103 J kg-1 ºC-1, Muatan haba tentu air/Specific heat capacity of water = 4.2 × 103 J kg-1 ºC-1, haba pendam tentu pelakuran ais specific latent heat of fusion of ice = 3.34 × 105 J kg-1 dan/and haba pendam tentu pengewapan air

specific latent heat of vaporisation of water

= 2.26 × 106 J kg-1]


Ais/Ice–15 °C

miciθi

Ais/Ice0 °C

mLf

Air/Water0 °C

mwcwθw

Air/Water100 °C

mLv

Stim/Steam100 °C

Jumlah tenaga habaTotal heat energy

= miciθi + mLf + mwcwθw + mLv

miciθi = (4 kg) × (2.1 × 103 J kg–1°C–1) × (15°C) = 1.26 × 105 JmLf = (4 kg) × (3.34 × 105 J kg–1) = 1.336 × 106 Jmwcwθw = (4 kg) × (4.2 × 103 J kg–1°C–1) × (100°C) = 1.68 × 106 JmLv = (4 kg) × (2.26 × 106 J kg–1) = 9.04 × 106 J

∴ Jumlah tenaga haba/Total heat energy= (1.26 × 105 J) + (1.336 × 106 J) + (1.68 × 106 J) + (9.04 × 106 J)= 1.218 × 107 J

KBAT

KBAT

Latihan/Exercises


4

UNIT



Penerangan Menggunakan Teori Kinetik JisimExplanation Using Kinetic Theory of Matter

Kuantiti/Quantity Penerangan/Explanation

Jisim (m)Mass (m)

Jumlah jirim dalam bekas yang tertutup, jumlah jirim adalah tetap. Jisim adalah malar.Amount of matter is in a closed container, the amount of matter is constant.The mass is constant.

Isi padu (V)Volume (V)

Isi padu gas adalah sama dengan isi padu bekas. Volume of gas is equal to the volume of container.

Tekanan (P)Pressure (P)

Kadar perlanggaran antara molekul-molekul gas dengan dinding bekas. Apabila kadar perlanggaran antara molekul gas dengan dinding bekas bertambah, tekanan gas juga bertambah. 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.

Suhu (T)Temperature (T)

Apabila molekul-molekul gas bergerak dengan halaju tinggi, tenaga kinetik bertambah, maka suhu gas juga bertambah.Kinetic energy of the gas molecules. When the gas moves at a higher speed, the kinetic energy increases; so the gas temperature also increases.

Hukum-hukum gas/Gas Laws:4 kuantiti fizikal yang berkaitan:The 4 physical quantities involved are:(i) Jisim (m) – sentiasa tetap dengan menggunakan bekas

tertutup Mass (m) – always kept constant by using a closed container(ii) Tekanan (P) Pressure (P)(iii) Isi padu (V) Volume (V)(iv) Suhu mutlak (T) Absolute temperature (T)

boleh berubahcan change

3 Hukum gas3 gas laws

HukumBoyleBoyle’s Law

Hukum CharlesCharles’ Law

Hukum tekananPressure Law

5 Sebuah blok pepejal 0.5 kg dipanaskan oleh pemanas elektrik 100 W. Graf menunjukkan bagaimana suhu berubah dengan masa. 0.5 kg of a solid block is heated by a 100 W heater. The graph shows how the temperature varies with time.

Suhu

/Tem

pera

ture

/ °C

Masa/Time / s0

0

20

40

60

80

100

100 200 300 400 500 600 700 800 900 10001100

Hitungkan haba pendam tentu pelakuran pepejal itu.Calculate the specific latent heat of fusion of the solid.

Penyelesaian/Solution: Pt = mLf

(100 W) × (1 050 – 300)s = 0.5 kg × Lf

Lf = 75 000 J0.5 kg

= 1.5 × 105 J kg–1

MEMAHAMI HUKUM-HUKUM GASUNDERSTANDING THE GAS LAWS4.4

KBAT


4

UNIT



Huk

um C

harl

es/C

harl

es’

Law

:(m

& P

– p

emal

ar/c

onst

ant:

V ∝

T)

V 1 , T

1V 2 ,

T2

Hab

a/H

eat

(i)

Pene

rang

an

men

ggun

akan

Teo

ri

Kin

etik

Ji

rim

:

Exp

lana

tion

usin

g K

inet

ic T

heor

y of

Mat

ter:

A

pabi

la g

as d

ipan

aska

n, s

uhu

akan

men

ingk

at

. (T

2 >

T 1)

W

hen

the

gas

is h

eate

d, te

mpe

ratu

re in

crea

ses

. (T 2

> T

1)

Is

i pa

du g

as j

uga

ber

tam

bah

men

yeba

bkan

ka

dar

perl

angg

aran

m

olek

ul g

as d

enga

n

dind

ing

beka

s t

idak

ber

ubah

.

The

volu

me

of t

he g

as a

lso

i

ncre

ases

t

o ke

ep t

he

rate

of

colli

sion

b

etw

een

the

gas

mol

ecul

es a

nd t

he

wal

ls o

f th

e co

ntai

ner

is

con

stan

t

.

Te

kana

n ga

s ad

alah

t

etap

.

Th

e ga

s pr

essu

re i

s

con

stan

t

.

T 2 >

T1

dan/

and

V2

> V

1

(ii)

Pe

rnya

taan

/Sta

tem

ent:

H

ukum

Cha

rles

men

yata

kan

baha

wa

untu

k jis

im

gas

yang

tet

ap,

isi

padu

gas

ber

kada

r la

ngsu

ng

deng

an s

uhu

mut

lakn

ya j

ika

teka

nan

gas

itu

adal

ah t

etap

.

