Physics NEET: Properties of Matter

Course Content

Total learning: 3 lessons / 4 quizzes

Thermal Properties of Matter


We all have common-sense notions of heat and temperature. Temperature is a measure of ‘hotness’ of a body. A kettle with boiling water is hotter than a box containing ice. In physics, we need to define the notion of heat, temperature, etc., more carefully. In this chapter, you will learn what heat is and how it is measured, and study the various processes by which heat flows from one body to another. Along the way, you will find out why blacksmiths heat the iron ring before fitting on the rim of a wooden wheel of a bullock cart and why the wind at the beach often reverses direction after the sun goes down. You will also learn what happens when water boils or freezes, and its temperature does not change during these processes even though a great deal of heat is flowing in to or freezes, and its temperature does not change during these processes even though a great deal of heat is flowing into or out of it.

Temperature and Heat

We can begin studying thermal properties of matter with definitions of temperature and heat. Temperature is a relative measure, or indication of hotness or coldness. A hot utensil is said to have a high temperature, and ice cube to have low temperature. An object that has a high temperature than another object is said to be hotter. Note that hot and cold are relative terms, like tall and short. We can perceive temperature by touch. However, this temperature sense is somewhat unreliable and its range is too limited to be useful for scientific purposes.

We know from experience that a glass of ice cold water left on a table on a hot summer day eventually warms up whereas a cup of hot tea on the same table cools down. It means that when the temperature of body, ice cold water or hot in this case, and its surrounding medium are different, heat transfer takes place between between the system and the surrounding medium are at the same temperature. We also know that in the case of glass tumbler of ice cold water, heat flows from the environment to the glass tumbler, whereas in the case of hot tea, it flows from the cup of hot tea to the environment. So, we can say that heat is the form of energy transferred between two (or more) systems or a system and its surroundings by virtue of temperature difference. The SI unit of heat energy transferred is exposed in joule (J) while SI unit of temperature is kelvin (K), and C is a commonly used unit of temperature. When an object is heated, many changes may take place. Its temperature may rise, it may expand or change state. We will study the effect of heat on different bodies in later sections.

Measurement of Temperature

A measure of temperature is obtained using a thermometer. Many physical properties of materials change sufficiently with temperature to be used as the basis for constructing thermometers. The commonly used property is variation of the volume of a liquid with temperature. For example, a common thermometer (liquid-in-glass type) with which you are familiar. Mercury and alcohol are the liquids used in most liquid-in-glass thermometers.

Thermometers are calibrated so that a numerical value may be assigned to a given temperature. For the definition of any standard scale, two fixed reference points are needed. Since all substances change dimensions with temperature, an absolute reference for expansion is not available. However, the necessary fixed points may be correlated to physical phenomena that always occur at the same temperature. The ice point and the steam point of water are two convenient fixed points and are known as the freezing and boiling points. These two points are the temperatures at which pure water freezes and boils under standard pressure. The two familiar temperature scales are the Fahrenheit temperature scale and the Celcius scale. The ice and steam point have values 32°F and 212°F respectively, on the Fahrenheit scale and 0°C and 100°C on the Celcius scale. On the Fahrenheit scale, there are 180 equal intervals between two reference points, and on the Celcius scale, there are 100.

Fig. 1 –
A plot of Fahrenheit temperature (tF
) versus Celsius temperature (tc).

A relationship for converting between the two scales may bee obtained from a graph of Fahrenheit temperature (tF) versus Celsius temperature (tc) in a straight line (Fig.1) whose equation is

(tF – 32)/180 = tc/100 ………………………………………………………..(1)

Ideal Gas Equation and Absolute Temperature

Liquid-in-glass thermometers show different readings for temperatures other than the fixed points because of differing expansion properties. A thermometer that uses a gas, however, gives the same readings regardless of which gas is used. Experiments show that all gases at low densities exhibit same expansion behaviour. The variables that describe the behaviour of a given quantity(mass) of gas are pressure, volume, and temperature (P, V and T) (where T = t + 273.15; t is the temperature in ºC). When temperature is held constant, the pressure and volume of a quantity of gas are related as PV = constant. This relationship is known as Boyle’s law, after Robert Boyle (1627-1691) the English Chemist who discovered it. When the pressure is held constant, the volume of a quantity of the gas is related to the temperature as V/T = constant. This relationship is known as Charle’s law, after the French scientist Jacques Charles (1747-1823). Low density gases obey these laws. which may be combined into single relationship.

Fig. 2 – Pressure versus temperature of a low density gas kept at constant volume.

Notice that since PV = constant and V/T = constant for a given quantity of gas, then PV/T should also be a constant. This relationship is known as ideal gas law. It can be written in a more general form that applies not just to a given quantity of a single gas but to any quantity of any dilute gas and is known as ideal-gas equation.

PV/T = µR

or PV = µRT ………………………………………………………….(2)

where, µ is the number of moles in the sample of gas and R is called universal gas constant:

R = 8.31 J mol-1 K-1

In equation (2), we have learnt that the pressure and volume are directly proportional to temperature: PV ∝ T. This relationship allows a gas to be used to measure temperature in a constant volume gas thermometer. Holding the volume of a gas constant, it gives P∝ T. Thus, with a constant volume gas thermometer, temperature is read in terms of pressure. A plot of pressure versus temperature gives a straight line in this case, as shown in Fig.2.

However, measurements on real gases deviate from the values predicted by the ideal gas law at low temperature. But the relationship is linear over a large temperature range, and it looks as though the pressure might reach zero with decreasing temperature if the gas continued to be a gas. The absolute minimum temperature for an ideal gas, therefore, inferred by extrapolating the straight line to the axis, as in Fig. 3. This temperature is found to be -273.15°C and is designated as absolute zero. Absolute zero is the foundation of the Kelvin temperature scale or absolute scale temperature named after the British scientist Lord Kelvin. On this scale, -273.15°C is taken as the zero point, that is 0°K (Fig. 4)

Fig. 3 –
A plot of pressure versus temperature and extrapolation of lines for low density gases indicates the same absolute zero temperature.

Fig. 4 – Comparison of the Kelvin, Celsius and Fahrenheit temperature scales.

