THERMOMETER

“The instrument which is used to measure the temperature is called a thermometer”

Thermometric Material:

“The material that is used in thermometer for measuring temperature is called thermometric material.”

Some substances have property that changes with temperature. Substance that show change with temperature can be used as thermometric material. For example some substances expand on heating, some change their colours, some change their electric resistance etc. Nearly all the substances expand on heating liquids also expand on heating and are suitable as thermometric materials.

Common thermometers are generally made using some suitable liquid as thermometric material.

Properties of Thermometric Properties:

A thermometric liquid should have the following properties:

It should be visible

It should have uniform thermal expansion

It should have a low freezing point

It should have high boiling point

It should not wet glass

It should be a good conductor of electricity

It should have small specific heat capacity

Liquid – In – glass Thermometer:

A liquid – in – glass thermometer has a bulb with a long capillary tube of uniform and fine bore. A suitable liquid is filled in the bulb. When the bulb contacts a hot object, the liquid in it expands and rises in the tube. The glass stem of a thermometer is thick and acts as a cylindrical lens. This makes it easy to see the liquid level in the glass tube as shown in the figure:

Mercury the Best Thermometric Material:

Mercury freezes at -39 ^oC and boils at 357 ^oC. It has all the thermometric properties listed above. Thus mercury is one of the most suitable thermometric materials. Mercury – in – glass thermometers are widely used in laboratories, clinics and houses to measure temperatures in range from -10 ^oC to 150 ^oC.

Reference Points: (GRW 2017)

A thermometer has a scale on its stem. This scale has two fixed points.

Lower Fixed Point:

The lower fixed point is marked to show the position of liquid in the thermometer when it is placed in ice.

Upper Fixed Point:

The upper fixed point is marked to show the position of liquid in the thermometer when it is placed in steam at standard pressure above boiling water.

Scales of Temperature:

The distance between two reference points is divided in different divisions. A scale is marked on the thermometer. The temperature of the body in contact with the thermometer can be read on that scale.

Types of Temperature Scale: (LHR 2017)

There are three types of temperature scale, which are as follows:

Celsius scale or centigrade scale

Fahrenheit scale

Kelvin scale

Fahrenheit and centigrade or Celsius scales are used to measure temperatures in ordinary life while Kelvin scale is in practice for scientific purposes. Various scales of temperature are shown below:

Celsius Scale:

On Celsius scale, for water the interval between lower and upper fixed point is divided into 100 equal divisions. The lower fixed point is marked as 0 ^oC and the upper fixed point is marked as 100 ^oC.

Fahrenheit Scale:

On Fahrenheit scale, the interval between lower and upper fixed points is divided into 180 equal divisions. The lower fixed point is marked as 32 ^oF and the upper fixed point is marked as 212 ^oF.

Kelvin Scale:

In SI units, the unit of temperature is Kelvin (K) and its scale is called Kelvin scale of temperature. The interval between the lower and upper fixed points is divided into 100 equal divisions. Thus a change in 1^oC is equal to a change of 1 K. the lower fixed point on the scale corresponds to 273 K and the upper fixed point is referred as 373 K. The zero on this scale is called the absolute zero and is equal to – 273 ^oC.

Scale Conversion Formulae:

Following are scale conversion formulae

From Celsius to Kelvin Scale:

The temperature T on Kelvin scale can be obtained by adding 273 in the temperature C on Celsius scale thus

T (K) = 273+C

From Kelvin to Celsius Scale:

The temperature on Celsius scale can be found by subtracting 273 from the temperature in Kelvin Scale. Thus

C = T (K) – 273

From Celsius to Fahrenheit Scale:

Since 100 divisions on Celsius scale are equal to 180 divisions on Fahrenheit scale. Therefore, each division on Celsius scale is equal to 1.8 divisions on Fahrenheit scale. Moreover, 0°C corresponds to 32°F.

F = 1.8 C + 32

Here F is the temperature on Fahrenheit scale and C is the temperature on Celsius scale

INTERNAL ENERGY

Introduction:

Heat is called as the energy in transit. Once heat enters a body, it becomes its internal energy and no longer exists as heat energy.

Definition:

The sum of kinetic energy and potential energy associated with the atoms, molecules and particles of a body is called its internal energy.

Dependence:

Internal energy of a body depends on many factors such as the mass of the body, kinetic and potential energies of molecules etc. Kinetic energy of an atom or molecule is due to its motion which depends upon the temperature. Potential energy of atoms or molecules is the stored energy due to intermolecular forces.

TEMPERATURE

Definition:

“Degree of coldness or hotness of the body is a measure of its temperature”

Quantity:

Temperature is a base and scalar quantity.

Unit:

SI unit of temperature is kelvin (K)

HEAT

Definition:

“Heat is the energy that is transferred from one body to the other in thermal contact with each other as a result of the difference of temperature”

Quantity:

Heat is a derived and scalar quantity.

Unit:

SI unit of heat is Joule (J)

THERMAL CONTACT

“Such a contact of bodies in which exchange of heat takes place is called thermal contact.”

Example:

To store ice in summer, people wrap it with cloth or keep it in wooden box or in thermos flask. In this way, they avoid the thermal contact of ice with its hot surroundings otherwise ice will soon melt away.

THERMAL EQUILIBRIUM

“The state of thermal contact at which two bodies attain same temperature and no exchange of heat takes place is called thermal equilibrium.”

Example:

TOUCH OF A HOT BODY

THERMOMETER

“The instrument which is used to measure the temperature is called a thermometer”

Thermometric Material:

“The material that is used in thermometer for measuring temperature is called thermometric material.”

Some substances have property that changes with temperature. Substance that show change with temperature can be used as thermometric material. Common thermometers are generally made using some suitable liquid as thermometric material.

