1.4 DIMENSIONS OF PHYSICAL QUANTITIES formula of the given physical quantity. For example, the dimensional formula of the volume The nature of a physical quantity is described is [M° L3 T°], and that of speed or velocity is by its dimensions. All the physical quantities [M° L T-1]. Similarly, [M° L T–2] is the dimensional represented by derived units can be expressed formula of acceleration and [M L–3 T°] that of in terms of some combination of seven mass density. fundamental or base quantities. We shall call An equation obtained by equating a physical these base quantities as the seven dimensions quantity with its dimensional formula is called of the physical world, which are denoted with the dimensional equation of the physical square brackets [ ]. Thus, length has the quantity. Thus, the dimensional equations are dimension [L], mass [M], time [T], electric current the equations, which represent the dimensions [A], thermodynamic temperature [K], luminous of a physical quantity in terms of the base intensity [cd], and amount of substance [mol]. quantities. For example, the dimensional The dimensions of a physical quantity are the equations of volume [V], speed [v], force [F ] and powers (or exponents) to which the base mass density [ρ] may be expressed as quantities are raised to represent that [V] = [M0 L3 T0] quantity. Note that using the square brackets [v] = [M0 L T–1] [ ] round a quantity means that we are dealing [F] = [M L T–2] with ‘the dimensions of’ the quantity. [ρ] = [M L–3 T0] In mechanics, all the physical quantities can be written in terms of the dimensions [L], [M] The dimensional equation can be obtained and [T]. For example, the volume occupied by from the equation representing the relations between the physical quantities. Thean object is expressed as the product of length, dimensional formulae of a large number andbreadth and height, or three lengths. Hence the wide variety of physical quantities, derived fromdimensions of volume are [L] × [L] × [L] = [L]3 = [L3]. the equations representing the relationships As the volume is independent of mass and time, among other physical quantities and expressed it is said to possess zero dimension in mass [M°], in terms of base quantities are given in zero dimension in time [T°] and three Appendix 9 for your guidance and ready dimensions in length. reference. Similarly, force, as the product of mass and acceleration, can be expressed as 1.6 DIMENSIONAL ANALYSIS AND ITS Force = mass × acceleration APPLICATIONS = mass × (length)/(time)2 The recognition of concepts of dimensions, which The dimensions of force are [M] [L]/[T]2 = guide the description of physical behaviour is [M L T–2]. Thus, the force has one dimension in of basic importance as only those physical Reprint 2025-26 8 PHYSICS quantities can be added or subtracted which such as angle as the ratio (length/length), have the same dimensions. A thorough refractive index as the ratio (speed of light in understanding of dimensional analysis helps us vacuum/speed of light in medium) etc., has no in deducing certain relations among different dimensions. physical quantities and checking the derivation, Now we can test the dimensional consistency accuracy and dimensional consistency or or homogeneity of the equation homogeneity of various mathematical 2expressions. When magnitudes of two or more x = x 0 + v 0 t + (1/2 ) a t physical quantities are multiplied, their units for the distance x travelled by a particle or body should be treated in the same manner as in time t which starts from the position x0 with ordinary algebraic symbols. We can cancel an initial velocity v0 at time t = 0 and has uniformidentical units in the numerator and acceleration a along the direction of motion. denominator. The same is true for dimensions The dimensions of each term may be written as of a physical quantity. Similarly, physical [x] = [L] quantities represented by symbols on both sides of a mathematical equation must have the same [x0 ] = [L] dimensions. [v0 t] = [L T–1] [T] = [L] [(1/2) a t2] = [L T–2] [T2]1.6.1 Checking the Dimensional Consistency of Equations = [L] As each term on the right hand side of this The magnitudes of physical quantities may be equation has the same dimension, namely that added together or subtracted from one another of length, which is same as the dimension of only if they have the same dimensions. In other left hand side of the equation, hence this words, we can add or subtract similar physical equation is a dimensionally correct equation. quantities. Thus, velocity cannot be added to It may be noted that a test of consistency of force, or an electric current cannot be subtracted dimensions tells us no more and no less than a from the thermodynamic temperature. This test of consistency of units, but has the simple principle called the principle of advantage that we need not commit ourselves homogeneity of dimensions in an equation is to a particular choice of units, and we need not extremely useful in checking the correctness of worry about conversions among multiples and an equation. If the dimensions of all the terms sub-multiples of the units. It may be borne in are not same, the equation is wrong. Hence, if mind that if an equation fails this consistency we derive an expression for the length (or test, it is proved wrong, but if it passes, it is distance) of an object, regardless of the symbols not proved right. Thus, a dimensionally correct appearing in the original mathematical relation, equation need not be actually an exact when all the individual dimensions are (correct) equation, but a dimensionally wrong simplified, the remaining dimension must be (incorrect) or inconsistent equation must be that of length. Similarly, if we derive an equation wrong. of speed, the dimensions on both the sides of ⊳equation, when simplified, must be of length/ Example 1.3 Let us consider an equation time, or [L T–1]. Dimensions are customarily used as a 1 2 m v = m g h preliminary test of the consistency of an 2 equation, when there is some doubt about the where m is the mass of the body, v its correctness of the equation. However, the velocity, g is the acceleration due to dimensional consistency does not guarantee gravity and h is the height. Check correct equations. It is uof dimensionless quantities or functions. The correct. arguments of special functions, such as the trigonometric, logarithmic and exponential Answer The dimensions of LHS are functions must be dimensionless. A pure [M] [L T–1 ]2 = [M] [ L2 T–2] number, ratio of similar physical quantities, = [M L2 T–2] Reprint 2025-26 UNITS AND MEASUREMENT 9 The dimensions of RHS are string, that oscillates under the action of [M][L T–2] [L] = [M][L2 T–2] the force of gravity. Suppose that the period = [M L2 T–2] of oscillation of the simple pendulum depends on its length (l), mass of the bobThe dimensions of LHS and RHS are the same and hence the equation is dimensionally correct. ⊳ (m) and acceleration due to gravity (g). Derive the expression for its time period ⊳ using method of dimensions. Example 1.4 The SI unit of energy is J = kg m2 s–2; that of speed v is m s–1 and Answer The dependence of time period T on of acceleration a is m s–2. Which of the the quantities l, g and m as a product may be formulae for