Chapter 4: Heat Form 4

CHAPTER 4: HEAT4.1 Understanding thermal equilibrium

The difference between temperature and heat

Temperature HeatIs the degree of hotness of a body

Is a form of energy

Is a base quantity Is a derived quantityDepends upon the kinetic energy of the molecules

Depend upon temperature, mass and type of material ( spesific heat capacity or specific latent heat) ( Q = mc or Q = ml) )

The S.I. unit is K or 0C The S.I. unit is Joule(J)Measured by thermometer

Measured by Joulemeter

Thermal Contact

Two substances are said to be in thermal contact,when heat flows from one substance to another it is in contact with. Heat flows according to temperature differences i.e.from substance hot to cold substance

The principle of thermal equilibrium

Two bodies in thermal contact are said to be in thermal equilibrium when its reach the same temperature and the net rate of heat transfer between the two bodies is zero.

How a liquid in glass thermometer works ?

When a thermometer is in thermal contact with a substance ( for example hot water) , heats flows from the hot water to the thermometer(mercury)

When thermal equilibrium is reached the net rate of heat transfer between the two substances is zero. The temperature of the thermometer is same as the temperature of the hot water.

Hence by showing its own temperature, the thermometer also reads the temperature of the hot water.

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Chapter 4: Heat Form 4

Basic principle to construct a thermometer.

Two important principles to construct a thermometer are.

1 Specific thermometric property i.e. a physical quantity which varies with temperature.The table shows four different types of thermometers.

Thermometer Thermometric propertyMercury thermometer

Volume of mercury varies with temperature

Resistance thermometer

Electrical resistance of a wire varies with temperature

Thermocouple thermometer

Electromotive force (e.m.f) varies with temperature

Gas thermometerGas pressure varies with temperature

2 Calibration of thermometer i.e the process of marking-up a scale on a thermometer.

To produce a scale for a thermometer, two fixed points of thermometer must first be selected.a) Lower fixed point (0oC)– is the melting temperature of pure ice

at standard atmospheric pressure.b) Upper fixed point (100 o C) – is the temperature of steam a

standard atmospheric pressure.

Calibration of a Mercury-in-glass Thermometer on the Celsius Scale

Freeze some pure water. Crush the ice into small and fill a funnel with them. When the ice begins to melt inset the bulb of a thermometer so that it is covered with

ice. When the mercury stops shrinking , mark the stem of the thermometer at the mercury

level, as 0oC. Now arrange the thermometer inside a flask so that its bulb is just above the surface of

boiling water. When the mercury stops expanding , mark its level on the thermometer stem ,as 100oC. Divide the distance between the marks 0oC and 100oC into 100 equal parts, marked as

a scale along the stem54

Chapter 4: Heat Form 4

The formula is used to calibrate a thermometer

= temperature of a substanceo = ice point100 = steam pointxo = the length of the mercury column at ice point x100 = the length of the mercury column at steam point x = the length of the mercury column

when the thermometer is placed in a substance

Mercury Thermometer

The specific thermometric property in used in this thermometer the changes of the volumes of mercury with the temperature i.e when the temperature increases ,the volume of the mercury increases.

The sensitivity of the thermometer can be increased by1 using a thinner-walled glass bulb2 reducing the diameter of the capillary tube

Mercury is used in the thermometer because1 has a higher boiling point2 does not stick to the glass3 is opaque and therefore it is easier to read.4 expands and contracts uniformly

4.2 UNDERSTANDING SPECIFIC HEAT CAPACITY, cDefinition and the S.I unit of Specific Heat Capacity ,c

The specific heat capacity of a material is defined as the amount of heat required to raise the temperature of 1 kg of the substance through a temperature of 1oC.

The S.I. unit for c is J kg-1 oC-1 A substance with a large specific heat capacity( able to store a

lot of heat ) will experience a smaller temperature rise when absorbing heat and vice versa.