ABSOLUTE ZERO

Absolute zero is the point at which the fundamental particles of nature have minimal vibrational motion, retaining only quantum mechanical, zero-point energy-induced particle motion.

By international agreement, absolute zero is defined as precisely; 0 K on the Kelvin scale, which is a thermodynamic (absolute) temperature scale; and –273.15 degrees Celsius on the Celsius scale.

CLINICAL THERMOMETER

A clinical thermometer is used to measure the temperature of human body. It has a narrow range from 35^oC to 42^oC. It has a constriction that prevents the mercury to return. Thus, its reading does not change until reset as shown in the figure:

CONVERSIONS OF SCALES

Following are scale conversion formulae.

Conversion of Celsius (centigrade) to Fahrenheit scale:

T_F =

Conversion of Fahrenheit to Celsius scale:

T_c =

Relationship between Kelvin and Celsius scales:

T_k = T_c + 273

Every thermometer makes use of some property of a material that varies with temperature. Name the property used in: (Mini exercise Pg. # 170)

Strip thermometers

Mercury thermometers

Ans. PROPERTIES OF MATERIALS

(a) In strip thermometers, colour variation is used.

(b) Uniform thermal expansion of liquids is used in mercury thermometer.

EXAMPLE 8.1

Given Data:

Temperature on Celsius scale = C = 20 ⁰C

To Find:

Temperature on Kelvin scale = T (K) = 20 ⁰C

Calculations:

We know

T (K) = 273 + C

Putting values

T (K) = 273 + 20 = 293 K

Result:

20 ^oC on Celsius scale is equal to 293 K.

EXAMPLE 8.2

Given Data:

Temperature on Kelvin scale = T (K) = 300K

To Find:

Temperature on Celsius scale = C =?

Calculations:

We know

T (K) = 273 + C

And C = T (K) -273

Putting values

C= 300 – 273 = 27 ^oC

Result:

300 K is equal to 27 ^oC.

EXAMPLE 8.3

Given Data:

Temperature on Celsius scale = C = 50 ⁰C

To Find:

Temperature on Fahrenheit scale = F =?

Calculations:

We know

Putting values

F = 122 ^oF

Result:

50 ^oC on Celsius scale is equal to 122 ^oF.

EXAMPLE 8.4

Given Data:

Temperature on Fahrenheit scale = F =100°F

To Find:

Temperature on Celsius scale = C =?

Calculations:

We know

Putting values

1.8 C = 68

Results:

100 ^oF is equal to 37.8 ^oC.

SPECIFIC HEAT CAPACITY

Definition:

“Specific heat of a substance is the amount of heat that is required to raise the temperature of 1 kg mass of that substance through 1K”.

Formula:

Unit:

SI unit of specific heat capacity is Jkg^-1K^-1.

Explanation:

Generally, when a body is heated, its temperature increases. Increase in the temperature of a body is found to be proportional to the amount of heat absorbed by it.

It has also been observed that the quantity of heat ∆Q required to raise the temperature ∆T of a body is proportional to the mass m of the body.

Thus ∆Q α m ∆T

or ∆Q = c m ∆T

Here ∆Q is the amount of heat absorbed by the body and c is the constant of proportionality called the specific heat capacity or simply specific heat.

So

Examples:

Specific heat capacity of water is 4200 Jkg^-1K^-1 and specific heat capacity of dry soil is 810 Jkg^-1K^-1

LARGE SPECIFIC HEAT OF WATER

Introduction:

Specific heat of water is 4200 Jkg^-1K^-1 and of dry soil is about 810 Jkg^-1K^-1. As a result the temperature of soil would increase five times more than the same mass of water by the same amount of heat.

Importance:

Water has a large specific heat capacity. For this reason, it is very useful in storing and carrying thermal energy due to its high specific heat capacity.

Some roles of water due to its large specific heat are given below

Keeping a Moderate Temperature:

The temperature of land rises and falls more rapidly than that of the sea. Hence, the temperature variations from summer to winter are much smaller at places near the sea than land far away from the sea. So climate of the regions near sea shore, like Karachi, remains moderate.

The presence of large water reservoir such as lakes and sees keep the climate of nearby land moderate due to the large heat capacity of these reservoirs.

Cooling System of Automobiles:

The cooling system of the automobiles uses water to carry away unwanted thermal energy. In an automobile, large amount of heat is produced by its engine due to which its temperature goes on increasing. The engine would cease unless it is not cooled down. Water circulating around the engine maintains the temperature. Water absorbs unwanted thermal energy of the engine and dissipates heat through its radiator as shown in the figure:

Water in Central Heating System:

In central heating systems hot water is used to carry thermal energy through pipes from boiler to radiators. Theses radiators are fixed inside the house at suitable places as shown in the figure:

HEAT CAPACITY

Definition:

“Heat capacity of a body is the quantity of thermal energy absorbed by it for one Kelvin (1K) increases in its temperature.”

Mathematical Form:

Thus, if the temperature of a body increases through ∆T on adding ∆Q amount of heat, then its heat capacity will be ∆Q/∆T. putting the value of ∆Q, we get

Heat capacity = =

Heat capacity = mc

The above equation shows that heat capacity of a body is equal to the product of its mass of the body and its specific heat capacity.

Unit:

SI unit of heat capacity is JK^-1

Example:

Heat capacity of 5 kg of water is (5 kg x 4200 Jkg^-1K^-1) 21000 Jkg^-1. That is 5 kg of water needs 21000 joules of heat for every 1 K rise in its temperature. Thus, larger is the quantity of a substance, larger will be its heat capacity.

SPECIFIC HEAT CAPACITY

Definition:

“Specific heat of a substance is the amount of heat that is required to raise the temperature of 1 kg mass of that substance through 1K”.