Cha

rles

' La

w s

tate

s th

at f

or a

fix

ed m

ass

of g

as,

the

volu

me

of th

e ga

s is

dir

ectly

pro

port

iona

l to

the

abso

lute

tem

pera

ture

of

the

gas

if th

e pr

essu

re i

s ke

pt c

onst

ant.

Huk

um T

ekan

an:

m &

V/P

ress

ure

Law

:(m

& V

– p

emal

ar/c

onst

ant

: P

∝ T

)

P 2 , T

2P 1 ,

T1

Hab

a/H

eat

(i)

Pene

rang

an

men

ggun

akan

Te

ori

Kin

etik

Ji

rim

:

Exp

lana

tion

usin

g K

inet

ic T

heor

y of

Mat

ter:

Apa

bila

gas

itu

dip

anas

kan,

suh

u m

enin

gkat

.(T

2> T

1)

Whe

n th

e ga

s is

hea

ted,

its

tem

pera

ture

inc

reas

es

.

(T2

> T

1)

Isi

padu

bek

as d

ikek

alka

n m

alar

, ja

di i

si p

adu

gas

adal

ah

tet

ap

.Th

e vo

lum

e of

the

con

tain

er i

s ke

pt c

onst

ant,

so t

he

volu

me

of t

he g

as i

s

con

stan

t

.

Kad

ar p

erla

ngga

ran

di a

ntar

a m

olek

ul g

as

deng

an d

indi

ng b

ekas

itu

m

enin

gkat

.

The

rate

of

colli

sion

bet

wee

n th

e ga

s m

olec

ules

and

the

wal

ls o

f th

e co

ntai

ner

i

ncre

ases

.

Te

kana

n ga

s

men

ingk

at

.

Th

e ga

s pr

essu

re

inc

reas

es

.

T 2 >

T1

dan

/and

P2

> P

1

(ii)

Pe

rnya

taan

/Sta

tem

ent:

H

ukum

Tek

anan

men

yata

kan

baha

wa

bagi

sua

tu

jisim

gas

yan

g te

tap,

teka

nan

gas

ad

alah

berk

adar

lan

gsun

g de

ngan

suh

u

mut

lak

gas

jika

isi

padu

gas

ada

lah

teta

p.

Pre

ssur

e La

w s

tate

s th

at f

or a

fix

ed m

ass

of g

as,

the

pres

sure

of

gas

is d

irec

tly p

ropo

rtio

nal

to t

he

abso

lute

tem

pera

ture

of

the

gas

if th

e vo

lum

e of

the

gas

is k

ept

cons

tant

.

Huk

um B

oyle

/Boy

le’s

Law

: (m

& T

– p

emal

ar/c

onst

ant:

P ∝

1 V)

P 1 , V

1

P 2 , V

2

Om

boh/

Pist

on

(i)

Pene

rang

an d

enga

n m

engg

unak

an T

eori

Kin

etik

Ji

rim

:

Exp

lana

tion

usin

g K

inet

ic T

heor

y of

Mat

ter:

A

pabi

la i

si p

adu

beka

s be

rkur

ang,

isi

pad

u ga

s

juga

b

erku

rang

.

(V2

< V

1)

Whe

n th

e vo

lum

e of

the

con

tain

er d

ecre

ases

, th

e

volu

me

of t

he g

as a

lso

decr

ease

s

. (V

2 <

V1)

K

adar

per

lang

gara

n an

tara

mol

ekul

gas

den

gan

dind

ing

beka

s

ber

tam

bah

.

Th

e ra

te o

f co

llisi

on b

etw

een

the

gas

mol

ecul

es a

nd

the

wal

ls o

f th

e co

ntai

ner

incr

ease

s

.

Te

kana

n ga

s tu

rut

b

erta

mba

h

.

Th

e ga

s pr

essu

re

in

crea

ses

.

Te

naga

kin

etik

gas

tet

ap,

mak

a su

hu g

as j

uga

t

etap

.

Th

e ki

netic

ene

rgy

of t

he g

as i

s co

nsta

nt,

so t

he

tem

pera

ture

of

the

gas

is

con

stan

t

.

V2

< V

1 da

n/an

d P 2

> P

1

(ii)

Pe

rnya

taan

/Sta

tem

ent:

Hukum

Boylemenyatakanbahawauntukjisim

gas

yang

tet

ap,

teka

nan

gas

berk

adar

son

gsan

g

deng

an i

si p

adu

gas

jika

suhu

ada

lah

mal

ar

.B

oyle

's L

aw s

tate

s th

at f

or a

fix

ed m

ass

of g

as,

the

gas

pres

sure

is

in

vers

ely

prop

ortio

nal

to

the

volu

me

of t

he g

as i

f th

e te

mpe

ratu

re i

s ke

pt c

onst

ant

.