The relationship between m,c, and Q

Where, Q = the heat energy transffered to the substance

m = the mass of the substancec = the specific heat capacity of the substance = the temperature change

Mixtures Whenever two materials are at different temperatures and in thermal contact, the hotter material will lose energy by transfer to the colder one until both come to the same temperature. They are now in thermal equilibrium.

To determine the specific heat capacity of a solid ( aluminium cylinder)55

Q = m c

Chapter 4: Heat Form 4

The electrical power of the heater is recorded = P The mass of the aluminium cylinder is recorded = m The initial temperature of the aluminium is read off the thermometer = 1 The electric heater is switched on and the stopwatch is started simultaneously. After heating for a t ime , t , the heater is switched off. The maximum reading on the thermometer is recorded = 2

Calculate the heat supplied by the heater = Pt Calculate the heat absorbed by the aluminum cylinder = mc ( 2-1) On the assumption that there is no heat loss to surroundings;,

Pt = mc (2-1) c = Pt . m (2-1)

Precautions (1) Insulates the aluminium cylinder with felt cloth or wool to decrease the loss of heat to

the surroundings.(2) Some oil is poured into the hole before the thermometer is inserted to ensure more

perfect heat conduction.Discussions

The value of the specific heat capacity of aluminium ,c determined in the experiment is larger than the standard value of c.This is because the experimental value of the temperature rise , less than the expected temperature rise due to some heat loss to the surroundings.The smaller the temperature , the greater thespecific heat capacity because c = Pt m

To determine the specific heat capacity of a solid ( aluminium cylinder)

The electrical power of the heater is recorded = P

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Chapter 4: Heat Form 4

The empty beaker is weighed and its mass is recorded = m1 The beaker is filled with water and the beaker is weighed again and it mass is recorded = m2

The initial temperature of the water is read off the thermometer = 1 The electric heater is switched on and the stopwatch is started simultaneously. After heating for a t ime , t , the heater is switched off. The maximum reading on the thermometer is recorded = 2

Calculate the heat supplied by the heater = Pt Calculate the heat absorbed by the water = ( m2 - m1)c (2-1) On the assumption that there is no heat loss to surroundings;,

Pt = ( m2 - m1)c (2-1) c = Pt . ( m2 - m1) (2-1)

Precautions (1) The water must be stirred continuously to ensure its temperature is uniform.(2) The beaker should be wrapped with a tissue or felt cloth and placed on a polystyrene

sheet to prevent heat loss to the surroundings and the table.

DiscussionsThe value of the specific heat capacity of water ,c determined in the experiment is larger than the standard value of c.This is because the experimental value of the temperature rise , less than the expected temperature rise due to some heat loss to the surroundings.The smaller the temperature , the greater thespecific heat capacity because c = Pt m

Applications of specific heat capacity of a substance.

Cooking instruments Cooking instruments such as frying pans, pots,kettles, electric iron and so on made of substances with low specific heat capacities . This is because they can quickly heated up when there is only small heat absorption.

The handle of the cooking instruments are made by the substances with high specific heat capacities. This is because these materials undergo a small heat change while heat is released or absorbed. So , the handles are not too hot to be held by the bare hands.

Water Water has a very high specific heat capacity. This makes it very useful for storing energy. For a given rise in temperature , water can store more energy than most other substances; as it cools , it releases this energy again. It is fortunate that water is cheap , safe and readily available.

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Chapter 4: Heat Form 4

In a central heating system water is good at storing energy and carrying it from the boiler to the radiators, as it cools there it releases a lot of energy into the room.The system is commonly used in cold countries to heat and keep houses warm.

In a car engine cooling system , water is circulated through pipes around the engine block to absorb energy from the hot engine and so to keep it cool. From the cylinder block ,the water passes into radiator where it is cooled by air drawn in by radiator fan.

The cool water is re-circulated through the engine to absorb the heat and this cycle is repeated continuously while the engine is running.