Formula:

Unit:

SI unit of specific heat capacity is Jkg^-1K^-1.

Examples:

Specific heat capacity of water is 4200 Jkg^-1K^-1 and specific heat capacity of dry soil is 810 Jkg^-1K^-1

HEAT CAPACITY

Definition:

“Heat capacity of a body is the quantity of thermal energy absorbed by it for one Kelvin (1K) increases in its temperature.”

Formula:

Heat capacity = =

Heat capacity = mc

Unit:

SI unit of heat capacity is JK^-1

Example:

Heat capacity of 5 kg of water is (5 kg x 4200 Jkg^-1K^-1) 21000 Jkg^-1. That is 5 kg of water needs 21000 joules of heat for every 1 K rise in its temperature. Thus, larger is the quantity of a substance, larger will be its heat capacity.

DIFFERENTIATION

Specific heat and heat capacity can be differentiated as:

EXAMPLE 8.5

Given Data:

Volume of water = V =2.5 litres

As 1 litre = 1 kg so

Mass of water = m = 2.5 kg

Specific heat of water = c = 4200 Jkg^-1K^-1

Initial temperature = T₁ = 20 ^oC

Finall temperature (As water is boiling) = T₂ = 100 ^oC

To Find:

Heat required to boil the water =

Calculations:

We know

Change in temperature

Putting values

Change in temperature

Now we find Heat required to boil the water

Putting values

Result:

Hence, the heat required to boil the water will be 84000 J or 840 kJ.

STATES OF MATTER

Matter exists in three states:

Solid

Liquid

Gas

Matter can be changed from one state to another. For such a change to occur, thermal energy is added to or removed from a substance as shown in the figure:

Activity:

Take a beaker and place it over a stand. Put small pieces of ice in the beaker and suspend a thermometer in the beaker to measure the temperature of ice.

Now place a burner under the beaker. The ice will start melting. The temperature of the mixture containing ice and water will not increase above 0 ^oC until all the ice melts and we get water at 0^o C If this water at 0^o C is further heated, its temperature will begin to increase above 0^o C as shown in figure.

Explanation of the Graph:

Different parts of the graph can be explained as:

Part AB: On this portion of the curve, the temperature of ice increases from -30^oC to 0^oC.

Part BC: when the temperature of ice reaches 0^oC, the ice water mixture remains at this temperature until all the ice melts.

Reason For Not Changing Temperature:

At this stage temperature does not increase for a while because whole thermal energy is being utilized in breaking intermolecular forces and converting solid state into liquid state.

Part CD: The temperature of the substance gradually increases from 0^oC to 100^oC. The amount of energy so added is used up in increasing the temperature of water.

Part DE: At 100^oC water begins to boil and changes into steam. The temperature remains 100°C until all the water changes into steam.

Reason For Not Changing Temperature:

At this stage again temperature does not increase for a while because whole thermal energy is being utilized in breaking intermolecular forces and converting liquid state into gaseous state.

LATENT HEAT OF FUSION

Introduction:

When a substance is changed from solid to liquid state by adding heat, the process is called melting or fusion. The temperature at which a solid starts melting is called its fusion point or melting point. When the process is reversed i.e. when a liquid is cooled, it changes into solid state. The temperature at which a substance changes from liquid to solid state is called its freezing point. Different substances have different melting points. However, the freezing point of a substance is the same as its melting point.

Definition:

“Heat energy required to change unit mass of a substance from solid to liquid state at its melting point without change in the temperature is called its latent heat of fusion”.

Mathematical Formula:

It is denoted by H_f.

H_f =

Or

∆Q_f = m H_f

Unit:

SI unit of Latent Heat of Fusion is Jkg^-1

Latent Heat of Fusion of Ice:

Ice changes at 0^o C into water. Latent heat of fusion of ice is 3.36×10⁵ Jkg^-1. That is; 3.36×10⁵ joules heat is required to melt 1 kg of ice into water 0^o C.

Experiment:

Take a beaker and place it over a stand. Put small pieces of ice in the beaker and suspend a thermometer in the beaker to measure the temperature. Place a burner under the beaker as shown in the figure:

The ice will start melting. The temperature of the mixture containing ice and water will not increase above 0°C until all the ice melts. Note the time which the ice takes to melt completely into water at 0°C.

Continue heating the water at 0°C in the beaker. Its temperature will begin to increase. Note the time which the water in the beaker takes to reach its boiling point at 100°C from 0°C. Draw a temperature-time graph such as shown in figure

Calculations:

Calculate the latent heat of fusion of ice from the data as follows:

Let the mass of ice =m

Finding the time from the graph

Time taken by ice to melt completely at 0^o C = t_f =t₂ – t₁ =3.6 minutes

Time taken by water to heat from 0^oC to 100^o C = t_o = t₃ – t₂ = 4.6 minutes

Specific heat of water c = 4200 JKg^-1 K^-1

Increase in the temperature of water ∆T = 100^o C = 100K

Heat required by water from 0^o C to 100^o C = ∆Q = ?

As we know that

∆Q = m c ∆T

= m x 4200 x 100

= m x 42000

= m x 4.2 x 10³ x 10²

= 4.2 x 10⁵ x m JKg^-1

Heat ∆Q is supplies to water in time t_o to raise the temperature of the water from 0^o C to 100^o C, Hence, the rate of absorbing heat by water in beaker can be given by

Rate of absorbing heat

= ∆Q/t_o

As we know that

∆Q_f = m x H_f

Putting the values of t_f and t_o which can be found though graph

Conclusion:

The latent heat of fusion of ice (H_f) found for above experiment is 3.29×10⁵ JKg^-1 however actual value is 3.36×10⁵ JKg^-1.