4

UNIT



(iii)

Hub

unga

n/R

elat

ions

hip:

Teka

nan

gas

(P)

be

rkad

ar s

ongs

ang

de

ngan

isi

pad

u ga

s (V

)

Gas

pre

ssur

e (P

) is

i

nver

sely

pro

port

iona

l

to

the

gas

volu

me

(V)

P ∝

1 V

(iv)

Per

sam

aan/

Equ

atio

n:

P =

k (

1 V)

; k a

dala

h m

alar

/is a

con

stan

t

k =

P V

= m

alar

/con

stan

t

∴

P1

V1

=

P2

V2

(v)

Gra

f/G

raph

:

(a)

(c)

(b)

P/Pa

P/Pa

PV 000

V1 __ V

1 __ V

1 __ V

V

P b

erka

dar

song

sang

de

ngan

VP

is in

vers

ely

prop

orti

onal

to V

P b

erka

dar

lang

sung

de

ngan

P is

dir

ectl

y pr

opor

tion

al to

PV

ada

lah

teta

pP

V is

con

stan

t

(iii)

Hub

unga

n/R

elat

ions

hip:

Isi

padu

gas

(V

) ad

alah

berk

adar

lan

gsun

g de

ngan

suh

u m

utla

k ga

s (T

)

Volu

me

of g

as (

V)

is di

rect

ly p

ropo

rtio

nal t

o th

e ab

solu

te t

empe

ratu

re o

f ga

s (T

)

V

∝ T

(iv)

Per

sam

aan/

Equ

atio

n:

V =

k (

T) ;

k ad

alah

mal

ar/is

a c

onst

ant

k =

V T =

pem

alar

/con

stan

t

∴

V1

T 1 =

V

2

T 2

(v)

Gra

f/G

raph

:

(c)

a

dala

h te

tap

is

con

stan

t

T0

(a)

V b

erka

dar

lang

sung

de

ngan

TV

is d

irec

tly

prop

orti

onal

to T

0T/

K

V V __ T

(b)

V b

erub

ah s

ecar

alin

ear

deng

an

V v

arie

s li

near

ly

wit

h

0–2

73/°

C

V

θ

θ

θ

V __ T

V __ T

(iii)

Hub

unga

n/R

elat

ions

hip:

Teka

nan

gas

(P)

be

rkad

ar l

angs

ung

den

gan

suhu

mut

lak

gas

(T).

Pre

ssur

e of

gas

(P

) is

d

irec

tly p

ropo

rtio

nal

to

the

abso

lute

tem

pera

ture

of

gas

(T).

P ∝

T

(iv)

Per

sam

aan/

Equ

atio

n:

P =

k (

T) ;

k ad

alah

mal

ar/is

a c

onst

ant

k =

P T =

pem

alar

/con

stan

t

∴

P1

T 1 =

P

2

T 2

(v)

Gra

f/G

raph

:

(a)

(c)

(b)

P b

erka

dar

lang

sung

de

ngan

TP

is d

irec

tly

prop

orti

onal

to T

T/K

T

P/Pa

0–2

73

00

/°CP

ber

ubah

sec

ara

linea

r de

ngan

P

var

ies

line

arly

w

ith

θ

θ

θ

P/Pa P T

a

dala

h te

tap

is

con

stan

tP __ T

P __ T


4

UNIT



Aplikasi Hukum-hukum Gas/Applications of Gas Laws

Hukum Boyle/Boyle’s Law

Bantal udaraAir pillow

• Penerangan: Hubungan antara tekanan udara dan isi padu mengguna teori kinetik gas (suhu malar)./Explanation: The relationship between the air pressure and the volume using the kinetic theory of gases (constant temperature).

• Apabilabudaklelakimeletakkankepalanyadiatasbantaludara,udaradidalambantal

itu dimampatkan .When the boy puts his head on the pillow, the air in the pillow is compressed .

• Isi padu udara dikurangkan./The volume of the air is decreased.

• Bilanganmolekul-molekulperunitisipadu meningkat .

The number of molecules per unit volume increases .

• Molekul-molekulberlanggardengandindingbantal lebih kerap .

The molecules collide more frequently with the wall of the pillow.

• Peningkatandalam kadar perlanggaran menyebabkan peningkatan tekanan

yang dikenakan oleh udara./The increase in the rate of collision results in an increase in the pressure exerted by the air.

• Tekananudara meningkat menyokong kepala budak lelaki itu.

The pressure of the air increases to support his head.

Hukum Tekanan/Pressure Law

Udara dalam tayar kereta dalam perjalanan jauhThe air in car tyre on a long journey

Sebelum perjalanan jauh Before long journey

Selepas perjalanan jauh After long journey

• Penerangan: Hubungan antara tekanan dan suhu menggunakan teori kinetik gas (isi padu malar).

Explanation: The relationship between the pressure and the temperature using the kinetic theory of gases (constant volume).

• Syarikattayarmencadangkanbahawatekanan‘sejuk’udaradalamtayarialah220kPa, walaupun tayar boleh berfungsi secara optimum pada 280 kPa./Tyre companies recommend that the 'cold' pressure of air in the tyre is 220 kPa, although the tyre can function at optimum at 280 kPa.

• Rajahmenunjukkanseoranglelakimengukurtekananudaratayarkeretanyasebelummemulakan perjalanan yang jauh. Selepas beberapa jam, dia mengukur tekanan semula dan tolok tekanan menunjukkan tekanan telah meningkat. Dia menyentuh tayar itu dan mendapati ianya panas./The diagram shows a man measuring the pressure of the air in his car tyre before starting on a long journey. After a few hours, he measured the pressure again and the pressure gauge showed that the pressure had increased. He touched the tyre and found that it was hot.

• Apabilaudaramenjadipanas, purata tenaga kinetik molekul-molekul meningkat.When the air becomes hot, the average kinetic energy of the molecules increases.

• Suhuudara meningkat ./The temperature of the air increases .