The size of the unit for Kelvin temperature is the same Celcius degree, so temperature on these scales are related by

T = tc + 273.15 …………………………………………………(3)

Thermal Expansion

You may have observed that sometimes sealed bottles with metallic lids are so tightly screwed that one has to put the lid in hot water for some time to open the lid. This would allow the metallic cover to expand, thereby loosening it unscrew easily. In case of liquids, you may have observed that mercury in thermometer rises, when the thermometer is put in a slightly warm water. If we take out the thermometer is put in a slightly warm water. If we take out the thermometer from the warm water the level of mercury falls again. Similarly, in the case of gases, a balloon the air inside.

It is our common experience that most substances expand on heating and contract on cooling. A change in temperature of a body causes change in its dimensions. The increase in the dimensions of a body due to the increase in its temperature is called thermal expansion. The expansion in length is called linear expansion. The expansion in area is called area expansion. The expansion in volume is called volume expansion (Fig. 5).

Fig. 5 – Thermal Expansion

If the substance is in the form of a long rod, then for small change in temperature ΔT, the fractional change in length, Δl/l, is directly proportional to ΔT.

Δl/l = α1ΔT ……………………………………………………………..(4)

where α1 is known as the coefficient of linear expansion and is characteristic of the material of the rod. In Table 1 are given typical average values of the coefficient of linear expansion for some materials in the temperature range 0ºC to 100ºC. From this Table, compare the value of α1 for glass and copper. We find that copper expands about five times more than glass for the same rise in temperature. Normally, metals expand more and have relatively high values of α1.

Table 1 – Values of coefficient of linear expansion of for some materials

Materials α1 ( 10-5K-1)
Aluminium 2.5
Brass 1.8
Iron 1.2
Copper 1.7
Silver 1.9
Gold 1.4
Glass (pyrex) 0.32
Lead 0.29

Similarly, we consider the fractional change in volume, ΔV/V, of a substance for temperature change ΔT and define the coefficient of volume expansion, αv as

αv = (ΔV/V)(1/ΔT) ………………………………………………………………………………………………(5)

Here αv is also a characteristic of the substance but is not strictly a constant. It depends in general on temperature (Fig. 6). It is seen that αv becomes constant only at a high temperature.

Fig.6 – Coefficient of volume expansion of copper as a function of temperature

Table 2 gives the values of co-efficient of volume expansion of some common substances in the temperature range 0-100ºC. You can see that thermal expansion of these substances (solids and liquids) is rather small, with materials like pyrex glass and invar (a special iron-nickel alloy) having particularly low values of αv. From this Table we find that the value of αv for alcohol (ethyl) is more than mercury and expands more than mercury for the same rise in temperature.

Table 2 – Values of coefficient of volume expansion for some substances

Materials αv ( K-1)
Aluminium 7 × 10-5
Brass 6 × 10-5
Iron 3.55 × 10-5
Paraffin 58.8 × 10-5
Glass (ordinary) 2.5 × 10-5
Glass (pyrex) 1 × 10-5
Hard rubber 2.4 × 10-4
Invar 2 × 10-6
Mercury 2.4 × 10-4
Invar 2 × 10-6
Water 20.7 × 10-5
Alcohol (ethyl) 110 × 10-5

Water exhibits an anomalous behaviour; it contracts on heating between 0ºC and 4ºC. The volume of a given amount of water decreases as it is cooled from room temperature, until its temperature reaches 4ºC, [Fig. 7(a)]. Below 4ºC, the volume increases, and therefore the density decreases [Fig. 7(b)].

This means that water has a maximum density at 4ºC. This property has an important environmental effect; Bodies of water, such as lakes and ponds, freeze at the top first. As a lake cools toward 4ºC, water near the surface loses energy to the atmosphere, becomes denser, and sinks; the warmer, less dense water near the bottom rises. However, once the colder water on top reaches temperature below 4ºC, it becomes less dense and remains at the surface,where it freezes. If water did not have this property, lakes and ponds would freeze from the bottom up, which would destroy much of their animal and plant life.

Gases at ordinary temperature expand more than solids and liquids. For, liquids, the coefficient of volume expansion is relatively independent of the temperature. However, for gases it is dependent on temperature. For an ideal gas, the coefficient of volume expansion at constant pressure can be found from the ideal gas equation:

PV = µRT

At constant pressure



i.e., αv = l/T for ideal gas …………………………………………………………………….(6)

At 0ºC, αv = 3.7 × 10-3 K-1, which is much larger than that for solids and liquids. Equation (6) shows the temperature dependence of αv; it decreases with increasing temperature. For a gas at room temperature and constant pressure αv is about 3300 × 10-6 K-1, as much as order(s) of magnitude larger than the coefficient of volume expansion of typical liquids.

Fig. 7b – Thermal expansion of water

There is a simple relation between the coefficient of volume expansion (αv) and coefficient of linear expansion (αl). Imagine a cube of length, l, that expands equally in all directions, when its temperature increases by ΔT. We have

Δl = αl l ΔT

so, ΔV = (l +Δl)³ – l³ ; 3l² Δl …………………………………………………..(7)

In equation (7), terms in (Δl)² and (Δl)³ have been neglected since Δl is small compared to l. So

ΔV = 3VΔl/l = 3lΔT ………………………………………………(8)

which gives

αv = 3 αl

What happens by preventing the thermal expansion of a rod by fixing its ends rigidly? Clearly, the rod acquires a compressive strain due to the external forces provided by the rigid support at the ends. The corresponding stress set up in the rod is called thermal stress. For example, consider a steel rail of length 5m and area of cross section 40 cm² that is prevented from expanding while the temperature rises by 10°C. The coefficient of linear expansion of steel is αl(steel) = 1.2 × 10-5K-1. Thus, the compressive strain is Δl/l = αl(steel)ΔT = 1.2× 10-5 × 10 = 1.2× 10-4.

Young’s modulus of steel is Y(steel)= 2× 1011 N m-2. Therefore, the thermal stress developed is ΔF/A = Y(steel)(Δl/l) = 2.4× 107× 40 × 10-4≅ 105N.

If two such steel rails, fixed at their outer ends, are in contact at their inner ends, a force of this magnitude can easily bend the rails.