LATENT HEAT OF VAPORIZATION

Introduction:

When heat is given to a liquid at its boiling point, its temperature remains constant. The heat energy given to liquid at its boiling point is used up in changing its state from liquid to gas without any increase in its temperature.

Definition:

“The quantity of heat that changes unit mass of a liquid completely into gas at its boiling point without any change in its temperature is called its latent heat of vaporization”.

Mathematical Form:

It is denoted by H_v

H_v =

OR

Δ Q_v = m H_v

Unit:

SI unit of Latent Heat of vaporization is Jkg^-1

Latent Heat of Vaporization of Water:

When water is heated, it boils at 100 ^oC under standard pressure. Its temperature remains 100^oC until it is changed into steam. Its latent heat of vaporization is 2.26 x 10⁶ Jkg^-1. That is; one kilogram of water requires 2.26 x 10⁶ joule heat to change it completely into gas (steam) at its boiling point.

Experiment:

Take a beaker and place it over a stand. Put small pieces of ice in the beaker and suspend a thermometer in the beaker to measure the temperature. Place a burner under the beaker as shown in the figure:

The ice will start melting and will convert into water. Continue heating water till all the water changes into steam. Note the time which the water in the beaker takes to change completely into steam at its boiling point 100°C.

Extend the temperature time graph such as shown in the figure:

Calculations:

Calculate the latent heat of vaporization of boiled water with the data as given:

Let

The mass of ice = m

Measuring the time from the graph

Time taken by water to heat from 0^o C to 100^o C = t_o = t₃ – t₂ = 4.6 minutes

Time taken by water 100^o C to get changed into steam = t_v = t₄ – t₃ = 24.4 minutes

Specific heat of water c = 4200 JKg^-1 K^-1

Increase in the temperature of water ∆T = 100^o C

Heat required to heat water from 0^o C to 100^o C =∆Q = m c ∆T

= m x 4200 x 100

= m x 4.2 x 10³ x 10²

= 4.2 x 10⁵ x m JKg^-1

To raise the temperature of the water from 0^o C to 100^o C, ∆Q is given to water. So the heat absorption rate of water in beaker can be given by

Rate of absorbing heat = ∆Q/t_o

Since heat absorption in time tv

As we know that ∆Q_v = m x H_v

Putting the values of t_v and t_o which can be found though graph

H_v = 2.23 x 10⁶ Jkg^-1

Results:

The latent heat of vaporization of boiled water (H_v) found for above experiment is 2.23 x 10⁶ JKg^-1 however actual value is 2.26 x 10⁶ JKg^-1.

LATENT HEAT OF FUSION

Definition:

“Heat energy required to change unit mass of a substance from solid to liquid state at its melting point without change in the temperature is called its latent heat of fusion”.

Mathematical Formula:

It is denoted by H_f.

H_f =

Or

∆Q_f = m H_f

Unit:

SI unit of Latent Heat of Fusion is Jkg^-1

Latent Heat of Fusion of Ice:

Ice changes at 0^o C into water. Latent heat of fusion of ice is 3.36 x 10⁵ Jkg^-1. That is; 3.36 x 10⁵ joules heat is required to melt 1 kg of ice into water 0^o C.

LATENT HEAT OF VAPORIZATION

Definition:

“The quantity of heat that changes unit mass of a liquid completely into gas at its boiling point without any change in its temperature is called its latent heat of vaporization”.

Mathematical Form:

It is denoted by H_v^.

H_v =

Or

Δ Q_v = m H_v

Unit:

SI unit of Latent Heat of vaporization is Jkg^-1

Latent Heat of Vaporization of Water:

FUSION POINT

“When a substance is changed from solid to liquid state by adding heat, the process is called melting or fusion. The temperature at which a solid starts melting is called its fusion point or melting point.”

Examples:

Melting point of ice is 0 ^oC

Melting point of Mercury is -39 ^oC

BOILING POINT

“When a substance is changed from liquid to gaseous state by adding heat, the process is called boiling. The temperature at which a liquid starts to convert into gas is called its boiling point.

Examples:

Boiling point of water is 100°C

Boiling point of Mercury is 357°C

FREEZING POINT

“When a liquid is cooled, it changes into solid state. The temperature at which a substance changes from liquid to solid state is called its freezing point. Different substances have different melting points. However, the freezing point of a substance is the same as its melting point.”

Examples:

Freezing point of water is 0^oC

Freezing point of Mercury is -39^oC

DIFFERENTIATION

Specific heat and latent heat of a material can be differentiated as:

TEMPERATURE AT CHANGING STATE

TABULATION

Table for above mentioned quantities of common substances is given below:

EVAPORATION

Definition:

“Evaporation is the changing of a liquid into vapors (gaseous state) from the surface of the liquid without heating it”.

Activity:

Take some water in a dish. The water in the dish will disappear after some time. It is because the molecules of water are in constant motion and possesses kinetic energy. Fast moving molecules escape out from the surface of water and goes into atmosphere this process is called evaporation as shown in the figure:

Comparison of Boiling and Evaporation:

Unlike boiling, evaporation takes place at all temperatures but only from the surface of a liquid. The process of boiling takes place at a certain fixed temperature which is the boiling point of that liquid. At boiling point, a liquid is changing into vapors not only from the surface but also within the liquid. These vapors are comes out of the boiling liquid as bubbles which breakdown on reaching the surface.

Evaporation Cause Cooling:

During evaporation fast moving molecules escape out from the surface of the liquid. Molecules that have lower kinetic energies are left behind. This lowers the average kinetic energy of the liquid molecules and the temperature of the liquid. Since temperature of a substance depends on the average kinetic energy of its molecules. Evaporation of perspiration helps to cool our bodies.