• Pergerakanpantasmolekul-molekulmenghentamdindingtayar lebih kerap .

The fast moving molecules strike the wall of the tyre more frequently .

• Molekul-molekulmengalami perubahan momentum yang besar apabila

melantun semula dari dinding tayar.

The molecules experience a large change of momentum when they bounce back from

the wall.

• Daya yang besar dikenakan ke atas dinding tayar.

A larger force is exerted on the wall of the tyre.

• Luaspermukaandindingkekaltetap,makatekanandikenakankeatasdindingmenjadi

tinggi ./The surface area of the wall remains constant, hence the pressure acting on the

wall becomes higher .

• Inimenyebabkantayarmenjadi panas .

This causes the tyre to become hot .


4

UNIT



Hukum Charles/Charles’ Law

Bola ping pongPing pong Ball

• Penerangan: Hubungan antara isi padu dan suhu gas menggunakan teori kinetik gas (tekanan malar).Explanation: The relationship between the volume and temperature of gas using the kinetic theory of gases (constant pressure).

• Apabilabolapingpongmenjadikemektanpabocor.When a ping pong ball gets dented without being punctured.

• Penyelesaianterbaikialahmerendamkannyadidalamairpanasbeberapaminit.The best solution is to dip it for a while in hot water.

• Sehinggaudaradidalambolamenyamakansuhunyadengansuhu air di luar.Until the air inside the ball tries to match its temperature to the temperature of the water outside.

• Apabilaudaradidalambola(objektertutup)dipanaskan, purata tenaga kinetik

molekul-molekul meningkat .

When the air inside the ball (closed container) is heated, the average kinetic energy of

the molecules increases .

• Suhu juga meningkat.

The temperature also increases.

• Udaradibenarkanuntuk mengembang dan meningkatkan isi padunya.

The air is allowed to expand and increase its volume.

• Molekul-molekulbergerakpantasdidalamruangyangbesar.The molecules move faster in a bigger space.

• Maka kadar perlanggaran antara molekul-molekul dengan dinding bola kekal

tetap menyebabkan tekanan tetap .

Therefore, the rate of collision between the molecules and the wall remains constant and

causes the pressure to be constant .

• Hasilpop,bahagiankemikkembalikebentukasal.As a result, the popping of the dented part restores the original shape of the ball.


4

UNIT



Tujuan Aim

Untuk menyiasat hubungan antara tekanan dan

isi padu untuk jisim gas yang tetap pada suhu

yang sama.

To investigate the relationship between the pressure

and volume for a fixed mass of gas at constant

temperature.

Untuk menyiasat hubungan antara isi padu dan

suhu untuk jisim gas yang tetap pada tekaan

yang sama.

To investigate the relationship between the volume

and temperature for a fixed mass of gas at constant

pressure.


Pemboleh ubah dimanipulasi:Manipulated variable:

Tekanan udara, P

Pressure of air, P

Pemboleh ubah bergerak balas:Responding variable:

Isi padu udara,V

Volume of air, V

Pemboleh ubah dimalarkanFixed variable:Jisim dan suhu udara

Mass and temperature of air

Pemboleh ubah dimanipulasi:Manipulated variable:

Suhu udara, T

Temperature of air, T

Pemboleh ubah bergerak balas:Responding variable:

Isi padu udara, V

Volume of air, V

Pemboleh ubah dimalarkanFixed variable:Jisim dan tekanan udara

Mass and pressure of air


Pam basikal, tiub kapilari, tiub getah dan tolok Bourdon.Bicycle pump, capillary tube, rubber tube and Bourdon gauge.

Tiub kapilari, asid sulfurik, pembaris meter separa, kaki retort, bikar, tungku kaki tiga, tiub getah, termometer dan penunu Bunsen.Capillary tube, sulphuric acid, half metre rule, retort stand, beaker, tripod stand, rubber tube, thermometer and Bunsen burner.


MinyakOil

TolokBourdonBourdongauge

Tangki minyakOil reservoir

PamPump

UdaraAir

Tiub kapilariCapillary tube

Skalaisi paduVolumescale

Tiub kapilariCapillary tube

Xxxxxxxxxxxx

AirWater Gas terpendam

Trappedair

BikarBeaker

AisIce

Pengacau/StirrerTermometerThermometer

Pembaris meter separuhHalf metre rule

(i) Mengkaji hubungan antara tekanan dan isi padu gas To investigate the relationship between the pressure and volume of a gas(ii) Mengkaji hubungan antara isi padu dan suhu gas To investigate the relationship between the volume and temperature of a gas



4

UNIT



ProsedurProcedure

1. Radas disediakan seperti rajah di atas.

The apparatus is set up as shown in the

diagram above.

2. Tekanan awal dan isi padu udara dalam

tiub kapilari direkodkan.

The initial pressure and volume of the air in

the capillary tube are recorded.

3. Tolak pam sehingga tekanan udara

adalah 100 kPa.

Push the pump until the pressure of air is 100 kPa.

4. Catatkan isi padu udara dalam tiub kapilari.

Record the volume of the air in the capillary tube.

5. Ulangi eksperimen dengan tekanan udara,

yang berbeza, iaitu, P = 120 kPa, 140 kPa,

160 kPa dan 180 kPa. / Repeat the

experiment with different pressures of air that is,

P = 120 kPa, 140 kPa, 160 kPa and 180 kPa.

6. Nilai-nilai tekanan, P, isi padu, V, dan 1 V

dijadualkan.

The values of pressure, P, volume, V, and 1 V

are tabulated.