kinetic energy (K) given below written as : can you rule out on the basis of T = k lx gy mz dimensional arguments (m stands for the mass of the body) : where k is dimensionless constant and x, y (a) K = m2 v3 and z are the exponents. (b) K = (1/2)mv2 By considering dimensions on both sides, we (c) K = ma have (d) K = (3/16)mv2 o o 1 1 x 1 –2 y 1 z [L M T ]=[L ] [L T ] [M ] (e) K = (1/2)mv2 + ma = Lx+y T–2y Mz Answer Every correct formula or equation must On equating the dimensions on both sides, have the same dimensions on both sides of the we have equation. Also, only quantities with the same x + y = 0; –2y = 1; and z = 0 physical dimensions can be added or 1 1 subtracted. The dimensions of the quantity on So that x = , y = – , z = 0 2 2the right side are [M2 L3 T–3] for (a); [M L2 T–2] for (b) and (d); [M L T–2] for (c). The quantity on the Then, T = k l½ g–½ right side of (e) has no proper dimensions since two quantities of different dimensions have been l or, T = kadded. Since the kinetic energy K has the g dimensions of [M L2 T–2], formulas (a), (c) and (e) Note that value of constant k can not be obtainedare ruled out. Note that dimensional arguments by the method of dimensions. Here it does notcannot tell which of the two, (b) or (d), is the matter if some number multiplies the right sidecorrect formula. For this, one must turn to the of this formula, because that does not affect itsactual definition of kinetic energy (see dimensions.Chapter 5). The correct formula for kinetic energy is given by (b). ⊳ l Actually, k = 2π so that T = 2π ⊳ 1.6.2 Deducing Relation among the g Physical Quantities The method of dimensions can sometimes be Dimensional analysis is very useful in deducing used to deduce relation among the physical relations among the interdependent physical quantities. For this we should know the quantities. However, dimensionless constants dependence of the physical quantity on other cannot be obtained by this method. The method quantities (upto three physical quantities or of dimensions can only test the dimensional linearly independent variables) and consider it validity, but not the exact relationship between as a product type of the dependence. Let us take physical quantities in any equation. It does not an example. distinguish between the physical quantities having same dimensions.⊳ Example 1.5 Consider a simple A number of exercises at the end of this pendulum, having a bob attached to a chapter will help you develop skill in dimensional analysis. Reprint 2025-26 10 PHYSICS SUMMARY 1. Physics is a quantitative science, based on measurement of physical quantities. Certain physical quantities have been chosen as fundamental or base quantities (such as length, mass, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity). 2. Each base quantity is defined in terms of a certain basic, arbitrarily chosen but properly standardised reference standard called unit (such as metre, kilogram, second, ampere, kelvin, mole and candela). The units for the fundamental or base quantities are called fundamental or base units. 3. Other physical quantities, derived from the base quantities, can be expressed as a combination of the base units and are called derived units. A complete set of units, both fundamental and derived, is called a system of units. 4. The International System of Units (SI) based on seven base units is at present internationally accepted unit system and is widely used throughout the world. 5. The SI units are used in all physical measurements, for both the base quantities and the derived quantities obtained from them. Certain derived units are expressed by means of SI units with special names (such as joule, newton, watt, etc). 6. The SI units have well defined and internationally accepted unit symbols (such as m for metre, kg for kilogram, s for second, A for ampere, N for newton etc.). 7. Physical measurements are usually expressed for small and large quantities in scientific notation, with powers of 10. Scientific notation and the prefixes are used to simplify measurement notation and numerical computation, giving indication to the precision of the numbers. 8. Certain general rules and guidelines must be followed for using notations for physical quantities and standard symbols for SI units, some other units and SI prefixes for expressing properly the physical quantities and measurements. 9. In computing any physical quantity, the units for derived quantities involved in the relationship(s) are treated as though they were algebraic quantities till the desired units are obtained. 10. In measured and computed quantities proper significant figures only should be retained. Rules for determining the number of significant figures, carrying out arithmetic operations with them, and ‘rounding off ‘ the u 11. The dimensions of base quantities and combination of these dimensions describe the nature of physical quantities. Dimensional analysis can be used to check the dimensional consistency of equations, deducing relations among the physical quantities, etc. A dimensionally consistent equation need not be actually an exact (correct) equation, but a dimensionally wrong or inconsistent equation must be wrong. EXERCISES Note : In stating numerical answers, take care of significant figures. 1.1 Fill in the blanks (a) The volume of a cube of side 1 cm is equal to .....m3 (b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ...(mm)2 (c) A vehicle moving with a speed of 18 km h–1 covers....m in 1 s (d) The relative density of lead is 11.3. Its density is ....g cm–3 or ....kg m–3. 1.2 Fill in the blanks by suitable conversion of units (a) 1 kg m2 s–2 = ....g cm2 s–2 (b) 1 m = ..... ly (c) 3.0 m s–2 = .... km h–2 (d) G = 6.67 × 10–11 N m2 (kg)–2 = .... (cm)3 s–2 g–1. Reprint 2025-26 UNITS AND MEASUREMENT 11 1.3 A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1J = 1 kg m2 s–2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time is γ s. Show that a calorie has a magnitude 4.2 α–1 β–2 γ 2 in terms of the new units. 1.4 Explain this statement clearly : “To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary : (a) atoms are very small objects (b) a jet plane moves with great speed (c) the mass of Jupiter is very large (d) the air inside this room contains a large number of molecules (e) a proton is much more massive than an electron (f) the speed of sound is much smaller than the speed of light. 1.5 A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance ? 1.6 Which of the following is the most precise device for measuring length : (a) a vernier callipers with 20 divisions on the sliding scale (b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale (c) an optical instrument that can measure length to within a wavelength of light ? 1.7 A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair ? 1.8 Answer the following : (a)You are given a thread and a metre scale. How will you estimate the diameter of the thread ? (b)A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale ? (c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only ? 1.9 The photograph of a house occupies an area of 1.75 cm2 on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m2. What is the linear magnification of the projector-screen arrangement. 