In refrigerator evaporation of liquefied gas produces cooling.

Factors:

Evaporation takes place at all temperatures from the surface of a liquid. The rate of evaporation is affected by various factors.

Temperature:

Wet clothes dry up more quickly in summer than in winter because at higher temperature, more molecules of a liquid are moving with high velocities. Thus, more molecules escape from its surface. Thus, evaporation is faster at high temperature that at low temperature.

Surface Area:

Large is the surface area of a liquid, greater number of molecules has the chance to escape from its surface that is why we spread wet clothes to increase their surface area and to increase rate of evaporation.

Wind:

Wind blowing over the surface of a liquid sweeps the liquid molecules that have just escaped out. This increases the chance for more liquid molecules to escape out.

Nature of the Liquid:

Evaporation depends on the nature of the liquid. The molecules having weaker intermolecular forces evaporate more quickly as compared to others, if we take spirit and water on our palm. As evaporation rate of spirit is greater than water, so we feel cooling effect due to evaporation of spirit.

DIFFERENTIATION

Evaporation and Boiling can be differentiated as:

DIFFERENTIATION

Evaporation and vaporization can be differentiated as:

COOLING EFFECT OF EVEPORATION

During evaporation fast moving molecules escape out from the surface of the liquid. Molecules that have lower kinetic energies are left behind. This lowers the average kinetic energy of the liquid molecules and the temperature of the liquid. Since temperature of a substance depends on the average kinetic energy of its molecules. Evaporation of perspiration helps to cool our bodies.

WET CLOTH

Wet clothes dry up more quickly in summer than in winter because at higher temperature, more molecules of a liquid are moving with high velocities. Thus, more molecules escape from its surface. Thus, evaporation is faster at high temperature that at low temperature.

LARGE AREA

Large is the surface area of a liquid, greater number of molecules has the chance to escape from its surface that is why we spread wet clothes to increase their surface area and to increase rate of evaporation.

RATE OF EVAPORATION

No, spirit and water does not evaporate at the same rate because evaporation depends on the nature of the liquid. The molecules having weaker intermolecular forces evaporate more quickly as compared to others, spirit has weaker intermolecular forces than water so spirit evaporates more quickly than water.

COOLING EFFECT

If we take ether or spirit on our palm. Ether or spirit having weaker intermolecular evaporates more quickly so we feel cooling effect due to evaporation of spirit or ether.

Ans. USES OF COOLING EFFECT

Following are uses of cooling effect produced by evaporation

Evaporation of perspiration helps to cool our body.

In refrigerator evaporation of liquefied gas produces cooling.

THERMAL EXPANSION

Definition:

“Increase in the length or volume of a substance due to heat is called thermal expansion.”

Most of the substances solids, liquids and gases expand on heating and contract on cooling.

Their thermal expansion and contractions are usually small and are not noticeable. However these expansions and contractions are important in our daily life.

Explanation on the Basis of Kinetic Molecular Theory:

The kinetic energy of the molecules of an object depends on its temperature. The molecules of a solid vibrate with large amplitude at high temperature than at low temperature. Thus, on heating, the amplitude of vibration of the atoms or molecules of an object increases. They push one another farther away as the amplitude of vibration increases as shown in the figure:

Thermal expansion results an increase in length, breadth and thickness of a substance.

LINEAR THERMAL EXPANSION

Definition:

“If a thin rod is heated, there is a prominent increase in its length as compared to its cross-sectional area. The expansion along length or in one dimension is called linear expansion”.

Dependence:

If we heat a metal rod the length of which is much larger than its thickness, then the increase in length depends on the following three factors:

Length of thin rod.

Change in temperature.

Nature of material of the rod.

Explanation:

Solids expand on heating and their expansion is nearly uniform over a wide range of temperature. Consider a metal rod of length L_o at certain temperature T_o. Let its length on heating to a temperature T becomes L as shown in the figure:

Thus

Increase in length of the rod = ∆L = L – L_o

Increase in temperature = ∆T = T – T_o

It is found that change in length ∆L of a solid is directly proportional to its original length L_o and the change in temperature ∆T. that is;

∆L  L_o ∆T

or ∆L =  L_o ∆T

or L – L_o =  L_o ∆T

L = L_o (1 + α ∆T)

Coefficient of Linear Expansion:

We know

∆L =  L_o ∆T

Where α is the proportionality constant and it is called co-efficient of linear expansion of the substance it can be defined as:

α =

Thus we can define coefficient of linear expansion α of a substance as “The fractional increase in its length per Kelvin rise in temperature.”

Unit:

Its unit is Per Kelvin (K^-1)

Value:

Its value depends on the nature of the material of the rod and

Relationship Between β and α:

Relationship Between β and α is given below:

β = 3α

Examples:

Some values for α and β are given:

VOLUME THERMAL EXPANSION

Definition:

“The volume of a solid also changes with the change in temperature and is called volume thermal expansion or cubical thermal expansion”.

Dependence:

If we heat a block then increase in volume of the block depends on the following three factors:

Original volume of block.

Change in temperature.

Nature of material of the block.

Explanation:

Consider a solid of initial volume V_o at certain temperature T_o. On heating the solid to a temperature T, let its volume becomes V as shown in the figure:

Then

Change in volume of a solid = ∆V = V – V_o

And Change in temperature = ∆T = T – T_o

Like linear expansion, the change in volume ∆V is found to be proportional to its original volume V_o and change in temperature ∆T. Thus

∆V  V_o

And ∆V  ∆T

∆V  V_o ∆T

∆V = βV_o ∆T

V – V_o = βV_o ∆T

V = V_o + βV_o ∆T

V = V_o (1 + β ∆T)

Coefficient of Volume Expansion:

We know

∆V = βV_o ∆T

Where β is the proportionality constant and is called the co-efficient of volume expansion. And it can be defined as

β =

“The fractional change in its volume per Kelvin change in temperature”.