Example 1
Show that the coefficient of area expansions,
(ΔA/A)/ΔT, of a rectangular sheet of the solid is twice its linear expansivity, αl.


Fig. 8

Consider a rectangular sheet of the solid material of length a and breadth b (Fig. 8). When the temperature increases by ΔT, a increases by Δa = αlaΔT and b increases by Δb = αlbΔT. From fig 8, the increase in area

ΔA = ΔA1 + ΔA2 + ΔA3

Δa = aΔb + bΔa + (Δa)(Δb)

= aαlbΔT + bα1aΔT +(αlab(ΔT)²

= αlabΔT (2+αlΔT ) = αlAΔT(2+αlΔT )

Since αl; 10-6K-1, fro Table 1, the product αlΔT for fractional temperature is small in comparison with 2 and may be neglected.


(ΔA/A)1/ΔT ≅ 2αl

Example 2
A blacksmith fixes iron ring on the rim of the wooden wheel of a bullock cart. The diameter of the rim and the iron ring are 5.243 m and 5.231 m respectively at 27°C. To what temperature should the ring be heated so as to fit the rim of the wheel?


Given, T1 = 27°C

LT1 = 5.231 m

LT2 = 5.243 m


LT2 = LT1[1+αl(T2-T1)]

5.243 m = 5.231 m[1+1.20×10-5K-1(T227°C)]

or T2= 218°C.

Specific Heat Capacity

Take some water in a vessel and start heating it on a barrier. Soon you will notice that bubbles begin to move upward. As the temperature is increased, the motion of water particles increase till it becomes turbulent as water starts boiling. What are the factors on which the quantity of heat required to raise the temperature of a substance depend? In order to answer this question in the first step, heat a given quantity of water to raise its temperature by, say 20°C and note the time taken. Again take the same amount of water to raise its temperature by 40°C using the same source of heat. Note the time taken by using a stopwatch. You will find it takes about twice the time and therefore, double the quantity of heat required raising twice the temperature of same amount of water.

In the second step, in place of water, now heat the same quantity of some oil, say mustard oil, and raise the temperature again by 20°C. Now note the time taken by the same stopwatch. You will find the time taken will be shorter and therefore, the quantity of heat required would be less than that required by the same amount of water for the same rise in temperature.

The above observations show that the quantity of heat required to warm a given substance depends on its mass, m, the change in temperature, ΔT and the nature of substance. The change in temperature of a substance, when a given quantity of heat is absorbed or rejected by it, is characterised by a quantity called the heat capacity of that substance. We define heat capacity, S of a substance as

S = ΔQ/ΔT ……………………………………………………………………………..(10)

where ΔQ is the amount of heat supplied to the substance to change its temperature from T to T + ΔT.

You have observed that if equal amount of heat is added to equal masses of different substances, the resulting temperature changes will not be the same. It implies that every substance has a unique value for the amount of heat absorbed or rejected to change the temperature of unit mass of it by one unit. This quantity is referred to as the specific heat capacity of the substance.

If ΔQ stands for the amount of heat absorbed or rejected by a substance of mass m when it undergoes a temperature change ΔT, then the specific heat capacity, of that substance is given by

s = S/m = (1/m)(ΔQ/ΔT) ……………………………………………………….(11)

The specific heat capacity is the property of the substance which determines the change in the temperature of the substance (undergoing no phase change) when a given quantity of heat is absorbed (or rejected) by it. It is defined as the amount of heat per unit mass absorbed(or rejected) by it. It is defined as the amount of heat per unit mass absorbed or rejected by the substance to change its temperature by one unit. It depends on the nature of the substance and its temperature. The SI unit of specific heat capacity is J kg-1K-1.

If the amount of substance is specified in terms of moles µ, instead of mass m in kg, we can define heat capacity per mole of the substance by

C = (S/µ) = (1/µ)(ΔQ/ΔT) …………………………………………………………..(12)

where C is known as molar specific heat capacity of the substance. Like S, C also depends on the nature of the substance and its temperature. The SI unit of molar specific heat capacity is J mol-1K-1.

However, in connection with specific heat capacity of gases, additional conditions may be needed to define C. In this case, heat transfer can be achieved by keeping either pressure or volume constant. If the gas is held under constant pressure during heat transfer, then its is called the molar specific heat capacity at constant pressure and is denoted by Cp. On the other hand, if the volume of the gas is maintained during the heat transfer, then the corresponding molar specific heat capacity is called molar specific heat capacity at constant volume and is denoted by Cv. Table 3 lists measured specific heat capacity of some substances at atmospheric pressure and ordinary temperature while Table 4 lists molar specific heat capacities of some gases. From Table 3 you can note that water has the highest specific heat capacity compared to other substances.

Table 3 – Specific heat capacity of some substances at room temperature and atmospheric pressure

Substance Specific heat capacity
( J kg-1K-1)
Substance Specific heat capacity
( J kg-1K-1)
Aluminium 900.0 Ice 2060
Carbon 506.5 Glass 840
Copper 386.4 Iron 450
Lead 127.7 Kerosene 2118
Silver 236.1 Edible oil 1965
Tungsten 134.4 Mercury 140
Water 4186.0

For this reason water is used as coolant in automobile radiators as well as a heater in hot water bags. Owing to its high specific heat capacity, the water warms up much more slowly than the land during summer and consequently wind from the sea has a cooling effect. Now, you can tell why in desert areas, the earth surface warms up quickly during the day and cools quickly at night.

Table 4 – Molar heat capacities of some gases

Gas Cp (J mol-1K-1) Cv (J mol-1K-1)
He 20.8 12.5
H2 28.8 20.4
N2 29.1 20.8
O2 29.4 21.1
CO2 37.0 28.5


A system is said to be isolated, if no exchange or transfer of heat occurs between the system and its surroundings. When different parts of an isolated system are at different temperature, a quantity of heat transfers from the part at higher temperature to the part at lower temperature. The heat lost by the part at higher temperature is equal to the heat gained by the part at lower temperature.