1.10 State the number of significant figures in the following : (a) 0.007 m2 (b) 2.64 × 1024 kg (c) 0.2370 g cm–3 (d) 6.320 J (e) 6.032 N m–2 (f) 0.0006032 m2 1.11 The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures. 1.12 The mass of a box measured by a grocer’s balance is 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures ? 1.13 A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ mo of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes : m 0 m = 2 1/2 . 1 − v ( ) Guess where to put the missing c. Reprint 2025-26 12 PHYSICS 1.14 The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: 1 Å = 10–10 m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m3 of a mole of hydrogen atoms ? 1.15 One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen ? (Take the size of hydrogen molecule to be about 1 Å). Why is this ratio so large ? 1.16 Explain this common observation clearly : If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you). 1.17 The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107 K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases ? Check if your guess is correct from the following data : mass of the Sun = 2.0 ×1030 kg, radius of the Sun = 7.0 × 108 m. Reprint 2025-26 CHAPTER TWO MOTION IN A STRAIGHT LINE 2.1 INTRODUCTION Motion is common to everything in the universe. We walk, run and ride a bicycle. Even when we are sleeping, air moves 2.1 Introduction into and out of our lungs and blood flows in arteries and 2.2 Instantaneous velocity and veins. We see leaves falling from trees and water flowing speed down a dam. Automobiles and planes carry people from one 2.3 Acceleration place to the other. The earth rotates once every twenty-four 2.4 Kinematic equations for hours and revolves round the sun once in a year. The sun uniformly accelerated motion itself is in motion in the Milky Way, which is again moving

11.6 SPECIFIC HEAT CAPACITY average energy of 2 × ½ kBT = kBT. In three dimensions, the average energy is 3 kBT.Suppose an amount of heat ∆Q supplied to a For a mole of a solid, the total energy is substance changes its temperature from T to T + ∆T. We define heat capacity of a substance U = 3 kBT × NA = 3 RT (∵kBT × NA = R) (see Chapter 10) to be Now, at constant pressure, ∆Q = ∆U + P ∆V ≅ ∆ Q ∆U, since for a solid ∆V is negligible. Therefore, S = (11.4) ∆ T ∆ Q ∆U C = = = 3 R (11.7) We expect ∆Q and, therefore, heat capacity S ∆ T ∆ T to be proportional to the mass of the substance. Table 11.1 Specific and molar heat capacities Further, it could also depend on the of some solids at room temperature, i.e., a different amount of heat may temperature and atmospheric be needed for a unit rise in temperature at pressure different temperatures. To define a constant –v Speci"c heat Molar speci"c characteristic of the substance and Substance –1 –1 –1 –1 (J kg K ) heat (J mol K ) independent of its amount, we divide S by the mass of the substance m in kg : S  1  ∆Q s = (11.5) m =  m  ∆T s is known as the specific heat capacity of the substance. It depends on the nature of the As Table 11.1 shows, the experimentally substance and its temperature. The unit of measured values which generally agrees withspecific heat capacity is J kg–1 K–1. Reprint 2025-26 232 PHYSICS predicted value 3R at ordinary temperatures. ideal gas, we have a simple relation. (Carbon is an exception.) The agreement is known to break down at low temperatures. Cp – Cv = R (11.8) Specific heat capacity of water where Cp and Cv are molar specific heat capacities of an ideal gas at constant pressure The old unit of heat was calorie. One calorie and volume respectively and R is the universal was earlier defined to be the amount of heat gas constant. To prove the relation, we begin required to raise the temperature of 1g of water with Eq. (11.3) for 1 mole of the gas : by 1°C. With more precise measurements, it was found that the specific heat of water varies ∆Q = ∆U + P ∆V slightly with temperature. Figure 11.5 shows If ∆Q is absorbed at constant volume, ∆V = 0this variation in the temperature range 0 to 100 °C.  ∆Q   ∆ U   ∆ U  (11.9) C v =  ∆T  v =  ∆T  v =  ∆T  where the subscript v is dropped in the last step, since U of an ideal gas depends only on temperature. (The subscript denotes the quantity kept fixed.) If, on the other hand, ∆Q is absorbed at constant pressure,  ∆ Q   ∆ U   ∆V  (11.10) C p =  ∆T  p =  ∆T  p + P  ∆ T  p The subscript p can be dropped from the Fig. 11.5 Variation of specific heat capacity of first term since U of an ideal gas depends only water with temperature. on T. Now, for a mole of an ideal gas For a precise definition of calorie, it was, PV = RTtherefore, necessary to specify the unit temperature interval. One calorie is defined which gives to be the amount of heat required to raise the temperature of 1g of water from 14.5 °C to  ∆ V  P = R (11.11)15.5 °C. Since heat is just a form of energy,  ∆ T  p it is preferable to use the unit joule, J. In SI units, the specific heat capacity of water Equations (11.9) to (11.11) give the desired is 4186 J kg–1 K–1 i.e. 4.186 J g–1 K–1. The so relation, Eq. (11.8). called mechanical equivalent of heat defined 11.7THERMODYNAMIC STATE VARIABLES as the amount of work needed to produce AND EQUATION OF STATE 1 cal of heat is in fact just a conversion factor between two different units of energy : calorie Every equilibrium state of a thermodynamic to joule. Since in SI units, we use the unit joule system is completely described by specific for heat, work or any other form of energy, the values of some macroscopic variables, also term mechanical equivalent is now called state variables. For example, an superfluous and need not be used. equilibrium state of a gas is completely As already remarked, the specific heat specified by the values of pressure, volume, capacity depends on the process or the temperature, and mass (and composition if conditions under which heat capacity transfer there is a mixture of gases). A thermodynamic takes place. For gases, for example, we can system is not always in equilibrium. For example, define two specific heats : specific heat a gas allowed to expand freely against vacuum capacity at constant volume and specific is not an equilibrium state [Fig. 11.6(a)]. During heat capacity at constant pressure. For an the rapid expansion, pressure of the gas may Reprint 2025-26 THERMODYNAMICS 233 not be uniform throughout. Similarly, a mixture temperature do not. To decide which variable is of gases undergoing an explosive chemical extensive and which intensive, think of a reaction (e.g. a mixture of petrol vapour and relevant system in equilibrium, and imagine that air when ignited by a spark) is not an it is divided into two equal parts. The variables that remain unchanged for each part areequilibrium state; again its temperature and intensive. The variables whose values get halvedpressure are not uniform [Fig. 11.6(b)]. in each part are extensive. It is easily seen, forEventually, the gas attains a uniform example, that internal energy U, volume V, total temperature and pressure and comes to mass M are extensive variables. Pressure P, thermal and mechanical equilibrium with its temperature T, and density ρ are intensive surroundings. variables. It is a good practice to check the consistency of thermodynamic equations using this classification of variables. For example, in the equation ∆Q = ∆U + P ∆V quantities on both sides are extensive*. (The product of an intensive variable like P and an extensive quantity ∆V is extensive.) 