Unit:

Its unit is Per Kelvin (K^-1)

Value:

Its value depends on the nature of the material of the rod and

Relationship Between β and α

The coefficients of linear and volume expansion are related by the following equation

β = 3α

Examples:

Some values for α and β are given:

CONSEQUENCES OF THERMAL EXPANSION

The expansions of solids many damage bridges, railway tracks and roads as they are constantly subjected to temperature changes.

Prevision is made during construction for expansion and contraction with temperature.

The expansion of solids may damage the bridges, railway tracks and roads as they are constantly subjected to temperature changes. So provision is made during construction for expansion and contraction with temperature. For example, railway tracks buckled on a hot summer day due to expansion if gaps are not left between sections as shown in the figure:

Bridges made of steel girders also expands during the day and contract during night. They will bend if their ends are fixed. To allow thermal expansion, one end is fixed while the other one of the girder rests on rollers in the gap left for expansion. Overhead transmission lines are also given a certain amount of sag so that they contract in winter without snapping as shown in the figure:

APPLICATIONS OF THERMAL EXPANSION

Thermal expansion is used in our daily life. In thermometers, thermal expansion is used in temperature measurements.

To open the cap of a bottle that is tight enough, immerse it in hot water for a minute or so. Metal cap expands and becomes loose. It would now be easy to turn it to open.

To join steel plates tightly together, red hot rivets are forced through holes in the plates as shown in figure. The end of hot rivet is then hammered. On cooling, the rivets contracts and bring the plates tightly griped.

Iron rims are fixed on wooden wheels of carts. Iron rims are heated. Thermal expansion allows them to slip over the wooden wheel. Water is poured on it to cool. The rim contracts and becomes tight over the wheel as shown in the figure:

Wires on electric poles are given some sag to prevent breaking in winter.

Thermal expansion concept is applied in Bimetal strip.

BIMETAL STRIP

A bimetal strip consists of two thin strips of different metals such as brass and iron joined together as shown in figure.

On heating the strip, brass expands more than iron. This unequal expansion causes bending of the strip as shown in figure:

Usage:

Bimetal strips are used for various purposes.

Bimetal thermometers are used to measure temperature especially in furnaces and ovens.

Bimetal strips are also used in thermo states.

Bimetal thermo state switch is used to control the temperature of heater coil in an electric iron.

THERMAL EXPANSION IN LIQUIDS

The molecules of liquids are free to move in all directions within the liquid. On heating a liquid, the average amplitude of vibration of its molecules increases. The molecules push each other and need more space to occupy. This accounts for the expansion of the liquid when heated. The thermal expansion in liquids is greater than solids due to the weak forces between their molecules. Therefore, the coefficient of volume expansion of liquids is greater than solids.

No Definite Shape of Liquids:

Liquids have no definite shape of their own. A liquid always attains shape of the container in which it is poured. Therefore, when a liquid is heated, both liquid and the container undergo a change in their volume.

Types of Thermal Expansion For Liquids:

There are two types of thermal expansion for liquids:

Real volume expansion

Apparent volume expansion

Activity:

Take a long-necked flask. Fill it with some colored liquid up to mark A on its neck as shown in figure.

Now start heating the flask from bottom. The liquid level first falls to B and then rises to C.

Relation between expansions:

We observe that there are two types of expansions appear as a result of heating a liquid in any container.

Real volume expansion

Apparent volume expansion

The heat first reaches the flask which expands and its volume increases. As a result liquid descends in the flask and its level falls to B. After sometime, the liquid begins to rise above B on getting hot. At certain temperature it reaches at C. The rise in level from A to C is due to the apparent expansion in the volume of the liquid. Actual expansion of the liquid is greater than that due to the expansion because of the expansion of the glass flask. Thus real expansion of the liquid is equal to the volume difference between A and C in addition to the volume expansion of the flask. Hence

Real expansion of liquid = Apparent expansion of liquid + Expansion of the flask

BC = AC + AB

The expansion of the volume of a liquid taking into consideration the expansion of the container also, is called the real expansion of the liquid.

Coefficients of volume expansions:

The real rate of volume expansion β_r of a liquid is defined as the actual change in unit volume of a liquid for 1K (or 1 ^oC) rise in its temperature. The real rate of volume expansion β_r is always greater than the rate of volume expansion β_a by an amount equal to the rate of volume expansion of the container β_g.

Thus β_r = β_a + β_g

It should be noted that different liquids have different coefficients of volume expansion.

Coefficients of liquid expansion:

In accordance with the apparent and real expansions of the liquids, their co-efficient of expansion are also measured in two ways:

Coefficient of apparent expansion

Coefficient of real expansion

It should be noted that different liquids have different coefficients of volume expansion

LINEAR THERMAL EXPANSION

Definition:

“If a thin rod is heated, there is a prominent increase in its length as compared to its cross-sectional area. The expansion along length or in one dimension is called linear expansion”.

Dependence:

If we heat a metal rod the length of which is much larger than its thickness, then the increase in length depends on the following three factors:

Length of thin rod.

Change in temperature.

Nature of material of the rod.

Formula:

∆L =  L_o ∆T

COEFFICIENT OF LINEAR THERMAL EXPANSION

Definition:

We know

∆L =  L_o ∆T

Where α is the proportionality constant and it is called co-efficient of linear expansion of the substance it can be defined as:

α =

“The fractional increase in its length per Kelvin rise in temperature.”