7. Graf P melawan V dan P melawan 1 V

diplotkan.

Plot a graph P against V and a graph of P

against 1 V .

1. Radas disediakan seperti rajah di atas.

The apparatus is set up as shown in the

diagram above.

2. Air di dalam bikar dipanaskan perlahan-

lahan dan dikacau dengan seragam dan

perlahan. Apabila bacaan termometer itu

adalah 30 °C, panjang turus udara yang

terperangkap, l cm, direkodkan.

The water in the beaker is heated slowly

and stirred unifromly and gently. When the

reading of the thermometer is 30 °C,

the length of the trapped air, l cm is recorded.

3. Ulangi eksperimen dengan suhu

berlainan iaitu θ = 40 °C, 50 °C, 60 °C

dan 70 °C.

Repeat the experiment with different

temperatures of water that is θ = 40 °C, 50° C,

60 °C and 70 °C.

4. Nilai-nilai suhu, θ, suhu mutlak, T, dan

panjang turus udara yang terperangkap, l,

dijadualkan.

The values of temperature, θ, absolute temperature,

T, and length of air trapped, l, are tabulated.

5. Plotkan graf l terhadap T (dalam Kelvin).

Plot a graph l against T (in Kelvin).


Tekanan gas, P/kPaPressure of gas, P/kPa

Isi padu gas/cm3

Volume of air/cm3 1V

/cm-3

V1 V2 Vavg

100

120

140

160

180

Suhu gas, θ/°CTemperature of gas, θ/°C

Suhu mutlak, T/KAbsolute temperature, T/K

Panjang udara terperangkap, l/cmLength of trapped air, l/cm

l1 l2 lavg

30

40

50

60

70


4

UNIT



Tujuan Aim

Untuk menyiasat hubungan antara tekanan dengan suhu untuk jisim gas yang ditetapkan

pada isi padu yang sama.

To investigate the relationship between the pressure and temperature for a fixed mass of gas at

constant volume.


Pemboleh ubah dimanupulasikan:Manipulated variable:Suhu udara, T/Temperature of air, T

Pemboleh ubah bergerak balas:Responding variable:Tekanan udara, P

Pressure of air, P

Pemboleh ubah dimalarkan:Constant variable:Jisim dan isi padu udara

Mass and volume of air


Kelalangdasarbulat,termometer,kakiretort,tolokBourdon,pengacau,ais,tiubgetah,tungkukakitiga,kasadawaidanpenunuBunsen.Round-bottomed flask, thermometer, retort stand, Bourdon gauge, stirrer, ice, rubber tube, tripod stand, wire gauze and Bunsen burner.


Tiub getahRubber tube

KasadawaiWiregauze

AirWater

UdaraAir

BikarBeaker

Kaki retortRetort stand


AisIce

Tolok BourdonBourdongauge

Kelalangdasar bulatRound-Bottomed Flask

Xxxxxxxxxxxxxxxxxx

Mengkaji hubungan antara tekanan dengan suhu gasTo investigate the relationship between the pressure and temperature of a gas

Analisis dataAnalysis of the data

P/ kPa

V/cm3

l/cm

T / K

l/cm

/ °C0

θ

P/ kPa

( ) /cm–31—V

0

0 0



4

UNIT



ProsedurProcedure

1. Radas disediakan seperti yang ditunjukkan dalam rajah di atas.

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

2. Kelalang dasar bulat direndam dalam sebuah bikar berisi air dan ais bersama.

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

3. Campuran air dan ais dikacau supaya udara di dalam kelalang mempunyai suhu yang

sama seperti air.

The mixture of water and ice is stirred so that the air in the flask has the same temperature as

the water.

4. Apabilabacaantermometeradalah30°C,bacaantekanan,P,padatolokBourdon

direkodkan.

When the reading of the thermometer is 30 °C, record the reading of the pressure, P on Bourdon

gauge.

5. Ulangi eksperimen dengan suhu air yang berlainan, iaitu θ = 40 °C, 50 °C, 60 °C dan

70 °C.

Repeat the experiment with different temperatures of water, that is θ = 40 °C, 50 °C, 60 °C and 70 °C.

6. Nilai-nilai suhu, θ, suhu mutlak, T dan tekanan udara terperangkap, P dijadualkan.

The values of temperature, θ, absolute temperature, T and pressure of the air trapped, P are tabulated.

7. Graf P melawan T (dalam Kelvin) diplotkan.

Plot a graph P against T (in Kelvin).


Suhu gas, θ/°CTemperature of air, θ/°C

Suhu mutlak, T/KAbsolute temperature, T/K

Tekanan udara, P/kpaAir pressure,P/kPa

30

40

50

60

70

Analisis dataAnalysis the data

P/ kPa

T / K

P/ kPa

00/ °Cθ


4

UNIT



1 Isi padu gelembung udara pada dasar laut sedalam 50 m ialah 250 cm3. Jika tekanan atmosfera ialah

10 m air, cari isi padu gelembung udara apabila ia tiba ke permukaan laut.The volume of an air bubble at a 50 m deep seabed is 250 cm3. If the atmospheric pressure is 10 m of water, find the volume of the air bubble when it reaches the surface of the sea.