11.8 THERMODYNAMIC PROCESSES 11.8.1 Quasi-static process Consider a gas in thermal and mechanical equilibrium with its surroundings. The pressure of the gas in that case equals the external Fig. 11.6 (a) The partition in the box is suddenly pressure and its temperature is the same as removed leading to free expansion of the that of its surroundings. Suppose that the gas. (b) A mixture of gases undergoing an external pressure is suddenly reduced (say by explosive chemical reaction. In both lifting the weight on the movable piston in the situations, the gas is not in equilibrium and container). The piston will accelerate outward. cannot be described by state variables. During the process, the gas passes through In short, thermodynamic state variables states that are not equilibrium states. The non- describe equilibrium states of systems. The equilibrium states do not have well-defined various state variables are not necessarily pressure and temperature. In the same way, if independent. The connection between the state a finite temperature difference exists between variables is called the equation of state. For the gas and its surroundings, there will be a example, for an ideal gas, the equation of state rapid exchange of heat during which the gas is the ideal gas relation will pass through non-equilibrium states. In due course, the gas will settle to an equilibrium P V = µ R T state with well-defined temperature and For a fixed amount of the gas i.e. given µ, there pressure equal to those of the surroundings. The are thus, only two independent variables, say P free expansion of a gas in vacuum and a mixture and V or T and V. The pressure-volume curve of gases undergoing an explosive chemical for a fixed temperature is called an isotherm. reaction, mentioned in section 11.7 are also Real gases may have more complicated examples where the system goes through non- equations of state. equilibrium states. The thermodynamic state variables are of two Non-equilibrium states of a system are difficult kinds: extensive and intensive. Extensive to deal with. It is, therefore, convenient to variables indicate the ‘size’ of the system. imagine an idealised process in which at every Intensive variables such as pressure and stage the system is an equilibrium state. Such a * As emphasised earlier, Q is not a state variable. However, ∆Q is clearly proportional to the total mass of system and hence is extensive. Reprint 2025-26 234 PHYSICS process is, in principle, infinitely slow, hence the A process in which the temperature of the name quasi-static (meaning nearly static). The system is kept fixed throughout is called an system changes its variables (P, T, V ) so slowly isothermal process. The expansion of a gas in that it remains in thermal and mechanical a metallic cylinder placed in a large reservoir of equilibrium with its surroundings throughout. fixed temperature is an example of an isothermal In a quasi-static process, at every stage, the process. (Heat transferred from the reservoir to difference in the pressure of the system and the the system does not materially affect the external pressure is infinitesimally small. The temperature of the reservoir, because of its very large heat capacity.) In isobaric processes thesame is true of the temperature difference pressure is constant while in isochoricbetween the system and its surroundings processes the volume is constant. Finally, if the(Fig.11.7). To take a gas from the state (P, T ) to another state (P ′, T ′ ) via a quasi-static process, system is insulated from the surroundings and no heat flows between the system and thewe change the external pressure by a very small surroundings, the process is adiabatic. The amount, allow the system to equalise its pressure definitions of these special processes are with that of the surroundings and continue the summarised in Table. 11.2 process infinitely slowly until the system achieves the pressure P ′. Similarly, to change Table 11.2 Some special thermodynamic the temperature, we introduce an infinitesimal processes temperature difference between the system and the surrounding reservoirs and by choosing reservoirs of progressively different temperatures T to T ′, the system achieves the temperature T ′. We now consider these processes in some detail : 11.8.2 Isothermal process For an isothermal process (T fixed), the ideal gas equation gives PV = constant i.e., pressure of a given mass of gas varies inversely as its volume. This is nothing but Boyle’s Law. Suppose an ideal gas goes isothermally (at temperature T ) from its initial state (P1, V1) to Fig. 11.7 In a quasi-static process, the temperature the final state (P2, V 2). At any intermediate stage of the surrounding reservoir and the with pressure P and volume change from V to external pressure differ only infinitesimally V + ∆V (∆V small) from the temperature and pressure of the system. ∆W = P ∆ V A quasi-static process is obviously a Taking (∆V → 0) and summing the quantity hypothetical construct. In practice, processes ∆W over the entire process, that are sufficiently slow and do not involve V2 accelerated motion of the piston, large W = ∫ P d V temperature gradient, etc., are reasonably V1 approximation to an ideal quasi-static process. V2 d V V2We shall from now on deal with quasi-static = µ RT = µRT In ∫ (11.12) Vprocesses only, except when stated otherwise. V1 V 1 Reprint 2025-26 THERMODYNAMICS 235 where in the second step we have made use of We can calculate, as before, the work done in the ideal gas equation PV = µ RT and taken the an adiabatic change of an ideal gas from the constants out of the integral. For an ideal gas, state (P1, V1, T1) to the state (P2, V2, T2). internal energy depends only on temperature. V 2Thus, there is no change in the internal energy W = ∫ P d Vof an ideal gas in an isothermal process. The V1First Law of Thermodynamics then implies that heat supplied to the gas equals the work done by the gas : Q = W. Note from Eq. (11.12) that for V2 > V1, W > 0; and for V2 < V1, W < 0. That is, in an isothermal expansion, the gas absorbs heat and does work while in an isothermal (11.15) compression, work is done on the gas by the environment and heat is released. From Eq. (11.14), the constant is P1V1γ or P2V2γ 11.8.3 Adiabatic process γ  γ  P2 V 2 P1V 1In an adiabatic process, the system is insulated W = 1 − 1   −1 γ γ − from the surroundings and heat absorbed or 1 − γ  V2 V1  released is zero. From Eq. (11.1), we see that work done by the gas results in decrease in its 1 µR(T1 − T2 ) = [ P2 V2 − P1V1 ] = (11.16)internal energy (and hence its temperature for 1 − γ γ − 1 an ideal gas). We quote without proof (the result that you will learn in higher courses) that for As expected, if work is done by the gas in an an adiabatic process of an ideal gas. adiabatic process (W > 0), from Eq. (11.16), P V γ = const (11.13) T2 < T1. On the other hand, if work is done on the gas (W < 0), we get T2 > T1 i.e., thewhere γ is the ratio of specific heats (ordinary temperature of the gas rises. or molar) at constant pressure and at constant volume. 