Sea breeze and Land breeze In daytime the sun warms the land to higher temperature than the sea.

It is because land has a lower specific heat capacity than sea-water.

The air above the land is heated and rises, and its place is taken by cooler air above the sea moving inland (convection currents)

Air higher in the atmosphere completes the circulation , and hence a sea-breeze is obtained.

At night the sea temperature drops only slightly , since it is warmed to a considerable depth during the day. On the other hand , the land temperature drops considerably at night. This time, therefore , a convection current is obtained in the opposite direction to daytime, and this is a land-breeze.

Understanding specific latent heat, l

What does the word “latent heat “ mean?

Latent heat means hidden heat. This heat energy changes the state of a substance (phase change). The heat cannot be ‘seen’ because there is no rise in temperature of the substance

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Chapter 4: Heat Form 4

Phase ChangeWhen a phase change has occurred , latent heat is absorbed or released

Latent heat and kinetic theory

In a solid, the molecules are linked to the neighbours by forces of attraction. As the solid is heated, the molecules vibrate more strongly. When the solid reaches its melting point, the vibrations have become so strong that the links begin to give way. Extra energy is needed to overcome these forces and separate the molecules. This is called the latent heat of fusion.

No temperature rise occurs during this process, because the latent

heat of fusion is used to overcome the intermolecular binding forces. The average translational kinetic energy does not change, so the temperature remains constant.

In a liquid, the molecules are free enough to slide around and change neighbours, but they are still almost as close to each other as in a solid. The links are weaker but still effective. As the liquid is heated further, the kinetic energy of the molecules increases more. At the boiling point, the molecules break free from each other and become a gas. Energy is needed to overcome the remaining links. This is called the latent heat of vaporisation.

No temperature rise occurs during this process, because the latent heat of vaporisation is used to overcome the intermolecular binding forces. The average translational kinetic energy does not change, so the temperature remains constant.

.

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Chapter 4: Heat Form 4

The Heating and Cooling Curve (Naphthalene)

Heating curve

Melting point = 80oC, Boiling point = 218oCAB = Solid , BC = solid + liquid CD = liquid DE = liquid + gas

EF = gas

At AB,CD dan EF :

The heat supplied increases the kinetic energy of naphthalene. So the temperature rises because the temperature is a measure of the average kinetic energy of molecules in a substance

At BC,DE :

At t1 and t2 phase change has occurred.The latent heat is absorbed to provide the energy to overcome the binding forces between the molecules. The energy absorbed does not increase the kinetic energy of the molecules, so the temperature remains constant.

Cooling curve

Freezing point = 80oC Condensation point = 218oCAB = Gas BC = Gas + Liquid CD = LiquidDE = Liquid + Solid EF = Solid

At AB,CD dan EF :

Heat is released to the surroundings and the kinetic energy of the molecules decreases, resulting in a fall in the temperature of the naphthalane because the temperature is a measure of the average kinetic energy of molecules in a substance

Pada BC,DE :

At t1 and t2 phase change has occurred. The latent heat is released to the

surroundings as the molecules become more closely pack..

The energy released does not decrease the kinetic energy of the molecules, so the temperature remains constant.

Definition and the S.I unit of Specific Latent Heat ,lThe specific latent heat of fusion , lf

Is the quantity heat of energy required to change 1 kg of a substance from the solid state to the liquid state , without a change in temperature

The specific latent heat of Vaporisation ,lv :

Is the quantity heat of energy required to change 1 kg of a substance from the liquid state to the gaseous state , without a change in temperature.

The S.I. unit of lf and lv is J kg-1

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Chapter 4: Heat Form 4

The relationship between m,l and Q Where,Q = the heat energy transferred to the

substancem = the mass of the substancel = the specific latent heat of the

substance To determine the latent heat of fusion of ice

The electrical power of the heater is recorded = PThe mass of each the two empty beakers is determined using the weighing balance.Mass of empty beaker A = m1 Mass of empty beaker B = m2

When water starts to drip from the filter funnels at a steady rate, the heater in Set A is switched on.