P I = (50 m + 10 m) air = 60 m air

V 1 = 250 cm3

P 2 = 10 m air

MenggunakanHukumBoyle,/Using Boyle's Law,

P1 V1 = P2 V2

(60 m air) (250 cm3) = (10 m air) × V2

V2 = (60 m air × (250 cm3)

10 m air = 1 500 cm3

2 Rajah menunjukkan tiub kaca yang mengandungi udara yang terperangkap di dalamnya. Pada 17 °C,

turus udara yang terperangkap ialah 29 cm.The diagram shows a glass tube containing some trapped air inside it. At 17 °C, the vertical column of trapped air is 29 cm.

5 cm

29 cm pada/at 17 °C

MerkuriMercury

UdaraAir

Berapakah panjang turus udara yang terperangkappada suhu 57 °C?What is the vertical column of trapped air at a temperature of 57 °C?

Penyelesaian/Solution:Menggunakan Hukum Charles,/Using Charles' Law,

V1

T1 =

V2

T2

A (L1)

T1 =

A (L2)T2

,

29 cm

(17 + 273) K =

L2

(57 + 273) K

L2 = 29 cm × 330 K

290 K = 33 cm

di mana A = luas keratan rentas tiub

where A = cross-sectional area of the tube

3 Satu campuran udara dan wap petrol disuntik ke dalam enjin silinder kereta apabila isi padu silinder itu ialah 100 cm3. Tekanan adalah 1.0 atm.

Injap ditutup dan campuran dimampatkan kepada 20 cm3. Cari tekanan sekarang.

A mixture of air and petrol vapour is injected into the cylinder of a car engine when the volume of the cylinder is 100 cm3. Its pressure is 1.0 atm. The valve is closed and the mixture is compressed to 20 cm3.Find the pressure now.

Penyelesaian/Solution:MenggunakanHukumBoyle,/Using Boyle's Law, P1 V1 = P2 V2

1 atm × 100 cm3 = P2 × 20 cm3

P2 = 100 cm3

20 cm3 × 1 atm = 5 atm

4 Isi padu gas 20 m3 pada suhu 37 °C dipanaskan sehingga menjadi 87 °C pada tekanan malar. Berapakahpeningkatanisipadunya?A gas of volume 20 m3 at 37 °C is heated until its temperature becomes 87 °C at constant pressure. What is the increase in volume?

Penyelesaian/Solution:Menggunakan Hukum Charles,/Using Charles' Law,

V1

T1 =

V2

T2

20 m3

(37 + 273) K =

V2

(87 + 273) K

V2 = 20 m3 × 360 K

310 K = 23.23 m3

Peningkatan isi padu/Increase in volume= (23.23 – 20) m3 = 3.23 m3

5 Tekanan udara di dalam bekas pada 33 °C adalah 1.4 × 105 N m-2. Bekas itu dipanaskan sehinggasuhu55°C.Berapakahtekananudaraakhirjikaisipadu bekas ditetapkan?The air pressure in a container at 33 °C is 1.4 × 105 N m-2. The container is heated until the temperature is 55 °C. What is the final air pressure if the volume of the container is fixed?

Penyelesaian/Solution:Menggunakan Hukum Tekanan,/Using Pressure Law,

P1

T1 =

P2

T2

1.4 × 105 N m–2

(33 + 273) K =

P2

(55 + 273) K

P2 = 1.4 × 105 N m–2 × 328 K

306 K = 1.5 × 105 N m-2

KBAT

KBAT

KBAT

Latihan/Exercises


4

UNIT



1 Rajah 1 menunjukkan satu sudu besi pada suhu bilik direndam dalam air panas pada 70 °C. Diagram 1 shows an iron spoon at room temperature being immersed in hot water at 70 ºC.

Sudu besiIron spoon

AirWater

Rajah 1 / Diagram 1

Keseimbangan termal tercapai apabilaThe termal equilibrium is reached when

A suhu air = suhu sudu. the temperature of the water = the temperature of the

spoon.B jisim sudu = jisim air yang tersesar. the mass of the spoon = the mass of water displaced.C isi padu sudu = isi padu air yang tersesar. the volume of the spoon = the volume of water

displaced.D muatan haba tentu sudu = muatan haba tentu air. thespecificheatcapacityofthespoon=thespecific

heat capacity of water.

2 Rajah 2 menunjukkan blok pepejal M dan N yang berjisim sama sedang dipanaskan. Suhu awal M dan N adalah sama dan dipanaskan dengan jumlah tenaga yang sama.Diagram 2 shows that solid blocks M and N of equal mass, are being heated. The initial temperature of M and N is the same and they are heated by the same amount of energy.

DapurStove

Pengadang habaHeat shield

M N

Rajah 2 / Diagram 2

Dapat diperhatikan bahawa M panas lebih cepat daripada N. Pemerhatian ini adalah disebabkan oleh perbezaan dalamIt is observed that M is hot faster than N. This observation is due to the difference in

A ketumpatan. / density.B takat lebur. / melting point.C haba pendam tentu pelakuran. latent heat of fusion.D muatan haba tentu. specificheatcapacity.

3 Rajah 3 menunjukkan susunan radas terdiri daripada penitis dan botol dengan lembaran getah.Diagram 3 shows the arrangement of apparatus made up of a dropper and bottle with a rubber sheet.

PenitisDropper Air

Water

BotolBottle

Lembaran getahRubber sheet

Rajah 3 / Diagram 3

Apabila lembaran getah ditekan,When the rubber sheet is pressed,A penitis ternaik. / the dropper rises.B air memasuki tiub penitis. water enters the dropper tube.C tekanan di A berkurang. the pressure at A decreases.D daya tujah penitis berkurang. the upthrust on the dropper decreases.