11.8.4 Isochoric process Cp In an isochoric process, V is constant. No work γ = Cv is done on or by the gas. From Eq. (11.1), the heat absorbed by the gas goes entirely to change Thus if an ideal gas undergoes a change in its internal energy and its temperature. The its state adiabatically from (P1, V1) to (P2, V2) : change in temperature for a given amount of γ = P2 V2γ (11.14) heat is determined by the specific heat of the P1 V1 gas at constant volume. Figure11.8 shows the P-V curves of an ideal gas for two adiabatic processes connecting two 11.8.5 Isobaric process isotherms. In an isobaric process, P is fixed. Work done by the gas is W = P (V2 – V1) = µ R (T2 – T1) (11.17) Since temperature changes, so does internal energy. The heat absorbed goes partly to increase internal energy and partly to do work. The change in temperature for a given amount of heat is determined by the specific heat of the gas at constant pressure. 11.8.6 Cyclic process In a cyclic process, the system returns to its initial state. Since internal energy is a stateFig. 11.8 P-V curves for isothermal and adiabatic variable, ∆U = 0 for a cyclic process. From processes of an ideal gas. Reprint 2025-26 236 PHYSICS Eq. (11.1), the total heat absorbed equals the 11.10 REVERSIBLE AND IRREVERSIBLE work done by the system. PROCESSES Imagine some process in which a thermodynamic11.9 SECOND LAW OF THERMODYNAMICS system goes from an initial state i to a final state The First Law of Thermodynamics is the f. During the process the system absorbs heat Q principle of conservation of energy. Common from the surroundings and performs work W on experience shows that there are many it. Can we reverse this process and bring both conceivable processes that are perfectly the system and surroundings to their initial allowed by the First Law and yet are never states with no other effect anywhere ? Experience observed. For example, nobody has ever seen suggests that for most processes in nature this a book lying on a table jumping to a height by is not possible. The spontaneous processes ofitself. But such a thing would be possible if nature are irreversible. Several examples can bethe principle of conservation of energy were cited. The base of a vessel on an oven is hotterthe only restriction. The table could cool than its other parts. When the vessel is removed,spontaneously, converting some of its internal heat is transferred from the base to the otherenergy into an equal amount of mechanical parts, bringing the vessel to a uniformenergy of the book, which would then hop to a height with potential energy equal to the temperature (which in due course cools to the mechanical energy it acquired. But this never temperature of the surroundings). The process happens. Clearly, some additional basic cannot be reversed; a part of the vessel will not principle of nature forbids the above, even get cooler spontaneously and warm up the base. though it satisfies the energy conservation It will violate the Second Law of Thermodynamics, principle. This principle, which disallows if it did. The free expansion of a gas is irreversible. many phenomena consistent with the First The combustion reaction of a mixture of petrol Law of Thermodynamics is known as the and air ignited by a spark cannot be reversed. Second Law of Thermodynamics. Cooking gas leaking from a gas cylinder in the The Second Law of Thermodynamics gives kitchen diffuses to the entire room. The diffusion a fundamental limitation to the efficiency of a process will not spontaneously reverse and bring heat engine and the co-efficient of the gas back to the cylinder. The stirring of a performance of a refrigerator. In simple terms, liquid in thermal contact with a reservoir will it says that efficiency of a heat engine can convert the work done into heat, increasing the never be unity. For a refrigerator, the Second internal energy of the reservoir. The process Law says that the co-efficient of performance cannot be reversed exactly; otherwise it would can never be infinite. The following two amount to conversion of heat entirely into work, statements, one due to Kelvin and Planck violating the Second Law of Thermodynamics. denying the possibility of a perfect heat engine, Irreversibility is a rule rather an exception and another due to Clausius denying the in nature. possibility of a perfect refrigerator or heat Irreversibility arises mainly from two causes: pump, are a concise summary of these one, many processes (like a free expansion, or observations. an explosive chemical reaction) take the system Kelvin-Planck statement to non-equilibrium states; two, most processes No process is possible whose sole result is the involve friction, viscosity and other dissipative absorption of heat from a reservoir and the effects (e.g., a moving body coming to a stop and complete conversion of the heat into work. losing its mechanical energy as heat to the floor and the body; a rotating blade in a liquid coming Clausius statement to a stop due to viscosity and losing its No process is possible whose sole result is the mechanical energy with corresponding gain in transfer of heat from a colder object to a the internal energy of the liquid). Since hotter object. dissipative effects are present everywhere and It can be proved that the two statements can be minimised but not fully eliminated, most above are completely equivalent. processes that we deal with are irreversible. Reprint 2025-26 THERMODYNAMICS 237 A thermodynamic process (state i → state f ) in a reversible heat engine operating between is reversible if the process can be turned back two temperatures, heat should be absorbed such that both the system and the surroundings (from the hot reservoir) isothermally and return to their original states, with no other released (to the cold reservoir) isothermally. We change anywhere else in the universe. From the thus have identified two steps of the reversible preceding discussion, a reversible process is an heat engine : isothermal process at temperature idealised notion. A process is reversible only if T1 absorbing heat Q1 from the hot reservoir, and another isothermal process at temperature T2it is quasi-static (system in equilibrium with the releasing heat Q2 to the cold reservoir. Tosurroundings at every stage) and there are no complete a cycle, we need to take the systemdissipative effects. For example, a quasi-static from temperature T1 to T2 and then back fromisothermal expansion of an ideal gas in a temperature T2 to T1. Which processes shouldcylinder fitted with a frictionless movable piston we employ for this purpose that are reversible?is a reversible process. A little reflection shows that we can only adopt Why is reversibility such a basic concept in reversible adiabatic processes for these thermodynamics ? As we have seen, one of the purposes, which involve no heat flow from any concerns of thermodynamics is the efficiency reservoir. If we employ any other process that is with which heat can be converted into work. not adiabatic, say an isochoric process, to take The Second Law of Thermodynamics rules out the system from one temperature to another, we the possibility of a perfect heat engine with 100% shall need a series of reservoirs in the efficiency. But what is the highest efficiency