Calorimetry means measurement of heat. When a body at higher temperature is brought in contact with another body at lower temperature, the heat lost by the hot body is equal to the heat gained by the colder body, provided no heat is allowed to escape to the surroundings. A device in which heat measurement can be made is called a calorimeter. It consists a metallic vessel and stirrer of the same material like copper or aluminium. The vessel is kept inside a wooden jacket which contains heat insulating materials like glass wool etc. The outer jacket acts as heat shield and reduces the heat loss from the inner vessel. There is an opening in the outer jacket through which a mercury thermometer can be inserted into the calorimeter. The following example provides a method by which the specific heat capacity of a given solid can be determined by using the principle, heat gained is equal to the heat lost.

Example 3
A sphere of aluminium of 0.047 kg placed for sufficient time in a vessel containing boiling water, so that the sphere is at 100°C. It is then immediately transferred to 0.14 kg copper calorimeter containing 0.25 kg of water at 20°C. The temperature of water rises and attains a steady state at 23°C. Calculate the specific heat capacity of aluminium.


In solving this example we shall use the fact that at a steady state, heat given by an aluminium sphere will be equal to the heat absorbed by the water and calorimeter.

Mass of aluminium sphere (m1) = 0.047 kg

Initial temp. of aluminium sphere = 100°C

Final temp. = 23°C

Change in temp.(ΔT) = (100°C – 23°C) = 77°C

Let specific heat capacity of aluminium be sAl

The amount of heat lost by the aluminium sphere= m1sAlΔT = 0.047kg ×sAl×77°C

Mass of water (m2) = 0.25 kg

Mass of calorimeter (m3) = 0.14 kg

Initial temp. of water and calorimeter = 20°C

Final temp. of water and calorimeter = 23°C

Change in temp. (ΔT2) = 23°C -20°C = 3°C

Specific heat capacity of water (sw) = 4.18 × 10³ J kg-1K-1-1-1-1-1

Specific heat capacity of copper calorimeter = 0.386 × 10³ J kg-1K-1

The amount of heat gained by water and calorimeter = (m2sw +m3scu)(ΔT2)

= (0.25 kg × 4.18 × 10³J kg-1K-1 + 0.14 kg × 0.386 × 10³ J kg-1K-1 )(23°C -20°C)

In the steady state heat lost by the aluminium sphere = heat gained by water + heat gained by calorimeter.

So, 0.047 kg × sAl × 77°C

= (0.25 kg × 4.18 × 10³ J kg-1K-1 + 0.14 kg × 0.386× 10³ J kg-1K-1) (3°C)

sAl = 0.911 kJkg-1K-1

Change of State

Matter normally exists in three states; solid, liquid, and gas. A transition from one of these states to another is called a change of state. Two common changes of states are solid to liquid and liquid to gas (and vice versa). These changes can occur when the exchange of heat takes place between the substance and its surroundings. To study the change of state on heating or cooling, let us perform the following activity.

Take some cubes of ice in a beaker. Note the temperature of ice (0°C) . Start heating it slowly on a constant heat source. Note the temperature after every minute. Continuously stir the mixture of water and ice. Draw a graph between temperature and time (Fig. 9). You will observe no change in the temperature so long as there is ice in the beaker. In the above process,the temperature of the system does not change even though heat is being continuously supplied. The heat supplied is being utilized in changing the state from solid (ice) to liquid (water).

Fig 9 – A plot of temperature versus time showing the changes in the state of ice on heating (not to scale)

The change of state from solid to liquid is called melting and from liquid to solid is called fusion. It is observed that the temperature remains constant until the entire amount of the solid substance melts. That is, both the solid and liquid states of the substance coexist in thermal equilibrium during the change of states from solid to liquid. The temperature at which the solid and the liquid states of the substance in thermal equilibrium with each other is called its melting point. It is characteristic of the substance. It also depends on pressure. The melting point of a substance at standard atmospheric pressure is called its normal melting point. Let us do the following activity to understand the process of melting of ice.

Take a slab of ice. Take a metallic wire and fix two blocks, say 5 kg each, at its ends. Put the wire over the slab as shown in Fig 10. You will observe that the wire passes through the ice slab. This happens due to the fact that just below the wire, ice melts at lower temperature due to increase in pressure. When the wire has passed, water above the wire freezes again. Thus the wire passes through the slab and the slab and the slab does not split. This phenomenon of refreezing is called regelation. Skating is possible on snow due to the formation of water below the skates. Water is formed due to the increase of pressure and it acts as a lubricant.

Fig 10

After the whole of ice gets converted into water and as we continue further heating, we shall see that temperature keeps on rising till it reaches nearly 100°C when it again becomes steady. The heat supplied is now being utilised to change water from liquid state to vapour or gaseous state.

The change of state from liquid to vapour (or gas) is called vaporization. It is observed that the temperature remains constant until the entire amount of the liquid is converted into vapour. That is, both the liquid and vapour states of the substance coexist is called its boiling point. Let us do the following activity to understand the process of boiling water.

Take a round bottom flask, more than half filled with water. Keep it over a burner and fix a thermometer and steam outlet through the cork of the flask (Fig. 11). As

Triple Point

The temperature of a substance remains constant during its change of state (phase change). A graph between the temperature T and the pressure P of the substance is called a phase diagram or P-T diagram. The following figure shows the phase diagram of water and CO2. Such a phase diagram divides the P-T plane in to a solid-region, the vapour region and the liquid region. The regions are separated by curves such as sublimation curve (SO), fusion curve (AO) and vaporization curve (CO). The points on sublimation curve represent states in which solid and vapour phases coexist. The point on the sublimation curve SO represent states in which the solid and vapour phase coexist. Points on the fusion curve AO represent states in which solid and liquid phase coexist. Points on the vaporization curve CO represent states in which the liquid and vapour phases coexist. The temperature and pressure at which the fusion curve, the vaporization curve and the sublimation curve meet and all the three phases of a substance coexist is called the triple point of the substance . For example the triple point of water is represented by the temperature 273.16 K and 6.11 × 10-3 Pa.

Pressure-temperature phase diagrams for (a) water and (b) CO2 (not to scale).