4 Graf pada Rajah 4 menunjukkan perubahan suhu ais apabila ia dipanaskan. / The graph in Diagram 4 shows the change in temperature of ice when it is heated.

Suhu

/Tem

pera

ture

/ °C

Masa/Time / s

AB C

D

Rajah 4 / Diagram 4

Antara titik A, B, C atau D, yang manakah air wujud sebagai campuran cecair dan gas?At which points, A, B, C or D, does water exist as mixture of a liquid and a gas?

5 Suhu air mendidih ialah 100 °C. Jika air mendidih pada suhu lebih tinggi daripada 100 °C, keadaan ini disebabkan apa? / The temperature at which water boils is 100 °C. if water boils at a temperature higher than 100 °C, what is the cause of this?A Air adalah tulen. / The water is pure.B Isi padu air besar. / The volume of the water is large.C Air mengandungi bendasing. The water contains impurities.D Air mendidih di dalam bekas logam yang

merupakan pengalir haba yang baik. The water is boiled in a metal container which is a

good conductor of heat.

Latihan Pengukuhan/Enrichment Exercises


4

UNIT



6 Rajah 6 menunjukkan satu omboh kedap udara.Diagram 6 shows an air tight piston.

KlipClip

Udara / Air

10 cmKedap udara

Airtight

Rajah 6 / Diagram 6

Ia adalah sukar untuk menolak omboh ke hadapan ke dalam picagari apabila ruangannya ditutup. Ini adalah keranaItisdifficulttopushthepistonforwardintosyringewhenits outlet is closed. This is becauseA rintangan geseran antara silinder dan omboh. the frictional resistance between the cylinder and the

pistonB molekul udara dalam silinder berlanggar dengan

omboh pada kadar yang lebih tinggi. the air molecules in the cylinder collide with the

piston at higher rate.C daya tolakan antara molekul-molekul. of the forces of repulsion between the molecules.D bilangan molekul-molekul di dalam silinder

bertambah. the number of air molecules in the cylinder increases.

7 Rajah 7 menunjukkan satu turus udara terperangkap di dalam tiub kapilari oleh benang merkuri. Tiub rerambut diletakkan di dalam tiga keadaan yang berbeza K, L dan M.Diagram 7 shows an air column trapped inside a capillary tube by a thread of mercury. The capillary tube is placed in three different situations K, L and M.

K M

L

Rajah 7 / Diagram 7

Tekanan udara di K, L dan M masing-masing P1, P2 dan P3. Antara perbandingan berikut, yang manakah adalah betul?The air pressure in K, L and M are P1, P2 and P3 respectively. Which of the following comparison is correct?

A P1 = P2 = P3

B P1 > P2 > P3

C P1 < P2 < P3 D P1 = P2 > P3

8 Rajah 8 menunjukkan pembentukan bayu darat.Diagram 8 shows the formation of a land breeze.

Udara panasHot air

Waktu malamNight time

Udara sejukCool air

Laut/SeaDarat/Land

Rajah 8 / Diagram 8

Antara pernyataan berikut, yang manakah betul?Which of the following statements is correct?

A Muatan haba air laut lebih tinggi daripada muatan haba darat.

The heat capacity of the sea water is higher than the heat capacity of the land.

B Tekanan laut adalah lebih tinggi daripada di atas darat.

The pressure of the sea is higher than that on the land.

C Suhu darat menjadi lebih tinggi daripada suhu air laut.

The temperature of the land becomes higher than the temperature of the sea water.

D Ketumpatan darat kurang daripada air laut. The density of the land is less than that of sea water.

9 Berapakahhabayangdiperlukanuntukmenukarkan0.2 kg air air mendidih kepada wap?How much heat is needed to change 0.2 kg of boiling water to vapour?

(Haba pendam tentu pengewapan air ialah / specificlatent heat of vaporisation of steam is 2.26 × 106 J kg–1)A 452 kJ C 113 kJB 500 kJ D 1.33 kJ

10 Sebuah pemanas berkuasa 1 200 W sedang memanaskan suatu cecair pada takat didihnya. Jika haba pendam tentu pengewapan cecair itu ialah 2.0 × 106 J kg–1, berapakah jisim cecair itu yang telah ditukarkan kepada wap setelah dipanaskan selama 10 minit?An electric heater with a power of 1 200 W is heating a liquidat itsboilingpoint. If thespecific latentheatofvaporisation of the liquid is 2.0 × 106 J kg–1, what is the mass of the liquid that has changed into vapour after 10 minutes?A 0.18 kgB 0.36 kgC 3.0 kgD 5.6 kg


4

UNIT



11 Rajah 11(a) dan Rajah 11(b) menunjukkan suatu eksperimen untuk menentukan haba pendam tentu pelakuran ais. Rajah 11(b) adalah eksperimen kawalan.Diagram 11(a) and Diagram 11(b) show an experiment beingusedtodeterminethespecificlatentheatoffusionofice. Diagram 11(b) is a control experiment.

AirWater

Pemanas rendamImmersion heater

AisIce

AirWater

AisIce

(a) (b)

Rajah 11 / Diagram 11

Eksperimen kawalan adalah untukThe control experiment is to

A mengawal kadar peleburan ais. control the rate of melting of ice.B memastikan air yang digunakan adalah tulen. ensure that the water used is pure. C menentukan nilai purata bagi haba pendam tentu

pelakuran ais. determinetheaveragevalueofthespecificlatentheat

of fusion of ice.D menentukan jisim ais yang melebur tanpa

pemanasan daripada pemanas rendam itu. determine the mass of ice which melts without heating

it with the heater.