temperature range T2 to T1 to ensure that at each possible for a heat engine working between two stage the process is quasi-static. (Remember reservoirs at temperatures T1 and T2? It turns again that for a process to be quasi-static and out that a heat engine based on idealised reversible, there should be no finite temperature reversible processes achieves the highest difference between the system and the reservoir.) efficiency possible. All other engines involving But we are considering a reversible engine that irreversibility in any way (as would be the case operates between only two temperatures. Thus for practical engines) have lower than this adiabatic processes must bring about the limiting efficiency. temperature change in the system from T1 to T2 and T2 to T1 in this engine.11.11 CARNOT ENGINE Suppose we have a hot reservoir at temperature T1 and a cold reservoir at temperature T2. What is the maximum efficiency possible for a heat engine operating between the two reservoirs and what cycle of processes should be adopted to achieve the maximum efficiency ? Sadi Carnot, a French engineer, first considered this question in 1824. Interestingly, Carnot arrived at the correct answer, even though the basic concepts of heat and thermodynamics had yet to be firmly established. We expect the ideal engine operating between two temperatures to be a reversible engine. Irreversibility is associated with dissipative Fig. 11.9 Carnot cycle for a heat engine with an effects, as remarked in the preceding section, ideal gas as the working substance. and lowers efficiency. A process is reversible if it is quasi-static and non-dissipative. We have A reversible heat engine operating between seen that a process is not quasi-static if it two temperatures is called a Carnot engine. We involves finite temperature difference between have just argued that such an engine must have the system and the reservoir. This implies that the following sequence of steps constituting one Reprint 2025-26 238 PHYSICS cycle, called the Carnot cycle, shown in Fig. 11.9. We have taken the working substance In  V 3  of the Carnot engine to be an ideal gas. T2   V4  = 1 −  T1   V2  (11.23)(a) Step 1 → 2 Isothermal expansion of the gas In  V1  taking its state from (P1, V1, T1) to (P2, V2, T1). Now since step 2 → 3 is an adiabatic process, The heat absorbed by the gas (Q1) from the γ −1 γ −1 reservoir at temperature T1 is given by T1 V 2 = T2 V3 Eq. (11.12). This is also the work done (W1 → 2) by the gas on the environment. 1 /( γ −1 ) V2  T2   V2  i.e. = (11.24) 3  T1  W1 → 2 = Q1 = µ R T1 ln  V1  (11.18) V (b) Step 2 → 3 Adiabatic expansion of the gas Similarly, since step 4 → 1 is an adiabatic from (P2, V2, T1) to (P3, V3, T2) process Work done by the gas, using γ −1 γ −1 Eq. (11.16), is T2 V 4 = T1 V1 µR ( T1 − T2 ) 1 /γ −1 W2 →=3 (11.19) V1  T2  γ − 1 i.e. = (11.25) V4  T1  (c) Step 3 → 4 Isothermal compression of the gas from (P3, V3, T2) to (P4, V4, T2). From Eqs. (11.24) and (11.25), 2 Heat released (Q2) by the gas to the reservoir V3 = V (11.26) at temperature T2 is given by Eq. (11.12). This V4 V1 is also the work done (W3 → 4) on the gas by the environment. Using Eq. (11.26) in Eq. (11.23), we get  V 3  2 W 3 → 4 = Q2 = µRT2 ln η = 1 −T (Carnot engine) (11.27)  V 4  (11.20) T1 (d) Step 4 → 1 Adiabatic compression of the gas from (P4, V4, T2) to (P1,V1, T1). We have already seen that a Carnot engine is a reversible engine. Indeed it is the only Work done on the gas, [using Eq.(11.16), is reversible engine possible that works between two reservoirs at different temperatures. Each  T1 − T2  W4 → 1 = µR (11.21) step of the Carnot cycle given in Fig. 11.9 can  γ -1  be reversed. This will amount to taking heat Q2 from the cold reservoir at T2, doing work W on From Eqs. (11.18) to (11.21) total work done by the gas in one complete cycle is the system, and transferring heat Q1 to the hot reservoir. This will be a reversible refrigerator. W = W1 → 2 + W2 → 3 – W3 → 4 – W4 → 1 We next establish the important result  V2   V3  (sometimes called Carnot’s theorem) that = µ RT1 ln  V1  – µ RT2 ln  V4  (11.22) (a) working between two given temperatures T1 and T2 of the hot and cold reservoirs respectively, The efficiency η of the Carnot engine is no engine can have efficiency more than that of the Carnot engine and (b) the efficiency of the W Q 2 η = = 1 − Carnot engine is independent of the nature of Q1 Q1 the working substance. Reprint 2025-26 THERMODYNAMICS 239 To prove the result (a), imagine a reversible reservoir and delivers the same amount of work (Carnot) engine R and an irreversible engine I in one cycle, without any change in the source working between the same source (hot reservoir) or anywhere else. This is clearly against the and sink (cold reservoir). Let us couple the Kelvin-Planck statement of the Second Law of engines, I and R, in such a way so that I acts Thermodynamics. Hence the assertion ηI > ηR like a heat engine and R acts as a refrigerator. is wrong. No engine can have efficiency greater Let I absorb heat Q1 from the source, deliver than that of the Carnot engine. A similar work W ′ and release the heat Q1- W′ to the sink. argument can be constructed to show that a We arrange so that R returns the same heat Q1 reversible engine with one particular substance to the source, taking heat Q2 from the sink and cannot be more efficient than the one using requiring work W = Q1 – Q2 to be done on it. another substance. The maximum efficiency of Now suppose ηR < ηI i.e. if R were to act a Carnot engine given by Eq. (11.27) is as an engine it would give less work output independent of the nature of the system performing the Carnot cycle of operations. Thus we are justified in using an ideal gas as a system in the calculation of efficiency η of a Carnot engine. The ideal gas has a simple equation of I state, which allows us to readily calculate η, but the final result for η, [Eq. (11.27)], is true for any Carnot engine. R This final remark shows that in a Carnot cycle, Q1 T1 = W (11.28) T2 Q 2 is a universal relation independent of the natureFig. 11.10 An irreversible engine (I) coupled to a reversible refrigerator (R). If W ′ > W, this of the system. Here Q1 and Q2 are respectively, would amount to extraction of heat the heat absorbed and released isothermally W′ – W from the sink and its full (from the hot and to the cold reservoirs) in a conversion to work, in contradiction with Carnot engine. Equation (11.28), can, therefore, the Second Law of Thermodynamics. be used as a relation to define a truly universal thermodynamic temperature scale that is than that of I i.e. W < W ′ for a given Q1. With R independent of any particular properties of the acting like a refrigerator, this would mean system used in the Carnot cycle. Of course, for Q2 = Q1 – W > Q1 – W ′. Thus, on the whole, an ideal gas as a working substance, this the coupled I-R system extracts heat universal temperature is the same as the ideal (Q1 – W) – (Q1 – W ′) = (W ′ – W ) from the cold gas temperature introduced in section 11.9. SUMMARY 1. The zeroth law of thermodynamics states that ‘two systems in thermal equilibrium with a third system separately are in thermal equilibrium with each other’. The Zeroth Law leads to the concept of temperature. 