Water gets heated in the flask, note first that the air which was dissolved in the water, will come out as small bubbles. Later, bubbles of steam will form at the bottom but they rise to the cooler water near the top, they condense and disappear. Finally, as the temperature of the entire mass of the water reaches 100°C, bubbles of steam reach the surface and boiling is said to occur. The steam in the flask may not be visible but as it comes out of the flask, it condenses as tiny droplets of water, giving a foggy appearance.

Fig. 11 – Boiling process

If now the steam outlet is closed for a few seconds to increase the pressure in the flask, you will notice that boiling stops. More heat would be required to raise the temperature (depending on the increase in pressure) before boiling begins again. Thus boiling point increases with increase in pressure.

Let us now remove the burner. Allow water to cool to about 80°C. Remove the thermometer and steam outlet. Close the flask with the airtight cork. Keep the flask turned upside down on the stand. Pour ice-cold water on the flask. Water vapors in the flask condense reducing the pressure on the water surface inside the flask. Water begins to boil again, now at a lower temperature. Thus boiling point decreases with decrease in pressure.

This explains why cooking is difficult on hills. At high altitudes, atmospheric pressure is lower, reducing the boiling point of water as compared to that at sea level. On the other hand, boiling point is increased inside a pressure cooker by increasing the pressure. Hence, cooking is faster. The boiling point of a substance at standard atmospheric pressure is called its normal boiling point.

However, all substances do not pass through the three states: solid-liquid-gas. There are certain substances which normally pass from the solid to the vapour state directly and vice versa. The change from solid state to vapour state without passing through the liquid state is called sublimation, and the substance is said to sublime. Dry ice (solid CO2) sublimes, so also iodine. During the sublimation process both the solid and vapour states of a substance coexist in thermal equilibrium.

Latent Heat

Earlier, we have learnt that certain amount of heat energy is transferred between a substance and its surroundings when it undergoes a change of state. The amount of heat per unit mass transferred during change of state of the substance is called latent heat of the substance for the process. For example, if heat is added to a given quantity of ice at -10°C, the temperature of ice increases until it reaches its melting point (0°C ). At this temperature, the addition of more heat does not increase the temperature but causes the ice to melt, or changes its state. Once the entire ice melts, adding more heat will cause the temperature of the water to rise. A similar situation occurs during liquid gas change of state at the boiling point. Adding more heat to boiling water causes vaporization, without increase in temperature.

Table 5 – Temperature of the change of state and latent heats for various substances at 1 atm pressure

Substance Melting Point (°C) Lf(105JKg-1) Boiling Point(°C) Lf(105JKg-1)
Ethyl alcohol -114 1.0 78 8.5
Gold 1063 0.645 2660 15.8
Lead 328 0.25 1744 8.67
Mercury -39 0.12 357 2.7
Nitrogen -210 0.26 -196 2.0
Oxygen -219 0.14 -183 2.1
Water 0 3.33 100 22.6

The heat required during a change of state depends upon the heat of transformation and the mass of the substance undergoing a change of state. Thus, if mass m of a substance undergoes a change from one state to another then the quantity of heat required is given by

Q = mL

or L = Q/m ……………………………………………………………………………(13)

where L is known as latent heat and is a characteristic of the substance. Its SI unit is JKg-1. The value of L also depends on the pressure. Its value is usually quoted at standard atmospheric pressure. The latent heat for a solid-liquid state change is called the latent heat of fusion(Lf) and that for a liquid-gas state change is called the latent heat of vaporization(Lv). These are often referred to as the heat of fusion and the heat of vaporization. A plot of temperature versus heat energy for a quantity of water is shown in Fig. 12. The latent heats of some substances, their freezing and boiling points are given in Table 5.

Fig. 12 – Temperature versus heat for water at 1 atm pressure (not to scale)

Note that when heat is added (or removed) during a change of state, the temperature remains constant. Note in Fig. 12 that the slopes of the phase lines are not all the same, which indicates that specific heats of the various states are not equal. For water, the latent heat of fusion and vaporization are Lf = 3.33 × 105 J kg-1 and Lv= 22.6 × 105 J kg-1 respectively. That is 3.33 × 105 J of heat are needed to melt 1 kg of ice at 0°C, and 22.6× 105 J of heat are needed to convert 1 kg of water to steam at 100°C. This is why burns from steam are usually more serious than those from boiling water.

Example 4
When 0.15 kg of ice of 0°C mixed with 0.30 kg of water at 50°C in a container, the resulting temperature is 6.7°C. Calculate the heat of fusion of ice. (swater = 4186 J kg-1K-1).


Heat lost by water = msw f θi)w

= (0.30 kg)(4186 J kg-1K-1) (50.0°C – 6.7°C)

= 54376.14 J

Heat required to melt ice = m2 Lf

= (0.15 kg)Lf

Heat lost required to raise temperature of ice water to final temperature = m1 sw f θi)I =(0.15 kg)(4186 J kg-1K-1) (6.7°C – 0°C) =4206.93 J

Heat lost = heat gained

54376.14 J = (0.15 kg)Lf + 4206.93 J

Lf = 3.34 × 105 J kg-1

Example 5
Calculate the heat required to convert 3 kg of ice at -12°C kept in a calorimeter to steam at 100°C at atmospheric pressure. Given specific heat capacity of ice = 2100 J kg-1K-1, specific heat capacity of water = 4186 J kg-1K-1, latent heat heat of fusion of ice = 3.35 ×105 J kg-1 and latent heat of steam = 2.256×106 J kg-1 .


We have

Mass of ice, m = 3 kg

specific heat capacity of ice, sice = 2100 J kg-1K-1

specific heat capacity of ice, swater = 4186 J kg-1K-1

latent heat of fusion of ice, Lfice = 3.35 × 105 J kg-1

latent heat of steam Lsteam = 2.256×106 J kg-1

Now, Q = heat required to convert 3 kg of ice at -12°C to steam at 100°C .

Q1 = heat required to convert ice at -12°C to ice at 0°C .

= m sice ΔT1 = (3kg)(2100 J kg-1K-1)[0-(-12)]°C= 75600J

Q2 = heat required to melt ice at 0°C to water at 0°C .

= m Lfice = (3kg)(3.35 × 105 J kg-1 )= 1005000J

Q3 = heat required to convert water at 0°C to water at 100°C .