12 Haba pendam tentu pengewapan air ialah haba yang diserap olehThespecificlatentheatofvaporisationofwateristheheatabsorbed byA 1 kg air untuk menaikkan suhu sebanyak 1 °C 1 kg of water to increase the temperature by 1 °CB 1 m3 air untuk menaikkan suhu sebanyak 1 °C. 1 m3 of water to increase the temperature by 1 °CC 1 kg air semasa mendidih untuk menukarkannya

kepada stim 1 kg of water during boiling to change it to steamD 1 m3 air semasa mendidih untuk menukarkannya

kepada stim 1 m3 of water during boiling to change it to steam

13 Antara graf berikut, yang mana menunjukkan hubungan antara isi padu, V, dengan suhu, T, bagi suatu gas yang berjisim tetap pada tekanan yang malar?Which of the following graphs shows the relationship between the volume, V, of a fixed mass of gas with itstemperature, T, at constant pressure?

A

0

00

0

V/m3

T/K

V/m3

T/K

V/m3

T/K

V/m3

T/K

B

0

00

0

V/m3

T/K

V/m3

T/K

V/m3

T/K

V/m3

T/K

C

0

00

0

V/m3

T/K

V/m3

T/K

V/m3

T/K

V/m3

T/K

D

0

00

0

V/m3

T/K

V/m3

T/K

V/m3

T/K

V/m3

T/K

14 Sebuah kolam berkedalaman 40 m. Suatu gelembung udara dengan isi padu, V cm3, pada dasar kolam itu menaik sehingga aras permukaan air. Tekanan atmosfera adalah bersamaan dengan 10 m air. Apabila gelembung udara itu telah menaik 20 m, isi padunya akan menjadiAn air bubble of volume, V cm3, at the bottom of a lake which is 40 m deep rises towards the surface of the water. Atmospheric pressure is equivalent to 10 m of water. When the air bubble has risen 20 m, its volume will become

A 0.50 VB 0.60 VC 1.67 VD 2.00 V


4

UNIT



1 Rajah 1.1 dan Rajah 1.2 menunjukkan gas, X, yang berjisim tetap terperangkap dalam dua kelalang serupa yang dipanaskan dengan kuantiti haba yang berbeza. / Diagram1.1andDiagram1.2showthatafixedmassoftrappedgas,X,intwoidenticalflasksheatedwithdifferentquantitiesofheat.

Rajah 1.1 / Diagram 1.1 Rajah 1.2 / Diagram 1.2

Tolok BourdonBourdon gauge

BlokkayuWooden block

UdaraAir

DapurStove

Air dan aisWater and ice

PengacauStirrer


KelalangFlask

UdaraAir

DapurStove

Air dan aisWater and ice

PengacauStirrer


Tolok BourdonBourdon gauge

BlokkayuWooden block

KelalangFlask

(a) NyatakankuantitifizikdanunitS.I.nyayangdiukurolehtolokBourdonitu. State the physical quantity and its S.I. unit measured by the Bourdon gauge.

Tekanan udara. Pascal atau N m–2

Air pressure. Pascal or N m–2

(b) Kedua-duadapurditutupselepas8minit.BerdasarkanRajah1.1danRajah1.2, Both stoves are turned off after 8 minutes. Based on Diagram 1.1 and Diagram 1.2,

(i) bandingkanbacaankedua-duatolokBourdonitu./compare the readings of the both Bourdon gauges.

BacaantolokBourdondalamRajah1.2lebihtinggidaripadabacaantolokBourdondalamRajah1.1.

The reading of Bourdon gauge in Diagram 1.2 is higher than that of Diagram 1.1.

(ii) bandingkan bacaan kedua-dua termometer. / compare the readings of both thermometers.

BacaantermometerdalamRajah1.2lebihtinggidaripadabacaantermometerdalamRajah1.1.

The reading of the thermometer in Diagram 1.2 is higher than that of Diagram 1.1.

(iii) nyatakan hubungan antara tekanan gas, X, dalam kelalang dengan suhu. statetherelationshipbetweenthepressureofgas,X,intheflaskandthetemperature.

Apabila suhu menaik, tekanan gas, X, dalam kelalang menaik.

Whenthetemperatureincreases,thepressureofgas,X,intheflaskincreases.

(c) Berdasarkanteorikinetik,terangkansebabbagijawapandi1(b)(iii). Based on kinetic theory, explain the reason for the answer in 1(b)(iii).

Apabila molekul-molekul gas dalam kelalang menerima haba, halaju molekul-molekul gas bertambah. Ini

menyebabkan tenaga kinetik molekul-molekul gas itu bertambah. Apabila tenaga kinetik molekul-molekul

gas bertambah, frekuensi perlanggaran di antara molekul-molekul gas dengan dinding kelalang bertambah,

maka tekanan gas dalam kelalang bertambah. Whenthegasmoleculesintheflaskreceiveheat,thevelocityofthesemoleculesincreases.Thiscausesthekinetic

energy of the gas molecules to increase.When the kinetic energy of the gas molecules increases, the frequency of

collisionbetweenthegasmoleculesandthewallsoftheflaskincreases,thusthegaspressureintheflaskincreases.

(d) Namakan hukum yang terlibat apabila isi padu gas, X, dimalarkan. Name the law involved when the volume of gas, X, is constant.

Hukum Tekanan / Pressure Law

Soalan Struktur/Structure Questions


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