The stopwatch is started and the empty beakers A and B are placed beneath the filter funnels.

After a period of t , the heater in Set A is switched off. The masses of both beakers of water , A and B are determined using the weighing balance.

Mass of beaker A + water = m3 Mass of beaker B + water = m4

Calculate mass of ice melted by the electric immersion heater, m = (m3 - m1) - (m4 - m2) Calculate the heat supplied by the heater = Pt Calculate the heat absorbed by the ice during melting = mlf

On the assumption that there is no heat loss to surroundings;, Pt = mlf

lf =

Precautions The immersion heater must be fully immersed in the ice cubes to avoid or reduce heat lossDiscussions

The value of the specific latent heat of fusion of ice ,lf determined in the experiment is larger than the standard value of lf.This is because the experimental value of the mass of ice melted ,m less than the expected m due to some heat loss to the surroundings.The smaller the mass m, the greater the specific latent heat of fusion of ice,lf, lf = Pt m

To determine the latent heat of vaporisation of water61

Q = m l

Chapter 4: Heat Form 4

The electrical power of the heater is recorded = P The electric heater is switched on the heat the water to its boiling point. When the water starts to boil at a steady rate , the stopwatch is started and the reading on

the balance is recorded = m1

After a time ,t the reading on the electronic balance is recorded again = m2 Calculate the mass of water evaporated, m = m1 - m2

Calculate the heat supplied by the heater = Pt Calculate the heat absorbed by the water during vaporisation = mlv

On the assumption that there is no heat loss to surroundings;,

Pt = mlv

lv = Pt m

Precautions The immersion heater must be fully immersed in the water to avoid or reduce direct heat loss to the surroundings.

Discussions The value of the specific latent heat of vaporization of water ,lv determined in the

experiment is larger than the standard value of lv. This is because the experimental value of the mass of water evaporated ,m less than

the expected m due to some heat loss to the surroundings. The smaller the mass m, the greater the specific latent heat of fusion of ice,lv,

lv = Pt m

. Applications of Specific Latent Heat in Everyday Life(1) When we are engaged in strenuous activities , sweating cools our bodies. The sweat

evaporates and the bodies heat is removed as the latent heat of vaporisation.thus our bodies temperature is decreased.

(2) Drinks can be cooled by adding in several cubes of ice. When the ice is melting , the latent heat of fusion is absorbed from the drinks. The temperature of the drinks is lowered.

(3) Food can be cooked by using steam. Food such as cakes, eggs, fish, buns and others receive a large amount of energy when the latent heat of vaporization of steam released from condensing steam.

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Chapter 4: Heat Form 4

4.4 Understanding the gas laws Gas pressure, temperature and volume in terms of gas molecules.

The kinetic theory of gases was proposed to explain the gas laws. The basic assumptions are:

1. All gases are made up of a very large number of tiny molecules.2. These molecules are constantly moving around randomly at high speeds.3. The molecules collide elastically with anything they meet. If they hit the inner walls of the

container , they bounce off again at the same speed.4. The molecules are so small and so far apart that they almost never collide with each other.

So the volume of the gas molecules themselves is negligible with the volume of container, that is, almost all the gas is empty space.

5. They do not exert any forces on each other ,but move randomly. There is no intermolecular attractive forces. Intermolecular forces of repulsion act only during collisions between molecules; the duration of collisions is negligible compare compared with the time interval between collisions.

How the gas pressure is produced?

Based on the assumptions of kinetic theory of gases , molecules of a gas will occupy

the entire available space and collisions occur between molecules and the walls of container.