= m swΔT2 = (3kg)(4186 J kg-1 )(100°C)= 12255800J

Q4 = heat required to convert water at 100°C to steam at 100°C .

= m Lsteam = (3kg)(2.256×106 J kg-1 )= 6768000J

So, Q = Q1 + Q2 +Q3 + Q4 =75600J + 1005000 J + 12255800J+6768000J

= 9.1 × 106 J

Heat Transfer

We have seen that heat is energy transfer from one system to another or from one part of a system to another part, arising due to temperature difference. What are the different ways by which this energy transfer takes place? There are three distinct modes of heat transfer: conduction, convection and radiation (Fig. 13).

Fig 13 – Heating by conduction, convection and radiation


Conduction is the mechanism of transfer of heat between two adjacent parts of a body because of their temperature difference. Suppose one end of a metallic rod is put in a flame, the other end of the rod will soon be so hot that you cannot hold it by your bare hands. Here heat transfer takes place by conduction from the hot end of the rod through its different parts to the other end. Gases are poor thermal conductors while liquids have conductivity intermediate between solids and gases.

Heat conduction may be described quantitatively as the time rate of heat flow in a material for a given temperature difference. Consider a metallic bar of length L and uniform cross section A with its two ends maintained at different temperatures. This can be done, for example, by putting the ends in thermal contact with large reservoirs at temperatures, say, TC and TD respectively(Fig. 14). Let us assume the ideal condition that the sides of the bar are fully insulated so that no heat is exchanged between the sides and the surroundings.

After some time, a steady state is reached; the temperature of the bar decreases uniformly with distance from TC to TD; (TC> TD). The reservoir at C supplies heat at a constant rate, which transfers through the bar and is given out at the same rate to the reservoir at D. It is found experimentally that in this steady state, the rate of flow of

Fig 14
Steady state heat flow by conduction in a bar with its two ends maintained at temperatures TC and TD; (TC> TD).

Heat (or Heat current) H is proportional to the temperature difference (TC– TD) and the area of cross section A and is inversely proportional to the length L:

H = KA (TC– TD)/L …………………………………………………………………….(14)

The constant of proportionality K is called the thermal conductivity of the material. The greater the value of K for a material, the more rapidly will it conduct heat. The SI unit of K is J S-1m-1K-1 or W m-1K-1. The thermal conductivity of various substances are listed in Table 5. These values vary slightly with temperature, but can be considered to be constant over a normal temperature range.

Compare the relatively large thermal conductivity of the good thermal conductors, the metals, with the relatively small thermal conductivity of some good thermal insulators, such as wood and glass wool. You may have noticed that some cooking pots have copper coating on the bottom. Being a good conductor of heat, copper promotes the distribution of heat over the bottom of a pot for uniform cooking. Plastic foams, on the other hand, are good insulators, mainly because they contain pockets of air. Recall that gases are poor conductors, and note the low thermal conductivity of air in Table 5. Heat retention and transfer are important in many other applications. Houses made of concrete roofs get very hot during summer days, because thermal conductivity of concrete (though much smaller than that of a metal) is still not small enough. Therefore, people usually prefer to give a layer of earth or foam insulation on the ceiling so that heat transfer is prohibited and keeps the room cooler. In some situations, heat transfer is critical. In a nuclear reactor, for example, elaborate heat transfer systems need to be installed so that the enormous energy produced by nuclear fission in the core transits out sufficiently fast, thus preventing the core form overheating.

Table 6 – Thermal conductivities of some materials

Materials Thermal conductivity (Js-1m-1>K-1)
Silver 406
Copper 385
Aluminium 205
Brass 109
Steel 50.2
Lead 34.7
Mercury 8.3
Insulating brick 0.15
Concrete 0.8
Body fat 0.20
Felt 0.04
Glass 0.8
Ice 1.6
Glass wool 0.04
Wood 0.12
Water 0.8
Air 0.024
Argon 0.016
Glass wool 0.14
Example 6
What is the temperature of the steel-copper junction in the steady state of the system shown in Fig. 15. Length of the steel rod = 15.0 cm, length of the copper rod = 10.0 cm, temperature of the furnace = 300°C, temperature of the other end = 0°C. The area of cross section of the steel rod is twice that of the copper rod , (Thermal conductivity of steel = 50.2 J s-1 m-1K-1; and of copper = 385 J s-1 m-1K-1)
Fig. 15

The insulating material reduces heat loss from the sides of the rods. Therefore, heat flows along the length of the rods. Consider any cross section of the rod. In the steady state, heat flowing in to the element must equal the heat flowing out of it; otherwise there would be a net gain or loss of heat by the element and its temperature would not be steady. Thus in the steady state, rate of heat flowing across a cross section of the rod is the same at every point along the length of the combined steel-copper junction in the steady state. Then,

K1 A1300 T/L1 = K2 A2 T – 0/L2

where 1 and 2 refer to the steel and copper rod respectively. For A1= 2 A2, L1 = 15.0 cm, L2 = 10.0 cm, K1= 50.2 Js-1 m-1K-1, K2= 385 Js-1 m-1K-1, we have

50.2 × 2 (300-T)/15 = 385 T/10

which gives T = 44.4°C


Convection is a mode of heat transfer by actual motion of matter. It is possible only in fluids. Convection can be natural or forced. In natural convection, gravity plays an important part. When a fluid is heated from below, the hot part expands and, therefore, becomes less dense. Because of buoyancy, it rises and the upper colder part replaces it. This again gets heated, rises up and is replaced by the colder part of the fluid. The process goes on. This mode of heat transfer is evidently different from conduction. Convection involves bulk transport of different parts of the fluid. In forced convection, material is forced to move by a pump or by some other physical means. The common examples of forced convection systems in home, the human circulatory system, and the cooling system of an automobile engine. In the human body, the heart acts as the pump that circulates blood through different parts of the body, transferring heat by forced and maintaining it at a uniform temperature.