2. Internal energy of a system is the sum of kinetic energies and potential energies of the molecular constituents of the system. It does not include the over-all kinetic energy of the system. Heat and work are two modes of energy transfer to the system. Heat is the energy transfer arising due to temperature difference between the system and the surroundings. Work is energy transfer brought about by other means, such as moving the piston of a cylinder containing the gas, by raising or lowering some weight connected to it. Reprint 2025-26 240 PHYSICS 3. The first law of thermodynamics is the general law of conservation of energy applied to any system in which energy transfer from or to the surroundings (through heat and work) is taken into account. It states that ∆Q = ∆U + ∆W where ∆Q is the heat supplied to the system, ∆W is the work done by the system and ∆U is the change in internal energy of the system. 4. The specific heat capacity of a substance is defined by 1 ∆Q s = m ∆T where m is the mass of the substance and ∆Q is the heat required to change its temperature by ∆T. The molar specific heat capacity of a substance is defined by 1 ∆ Q C = µ ∆T where µ is the number of moles of the substance. For a solid, the law of equipartition of energy gives C = 3 R which generally agrees with experiment at ordinary temperatures. Calorie is the old unit of heat. 1 calorie is the amount of heat required to raise the temperature of 1 g of water from 14.5 °C to 15.5 °C. 1 cal = 4.186 J. 5. For an ideal gas, the molar specific heat capacities at constant pressure and volume satisfy the relation Cp – Cv = R where R is the universal gas constant. 6. Equilibrium states of a thermodynamic system are described by state variables. The value of a state variable depends only on the particular state, not on the path used to arrive at that state. Examples of state variables are pressure (P ), volume (V ), temperature (T ), and mass (m ). Heat and work are not state variables. An Equation of State (like the ideal gas equation PV = µ RT ) is a relation connecting different state variables. 7. A quasi-static process is an infinitely slow process such that the system remains in thermal and mechanical equilibrium with the surroundings throughout. In a quasi-static process, the pressure and temperature of the environment can differ from those of the system only infinitesimally. 8. In an isothermal expansion of an ideal gas from volume V1 to V2 at temperature T the heat absorbed (Q) equals the work done (W ) by the gas, each given by  V 2   Q = W = µ R T ln   V1  9. In an adiabatic process of an ideal gas γ PV = constant C p where γ = C v Work done by an ideal gas in an adiabatic change of state from (P1, V1, T1) to (P2, V2, T2) is µR ( T1 − T2 ) W = γ – 1 Reprint 2025-26 THERMODYNAMICS 241 10. The second law of thermodynamics disallows some processes consistent with the First Law of Thermodynamics. It states Kelvin-Planck statement No process is possible whose sole result is the absorption of heat from a reservoir and complete conversion of the heat into work. Clausius statement No process is possible whose sole result is the transfer of heat from a colder object to a hotter object. Put simply, the Second Law implies that no heat engine can have efficiency η equal to 1 or no refrigerator can have co-efficient of performance α equal to infinity. 11. A process is reversible if it can be reversed such that both the system and the surroundings return to their original states, with no other change anywhere else in the universe. Spontaneous processes of nature are irreversible. The idealised reversible process is a quasi-static process with no dissipative factors such as friction, viscosity, etc. 12. Carnot engine is a reversible engine operating between two temperatures T1 (source) and T2 (sink). The Carnot cycle consists of two isothermal processes connected by two adiabatic processes. The efficiency of a Carnot engine is given by T 2 η= 1 − (Carnot engine) T1 No engine operating between two temperatures can have efficiency greater than that of the Carnot engine. 13. If Q > 0, heat is added to the system If Q < 0, heat is removed to the system If W > 0, Work is done by the system If W < 0, Work is done on the system Quantity Symbol Dimensions Unit Remark Co-efficienty of volume αv [K–1] K–1 αv = 3 α1 expansion Heat supplied to a system ∆Q [ML2 T–2] J Q is not a state variable Specific heat capacity s [L2 T–2 K–1] J kg–1 K–1 dt Thermal Conductivity K [MLT–3 K–1] J s–1 K–1 H = – KA dx POINTS TO PONDER 1. Temperature of a body is related to its average internal energy, not to the kinetic energy of motion of its centre of mass. A bullet fired from a gun is not at a higher temperature because of its high speed. 2. Equilibrium in thermodynamics refers to the situation when macroscopic variables describing the thermodynamic state of a system do not depend on time. Equilibrium of a system in mechanics means the net external force and torque on the system are zero. Reprint 2025-26 242 PHYSICS 3. In a state of thermodynamic equilibrium, the microscopic constituents of a system are not in equilibrium (in the sense of mechanics). 4. Heat capacity, in general, depends on the process the system goes through when heat is supplied. 5. In isothermal quasi-static processes, heat is absorbed or given out by the system even though at every stage the gas has the same temperature as that of the surrounding reservoir. This is possible because of the infinitesimal difference in temperature between the system and the reservoir. EXERCISES 11.1 A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 × 104 J/g ? 11.2 What amount of heat must be supplied to 2.0 × 10–2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure ? (Molecular mass of N2 = 28; R = 8.3 J mol–1 K–1.) 11.3 Explain why (a) Two bodies at different temperatures T1 and T2 if brought in thermal contact do not necessarily settle to the mean temperature (T1 + T2 )/2. (b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat. (c) Air pressure in a car tyre increases during driving. (d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude. 11.4 A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume ? 11.5 In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case ? (Take 1 cal = 4.19 J) 11.6 Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following : (a) What is the final pressure of the gas in A and B ? (b) What is the change in internal energy of the gas ? (c) What is the change in the temperature of the gas ? (d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface ? 