Unit:

Its unit is Per Kelvin (K^-1)

Value:

Its value depends on the nature of the material of the rod and

Relationship Between β and α

β = 3α

VOLUME THERMAL EXPANSION

Definition:

“Heating a block causes an increase in length, breadth and thickness, i.e., volume of the block increases that is known as volume expansion. It is also called cubic thermal expansion”.

Dependence:

If we heat a block then increase in volume of the block depends on the following three factors:

Original volume of block.

Change in temperature

Nature of material of the block.

Formula:

∆V = βV_o ∆T

COEFFICIENT OF VOLUME THERMAL EXPANSION

Definition:

We know

∆V = βV_o ∆T

Where β is the proportionality constant and is called the co-efficient of volume expansion. And it can be defined as

β =

“The fractional change in its volume per Kelvin change in temperature”.

Unit:

Its unit is Per Kelvin (K^-1)

Value:

Its value depends on the nature of the material of the rod and

Relationship Between β and α:

Relationship Between β and α is below:

β = 3α

VALUES OF COEFFICIENT OF THERMAL EXPANSION

Following vales have been tabulated by applying relation

GAPS IN RAILWAY TRACKS

Gaps are left in railway tracks to compensate thermal expansion during hot season. Railway tracks buckled on a hot summer day due to expansion if gaps are not left between sections

ANOMALOUS EXPANSION

Water on cooling below 4^oC begins to expand until 0^oC. On further cooling its volume increases suddenly as it changes into ice at 0^oC. When ice is cooled below 0^oC, it contracts i.e. its volume decreases like solids. This unusual expansion of water is called the anomalous expansion of water.

EXAMPLE 8.6

Given Data:

Initial length of the brass rod = L_o = 1m

Initial temperature of the brass rod = T_o = 0^oC = 0+273 = 273 K

Final temperature of the brass rod = T = 30^oC = 30 +273 = 303 K

Change in temperature

Change in temperature

Coefficient of linear expansion of brass =

To Find:

Final length of the brass rod = L =?

Calculations:

We know

Putting values

L = 1.00057 m

Result:

Hence, the length of the brass bar at 30°C will be 1.00057 m.

EXAMPLE 8.7

Given Data:

Initial length of the a side of brass = L_o = 0.1m

Initial volume of the brass cube = V_o = (L_o)³ = (0.1m)³ = 0.001m³ =10^-3m³

Initial temperature of the brass cube= T_o = 0 ^oC = 0+273 = 273 K

Final temperature of the brass cube = T = 100 ^oC = 100 +273 = 373 K

Change in temperature

Change in temperature

Coefficient of linear expansion of brass =

To Find:

Final volume of the brass cube = L =?

Calculations:

We know

Coefficient of linear expansion of brass =

We can find as

Putting values

Putting values

L = 1.0057 × 10^–5 m³

Result:

Hence, the volume of brass cube at 100^oC will be 1.0057×10^-3m³.

HEAT

Definition:

“Heat is the energy that is transferred from one body to the other in thermal contact with each other as a result of the difference of temperature”.

Quantity:

Heat is a derived and scalar quantity.

Unit:

SI unit of heat is Joule (J)

TEMPERATURE

Definition:

“Degree of coldness or hotness of the body is a measure of its temperature”

Quantity:

Temperature is a base and scalar quantity.

Unit:

SI unit of temperature is kelvin (K)

THERMOMETER

Definition:

“The instrument which is used to measure the temperature is called a thermometer.”

Mercury The Best Thermometric Material:

Mercury is preferred as thermometric substance because

It has high boiling point i.e. 357°C

It has low melting point i.e. -39°C

It does not wet glass

It is good conductor

It is opaque

It has low heat capacity

See Q. 3 Long Question TOPIC 8.8

See Q. 1 ‘Long Question TOPIC 8.3

See Q. 2 Long Question TOPIC 8.5

See Q. 3 Long Question TOPIC 8.6

See Q. 1 Long Question TOPIC 8.7

FLOW OF HEAT

Molecules of hot body have greater kinetic energy than the molecules of cold body. Therefore, fast moving molecules give their energy to cold body. So we can say that heat flows from hot body to the cold body. In other word we can say heat flows from hot body to cold body due to temperature difference between them.

INTERNAL ENERGY

Definition:

“The sum of kinetic energy and potential energy associated with the atoms, molecules and particles of a body is called its internal energy”.

Dependence:

Internal energy of a body depends on many factors such as the mass of the body, kinetic and potential energies of molecules etc. Kinetic energy of an atom or molecule is due to its motion which depends upon the temperature. Potential energy of atoms or molecules is the stored energy due to intermolecular forces.

EFEECT OF HEAT ON GAS

The kinetic energy of gas molecules goes on increasing if a gas is heated continuously. This causes the gas molecules move faster and faster. The collisions between atoms and molecules of the gas become so strong that they tear off the atoms. Atoms lose their electrons and become positive ions. This ionic state of matter is called plasma.

In short on heating the gas, the motion of the molecules becomes faster. As a result average K.E and temperature of gas increases.

Given Data:

Temperature in Celsius = T_c = 50⁰C

To Find:

Temperature in Fahrenheit = T_f = ?

Calculations:

Given Data:

Normal human Temperature in Fahrenheit = T_f = 98.6⁰ F

To Find:

Temperature in Celsius = T_c =?

Temperature in Kelvin = T_k =?

Calculations:

As we know that

C =

By putting the values, we have

C =

C =

C = 37 ^oC

As we know that

T_K = C + 273

By putting the values, we have

T_K = 37 + 273 = 310 K

T_K = 310 K

Result:

Hence, the Temperature of normal human body in Celsius will be 37 ^oC and the temperature of normal human body in Kelvin will be 310 K.