Fig. 17 – Convection cycles

Natural convection is responsible for many familiar phenomena. During the day, the ground heats up more quickly than large bodies of water do. This occurs both because mixing currents disperse the absorbed heat throughout the great volume of water. The air in contact with the warm ground is heated by conduction. It expands, becoming less dense than the surrounding cooler air. As a result, the warm air rises (air currents) and other air moves (winds) to fill the space-creating a sea breeze near a large body of water. Cooler air descends, and a thermal convection cycle is set up, which transfers heat away from the land. At night, the ground loses its heat more quickly, and the water surface is warmer than the land. As a result, the cycle is reversed. (Fig. 17)

The other example of natural convection is the steady surface wind on the earth blowing in from north-east towards the equator, the so called trade wind. A reasonable explanation is as follows: the equatorial and polar regions of the earth receive unequal solar heat. Air at the earth’s surface near the equator is hot while the air in the upper atmosphere of the poles is cool. In the absence of any other factor, a convection current would be set up, with the air at the equatorial surface rising and moving out towards the poles, descending and streaming in towards the equator. The rotation of the earth, however, modifies this convection current. Because of this, air close to the equator has an eastward speed 1600 km/h, while it is zero close to the poles. As a result, the air descends not at the poles but at 30°N (North) latitude and returns to the equator. This is called trade wind.


Conduction and convection require some material as a transport medium. These modes of heat transfer cannot operate between bodies separated by a distance in vacuum. But the earth does receive heat from the sun across a huge distance and we quickly feel the warmth of the fire nearby even though air conducts poorly and before convection can set in. The third mechanism for heat transfer needs no medium; it is called radiation and the energy so radiated by electromagnetic wave electric and magnetic fields oscillate in space and time. Like any wave, electromagnetic waves can have different wavelengths and can travel in vacuum with the same speed, namely the speed of light i.e., 3 × 108ms-1. You will learn these in more details later, but you now know why heat transfer by radiation does not need any medium and why is so fast. This is how heat is transferred to the earth from the sun through empty space. All bodies emit radiant energy, whether they are solid, liquid or gases. The electromagnetic radiation emitted by a body by virtue of its temperature like the radiation by a red hot iron or light from a filament lamp is called thermal radiation.

When this thermal radiation falls on other bodies, it is partly reflected and partly absorbed. The amount of heat that a body can absorb by radiation depends on the color of the body.

We find that black bodies absorb and emit radiant energy better than bodies of lighter colors. This fact finds many applications in our daily life. We wear white or light colored clothes in summer so that they absorb the least heat from the sun and keep our body warm. The bottoms of the utensils for cooking food are blackened so that they absorb maximum heat from the fire and give it to the vegetables to be cooked.

Similarly, a Dewar flask or thermos bottle is a device to minimise heat transfer between the contents of the bottle and outside. It consists of a double-walled glass vessel with the inner and outer walls covered with silver. Radiation from the inner wall is reflected back in to the contents of the bottle. The outer wall similarly reflects back any incoming radiation. The space between the walls is evacuated to reduce conduction and convection losses and the flask is supported on an insulator like cork. The device is, therefore, useful for preventing hot contents (like milk) from getting cold, or alternatively to store cold contents (like ice).

Blackbody Radiation

We have so far not mentioned the wavelength content of thermal radiation. The important thing about thermal radiation at any temperature is that it is not of one (or a few) wavelength(s) but has a continuous spectrum from the small to the long wavelengths. The energy content of radiation, however, varies for different wavelengths. Figure 18 gives the experimental curves for radiation energy per unit area per unit wavelength emitted by a blackbody versus wavelength for different temperatures.

Fig. 18 – : Energy emitted versus wavelength for a blackbody at different temperatures

Notice that the wavelength λm for which energy is the maximum decreases with increasing temperature. The relation between λm and T is given by what is known as Wien’s Displacement Law:
λm T = constant …………………………………………………………………. (15)
The value of the constant (Wien’s constant) is 2.9 × 10–3 m K. This law explains why the color of a piece of iron heated in a hot flame first becomes dull red, then reddish yellow, and finally white hot. Wien’s law is useful for estimating the surface temperatures of celestial bodies like, the moon, Sun and other stars. Light from the moon is found to have a maximum intensity near the wavelength 14 μm. By Wien’s law, the surface of the moon is estimated to have a temperature of 200 K. Solar radiation has a maximum at λm = 4753 Å. This corresponds to T = 6060 K. Remember, this is the temperature of the surface of the sun, not its interior.

The most significant featurse of the blackbody radiation curves in Fig. 18 is that they are universal. They depend only on the temperature and not on the size, shape or material of the blackbody. Attempts to explain blackbody radiation theoretically, at the beginning of the twentieth century, spurred the quantum revolution in physics, as you will learn in later courses.

Energy can be transferred by radiation over large distances, without a medium (i.e., in vacuum). The total electromagnetic energy radiated by a body at absolute temperature T is proportional to its size, its ability to radiate (called emissivity) and most importantly to its temperature. For a body, which is a perfect radiator, the energy emitted per unit time (H) is given by
H = AσT4 …………………………………………………………………………………………….(16)
where A is the area and T is the absolute temperature of the body. This relation obtained experimentally by Stefan and later proved theoretically by Boltzmann is known as Stefan-
Boltzmann law and the constant σ is called Stefan-Boltzmann constant. Its value in SI units is 5.67 × 10-8 W m-2 K-4. Most bodies emit only a fraction of the rate given by Eq. 11.16. A substance like lamp black comes close to the limit. One, therefore, defines a dimensionless fraction e called emissivity and writes,

H = AεσT4 ……………………………………………………………………………………….(17)

Here, ε = 1 for a perfect radiator. For a tungsten lamp, for example, ε is about 0.4. Thus, a tungsten lamp at a temperature of 3000 K and a surface area of 0.3 cm2 radiates at the rate H = 0.3 × 10-4 × 0.4 × 5.67 × 10-8 × (3000)4 = 60 W.