Imagine a molecule of mass m approaching one wall with velocity , v . Its momemtum = mv. It rebounds with velocity (-v) because it experiences an elastic collision. Its momentum now is - mv. So the change of momentum = 2mv According to Newton’s second law of motion , force is exerted on the wall of container

because force is the rate change of momentum F = change of momentum time As the result gas pressure is produced because by the definition of pressure;

Pressure is Force per unit area P =

Hence the gas pressure in the container is the total force , produced by the collision

between molecules and the walls of container. The higher the average velocity of the molecules in the gas, the greater pressure

exerted by the gas.

What happen when a gas is heated?

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Chapter 4: Heat Form 4

As a gas is heated , the molecules move faster because the kinetic energy of the gas molecules is proportional to the temperature of the gas. As the result the pressure of the gas increases if the volume of the gas is fixed.

Therefore a fixed mass of a gas in a container has three characteristics , i.e pressure, volume and temperature. The relationship between these characteristics can be explained by the three gas laws.

The three gas laws are shown in the following table.

Gas law Relationship Constant Application

Boyle

P 1 V

orP1V1 = P2V2

T

1. The bubbles formed by a fish expand as they floats towards the surface.

2. Bicycle pump

Charles

V T

P 1. Hot- air balloon

PressureP T

V1.Car tyres after a long drive become very firm.

Boyle’s Law Boyle’s law states that “ For a fixed mass at constant temperature, the pressure of gas is inversely proportional to its volume

Equation Or P 1 if T constant V Where P = pressure

V = volume T = temperatureOr P = k V PV = k

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P1 V1 = P2 V2

Chapter 4: Heat Form 4

Graph

Boyle’s Law and the kinetic theory of gases.

Imaging a sample of gas being compressed , with the temperature staying constant.

The average kinetic energy of the molecules of the gas remains unchanged but they are now confined to a smaller space.

The molecules are squeezed closer together. As a result , the frequency of collisions between the molecules and the walls of the container increases .

Therefore , the force increases resulting in a corresponding increase in the pressure of the gas.

P= F Ai.e. as volume decreases , pressure increases

Charle’s Law

Charles’s law states that “ For a fixed mass at constant pressure, the volume of gas is directly proportional to its absolute temperature”

V T Where V = VolumeV = KT T = Absolute temperatureV = KT

Charle’s Law and the kinetic theory of gases

In fixed mass of gas at constant pressure, the frequency of collisions between the gas molecules and the walls container is constant.

As the gas is heated , the molecules move faster. They collide with the walls more frequently and at greater speed.

So they exert a larger pressure on the walls of the containers. As a result , the gas will expand if it is able to. This allows molecules

to spread out a little which reduces the number of collisions per second with each unit area of the walls.

The gas continues to expand until the pressure is back to its original value.

i.e. if temperature is increased but pressure stays the same, the volume must increase.

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V1 = V2

T1 T2

Chapter 4: Heat Form 4

The Pressure Law The pressure law states that “ For a fixed mass at constant volume, the pressure of gas is directly proportional to its absolute temperature”

P T Where P = PressureP = KT T = Absolute temperatureP = KT

The pressure Law and the kinetic theory of gases

The sample of gas is kept at constant volume. As the temperature of the gas rises, its molecules move more rapidly. As the result, they collide with the walls of the container at higher

frequency , the change of momentum is greater ,and so the force they exert on the walls is larger.

As a consequence, the force and hence the pressure increases. P= F A

i.e. as temperatures increases , pressure increases

Absolute Temperature,T The Kelvin scale is known as the absolute temperature scale.

o C = ( + 273 ) K

The absolute zero temperature of -273 oC or 0 K is the lowest possible temperature that could be attained. Based on Charle’s Law and the pressure law, at the

absolute zero temperature the volume and the pressure of the gas become zero.

If absolute zero temperature is related to the kinetic energy of molecules, then we might expect that there would be a temperature where the molecules would be stationary and their kinetic energy would be zero.

At absolute zero the kinetic energy of molecules is a minimum. No object can be cooled to a lower temperature than this.

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P1 = P2

T1 T2