Given Data:

Length of aluminum bar = L₁ = 2 m

Initial temperature = T₁ = 0^oC = (0 + 273) K = 273 K

Final temperature = T₂ = 20^oC = (20 + 273) K = 293 K

Coefficient of linear expansion of aluminum = α = 2.5 x 10^-5 K^-1

To Find:

Increase in length = L – L_o =?

Calculations:

As we know that

L – L_o = α L_o (T₂ – T₁)

By putting the values, we have

L – L_o = 2.5 x 10^-5 x 2 x (293 – 273)

L – L_o = 5 x 10^-5 (20)

L – L_o = 100 x 10^-5

L – L_o = 1 x 10^-3 m = 0.1 cm = 1 mm

Result:

Hence, the increase in length of Aluminum bar will be 1×10^-3 m = 0.1 cm = 1 mm.

Given Data:

Initial volume of air in balloon = V₁ = 1.2 m³

Initial temperature = T₁ = 15⁰ C = (15 + 273) K = 288 K

Final temperature = T₂ = 40⁰ C = (40 + 273) K = 313 K

Coefficient of volume expansion = β = 3.67 x 10^-3 K^-1

To Find:

Final volume of gas = V₂ = ?

Calculations:

As we know that

V = V_o (1 + β(T₂ – T₁))

By putting the values, we have

V = 1.2 (1 + 3.67 x 10^-3 x (313 – 288))

V = 1.2 (1 + 3.67 x 10^-3 (25))

V = 1.2 (1 + 91.75 x 10^-3)

V = 1.2 (1 + 0.091)

V = 1.2 + 0.108 = 1.308 = 1.3 m³

Result:

Hence, the final volume of gas will be 1.3 m³.

Given Data:

Mass of water = m = 0.5 kg

Initial temperature = T₁ = 10⁰ C

Final temperature = T₂ = 65⁰ C

Change in Temperature:

To Find:

Heat required = Q =?

Calculations:

As we know that

ΔQ = mcΔT

By putting the values, we have

ΔQ = 0.5 x 4200 x 55

ΔQ = 115500 J

Result:

Hence, the heat required will be 115500 J.

Given Data:

Rate of heat supplied by heat = P = 1000 Js^-1

Mass of water = m = 200 g = 0.2 kg

Specific heat of water = c = 4200J

Initial temperature = T₁ = 20⁰ C

Final temperature = T₂ = 90⁰ C

Change in temperature = ΔT = 90 – 20 = 70^o C = 70K

To Find:

Heat required = Q =?

Time = t =?

Calculations:

As we know that

Q = cm ΔT

Q = 0.2 x 4200 x 70

Q = 58800 J

As we also know that

P×t = Q

t = Q/P

t = 588000/1000

t = 58.8 s

Result:

Hence, the heat required will be 58800 J and the time taken will be 58.8 s.

Given Data:

Heat supplied to ice = ΔQ_f = 50000 J

Latent heat of fusion of ice = H_f = 336000 Jkg^-1

To Find:

Mass of ice = m = ?

Calculations:

As we know that

ΔQ = m x H_f

So m =

By putting the values, we have

m =

m = 0.15 kg = 150 g

Result:

Hence, the mass of ice will be 150 g.

Given Data:

Mass of ice = m = 100 g = 0.1 kg

Specific heat of ice = 2100 JKg^-1K^-1

Specific heat of water = 4200 JKg^-1K^-1

Latent heat of fusion of ice = 336000 JKg^-1K^-1

Initial temperature of ice = T₁ = -10⁰ C

Final temperature = T₂ = 10⁰ C

To Find:

Heat required to raise the temperature of ice from –10°C to 10°C = Q = ?

Calculations:

Step-I:

Heat required to raise the temperature of ice from –10°C to 0°C = Q₁ = ?

T₁ = –10°C

T₂ = 0°C

T = 0°C–(–10)°C = 10°C = 10 K

Q = cmT

Q₁ = 2100 ×0.1×10

Q₁ = 2100 J

Step-II:

Heat required to convert ice at 0°C into water at 0°C = Q₂ =?

We know that

Q = mL_f

Q₂= 0.1 ×336000

Q₂ = 33600 J

Step-III:

Heat required to raise temperature water from 0°C to 10°C = Q₃ =?

T1 = 0°C

T2 = 10°C

T = 10°C –0°C = 10°C = 10K

We know that

Q = cmT

Q₃ = 4200 ×0.1×10

Q₃ = 4200 J

Total heat required = Q = Q₁+Q₂+Q₃

Q = 2100+33600+4200

Q = 39900 J

Result:

Hence, the total heat required will be 39900 J.

Given Data:

Mass of water = m = 100 g = 0.1 kg

Temperature of water = T₁ = 100⁰ C

Temperature of steam = T₂ = 100⁰ C

Latent heat of vaporization of water = H_v = 2.26 x 10⁶ Jkg^-1

To Find:

Heat required to change water into steam = Q_v =?

Given Data:

Mass of water = m₁ = 500 g = 0.5 kg

Mass of steam = m₂ = 5 g = 0.005 kg

Temperature of water = T₁ = 10⁰ C

Temperature of steam = T₂ = 100⁰ C

Specific heat of water = c = 4200 Jkg^-1K^-1

Latent heat of vaporization of vaporization = H_v = 2.26×106 Jkg^-1

To Find:

Final temperature of water = T =?

Calculations:

According to law of heat exchange

Heat lost by steam = Heat gain by water

mH_v + cmT = cmT

(0.005)(2.26×10⁶) + (4200)(0.005)(100–T) = (4200)(0.5) (T–10)

11300+21(100–T) = 2100(T–10)

11300+2100–21T = 2100T–21000

11300+2100+21000 = 2100T+21T

344400 = 2121T

T =

T = 16.2°C

Result:

Hence, the final temperature of water will be 16.2^oC.