A body at temperature T, with surroundings at temperatures Ts, emits, as well as, receives energy. For a perfect radiator, the net rate of loss of radiant energy is

H = σA (T4 – Ts4) For a body with emissivity e, the relation modifies to

H = εσ A (T4 – Ts4) ………………………………………………………..(18)

As an example, let us estimate the heat radiated by our bodies. Suppose the surface area of a person’s body is about 1.9 m2 and the room temperature is 22°C. The internal body temperature, as we know, is about 37 °C. The skin temperature may be 28°C (say). The emissivity of the skin is about 0.97 for the relevant region of electromagnetic radiation. The
rate of heat loss is:
H = 5.67 × 10-8 × 1.9 × 0.97 × {(301)4 – (295)4}
= 66.4 W
which is more than half the rate of energy production by the body at rest (120 W). To prevent this heat loss effectively (better than ordinary clothing), modern arctic clothing has an additional thin shiny metallic layer next to the skin, which reflects the body’s radiation.

Greenhouse Effect

The earth’s surface is a source of thermal radiation as it absorbs energy received from the Sun. The wavelength of this radiation lies in the long wavelength (infrared) region. But a large portion of this radiation is absorbed by greenhouse gases, namely, carbon dioxide (CO2); methane (CH4); nitrous oxide (N2O);chlorofluorocarbon (CFxClx); and tropospheric ozone (O3). This heats up the atmosphere which, in turn, gives more energy to earth, resulting in warmer surface. This increases the intensity of radiation from the surface. The cycle of processes described above is repeated until no radiation is available for absorption. The net result is heating up of earth’s surface and atmosphere. This is known as Greenhouse Effect. Without the Greenhouse Effect, the temperature of the earth would have been –18°C. Concentration of greenhouse gases has enhanced due to human activities, making the earth warmer. According to an estimate, average temperature of earth has increased by 0.3 to 0.6°C, since the beginning of this century because of this enhancement. By the middle of the next century, the earth’s global temperature may be 1 to 3°C higher than today. This global warming may cause problem for human life, plants and animals. Because of global warming, ice caps are melting faster, sea level is rising,and weather pattern is changing. Many coastal cities are at the risk of getting submerged. The enhanced Greenhouse Effect may also result in expansion of deserts. All over the world, efforts are being made to minimise the effect of global warming.

Newton’s Law of Cooling

We all know that hot water or milk when left on a table begins to cool gradually. Ultimately it attains the temperature of the surroundings. To study how a given body can cool on exchanging heat with its surroundings, let us perform the following activity.

Take some water, say 300 ml, in a calorimeter with a stirrer and cover it with two holed lid. Fix a thermometer through a hole in the lid and make sure that the bulb of thermometer is immersed in the water. Note the reading of the thermometer. This reading T1 is the temperature of the surroundings. Heat the water kept in the calorimeter till it attains a temperature (i.e., temperature of the surroundings). Then stop heating the water by removing the heat source. Start the stop watch and note the reading of the thermometer after fixed interval of time, say after every one minute of stirring gently with the stirrer. Continue to note the temperature (T2) of water till it attains a temperature about 5°C above that of the surroundings. Then plot a graph by taking each value of temperature ΔT = T2 – T1 along y axis and the corresponding value of t along x-axis (Fig. 19).

Fig. 19 – Curve showing cooling of hot water with time

From the graph you will infer how the cooling of hot water depends on the difference of its temperature from that of the surroundings. You will also notice that initially the rate of cooling is higher and decreases as the temperature of the body falls.

The above activity shows that a hot body loses heat to its surroundings in the form of heat radiation. The rate of loss of heat depends on the difference in temperature between the body and its surroundings. Newton was the first to study, in a systematic manner, the relation between the heat lost by a body in a given enclosure and its temperature.

According to Newton’ law of cooling, the rate of loss of heat, dQ/dt of the body is directly proportional to the difference of the temperature ΔT = T2 – T1 of the body and the surroundings. The law holds good only for small difference of temperature. Also, the loss of heat by radiation depends upon the nature of the surface of the body and the area of the exposed surface. We can write

dQ/dt = k(T2 – T1) …………………………………………………………………..(19)

where k is a positive constant depending upon the area and nature of the surface of the body. Suppose a body of mass m and specific heat capacity s is at temperature T2. Let T1 be the temperature of the surroundings. If the temperature falls by a small amount dT2 in time dt, then the amount of heat lost is

dQ = ms dT2

∴ Rate of loss of heat is given by

dQ/dt = ms dT2/dt ………………………………………………(20)

From Eqs. (15) and (16) we have

msdT2/dt = k(T2 – T1) ………………………………………………..(21)

where K = k/ms

On integrating,

loge(T2 – T1) = -Kt + c ………………………………………………..(22)

or T2 = T1 +C′ e-Kt; where C′ = ec………………………………………….(23)

Equation (23) enables you to calculate the time of cooling of a body through a particular range of temperature.

For small temperature differences, the rate of cooling, due to conduction, convection, and radiation combined, is proportional to the difference in temperature. It is a valid approximation in the transfer of heat from a radiator to a room, the loss of heat through the wall of a room, or the cooling of a cup of tea on the table.

Fig 20 – Verification of Newton’s Law of cooling

Newton’s law of cooling can be verified with the help of the experimental set-up shown in Fig. 20(a). The set-up consists of a double walled vessel (V) containing water in between the two walls. A copper calorimeter (C) containing hot water is placed inside the double walled vessel. Two thermometers through the corks are used to note the temperatures T2 of water in calorimeter and T1 of hot water in between the double walls respectively. Temperature of hot water in the calorimeter is noted after equal intervals of time. A graph is plotted between loge(T2 – T1) and time (t). The nature of the graph is observed to be straight line having a negative slope as shown in Fig. 20(b). This is in support of Eq. (22).

Example 7
A pan filled with hot food cools from 94°C to 86°C in 2 minutes when the room temperature is at 20°C. How long will it take it to cool from 71°C to 69°C?


The average temperature of 94°C and 86°C is 90°C, which is 70°C above the room temperature. Under these conditions the pan cools 8°C in 2 minutes.

Using Eq. (17), we have

Change in temperature/Time = KΔT

(8°C/2 min) =K (70°C)

The average of 69°C and 71°C is 70°C, which is 50°C above room temperature. K is the same for this situation as for the original.

2°C/Time =K (50°C)

When we divide above two equations, we have

(8°C/2 min)/(2°C/Time) = K (70°C)/K (50°C)

Time = 0.7 min

= 42 s