11.7 An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing? Reprint 2025-26 THERMODYNAMICS 243 11.8 A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (11.13) Fig. 11.11 Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F Reprint 2025-26 CHAPTER TWELVE KINETIC THEORY 12.1 INTRODUCTION Boyle discovered the law named after him in 1661. Boyle, Newton and several others tried to explain the behaviour of 12.1 Introduction gases by considering that gases are made up of tiny atomic 12.2 Molecular nature of matter particles. The actual atomic theory got established more than 12.3 Behaviour of gases 150 years later. Kinetic theory explains the behaviour of gases 12.4 Kinetic theory of an ideal gas based on the idea that the gas consists of rapidly moving 12.5 Law of equipartition of energy atoms or molecules. This is possible as the inter-atomic forces, 12.6 Specific heat capacity which are short range forces that are important for solids and liquids, can be neglected for gases. The kinetic theory12.7 Mean free path was developed in the nineteenth century by Maxwell, Summary Boltzmann and others. It has been remarkably successful. It Points to ponder gives a molecular interpretation of pressure and temperature Exercises of a gas, and is consistent with gas laws and Avogadro’s hypothesis. It correctly explains specific heat capacities of many gases. It also relates measurable properties of gases such as viscosity, conduction and diffusion with molecular parameters, yielding estimates of molecular sizes and masses. This chapter gives an introduction to kinetic theory. 12.2 MOLECULAR NATURE OF MATTER Richard Feynman, one of the great physicists of 20th century considers the discovery that “Matter is made up of atoms” to be a very significant one. Humanity may suffer annihilation (due to nuclear catastrophe) or extinction (due to environmental disasters) if we do not act wisely. If that happens, and all of scientific knowledge were to be destroyed then Feynman would like the ‘Atomic Hypothesis’ to be communicated to the next generation of creatures in the universe. Atomic Hypothesis: All things are made of atoms - little particles that move around in perpetual motion, attracting each other when they are a little distance apart, but repelling upon being squeezed into one another. Speculation that matter may not be continuous, existed in many places and cultures. Kanada in India and Democritus Reprint 2025-26 KINETIC THEORY 245 Atomic Hypothesis in Ancient India and Greece Though John Dalton is credited with the introduction of atomic viewpoint in modern science, scholars in ancient India and Greece conjectured long before the existence of atoms and molecules. In the Vaiseshika school of thought in India founded by Kanada (Sixth century B.C.) the atomic picture was developed in considerable detail. Atoms were thought to be eternal, indivisible, infinitesimal and ultimate parts of matter. It was argued that if matter could be subdivided without an end, there would be no difference between a mustard seed and the Meru mountain. The four kinds of atoms (Paramanu — Sanskrit word for the smallest particle) postulated were Bhoomi (Earth), Ap (water), Tejas (fire) and Vayu (air) that have characteristic mass and other attributes, were propounded. Akasa (space) was thought to have no atomic structure and was continuous and inert. Atoms combine to form different molecules (e.g. two atoms combine to form a diatomic molecule dvyanuka, three atoms form a tryanuka or a triatomic molecule), their properties depending upon the nature and ratio of the constituent atoms. The size of the atoms was also estimated, by conjecture or by methods that are not known to us. The estimates vary. In Lalitavistara, a famous biography of the Buddha written mainly in the second century B.C., the estimate is close to the modern estimate of atomic size, of the order of 10 –10 m. In ancient Greece, Democritus (Fourth century B.C.) is best known for his atomic hypothesis. The word ‘atom’ means ‘indivisible’ in Greek. According to him, atoms differ from each other physically, in shape, size and other properties and this resulted in the different properties of the substances formed by their combination. The atoms of water were smooth and round and unable to ‘hook’ on to each other, which is why liquid /water flows easily. The atoms of earth were rough and jagged, so they held together to form hard substances. The atoms of fire were thorny which is why it caused painful burns. These fascinating ideas, despite their ingenuity, could not evolve much further, perhaps because they were intuitive conjectures and speculations not tested and modified by quantitative experiments - the hallmark of modern science. in Greece had suggested that matter may consist of matter. The theory is now well accepted by of indivisible constituents. The scientific ‘Atomic scientists. However even at the end of the Theory’ is usually credited to John Dalton. He nineteenth century there were famous scientists proposed the atomic theory to explain the laws who did not believe in atomic theory ! of definite and multiple proportions obeyed by From many observations, in recent times we elements when they combine into compounds. now know that molecules (made up of one or The first law says that any given compound has, more atoms) constitute matter. Electron a fixed proportion by mass of its constituents. microscopes and scanning tunnelling The second law says that when two elements microscopes enable us to even see them. The form more than one compound, for a fixed mass size of an atom is about an angstrom (10 -10 m). of one element, the masses of the other elements In solids, which are tightly packed, atoms are are in ratio of small integers. spaced about a few angstroms (2 Å) apart. In To explain the laws Dalton suggested, about liquids the separation between atoms is also 200 years ago, that the smallest constituents about the same. In liquids the atoms are not of an element are atoms. Atoms of one element as rigidly fixed as in solids, and can move are identical but differ from those of other around. This enables a liquid to flow. In gases elements. A small number of atoms of each the interatomic distances are in tens of element combine to form a molecule of the angstroms. The average distance a molecule compound. Gay Lussac’s law, also given in early can travel without colliding is called the mean 19th century, states: When gases combine free path. The mean free path, in gases, is of chemically to yield another gas, their volumes the order of thousands of angstroms. The atoms are in the ratios of small integers. Avogadro’s are much freer in gases and can travel long law (or hypothesis) says: Equal volumes of all distances without colliding. If they are not gases at equal temperature and pressure have enclosed, gases disperse away. In solids and the same number of molecules. Avogadro’s law, liquids the closeness makes the interatomic force when combined with Dalton’s theory explains important. The force has a long range attraction Gay Lussac’s law. Since the elements are often and a short range repulsion. The atoms attract in the form of molecules, Dalton’s atomic theory when they are at a few angstroms but repel when can also be referred to as the molecular theory they come closer. The static appearance of a gas Reprint 2025-26 246 PHYSICS is misleading. The gas is full of activity and the is 6.02 × 1023. This is known as Avogadro number equilibrium is a dynamic one. In dynamic and is denoted by NA. The mass of 22.4 litres of equilibrium, molecules collide and change their any gas is equal to its molecular weight in grams speeds during the collision. Only the average at S.T.P (standard temperature 273 K and properties are constant. pressure 1 atm). This amount of substance is Atomic theory is not the end of our quest, but called a mole (see Chapter 1 for a more precise the beginning. We now know that atoms are not definition). Avogadro had guessed the equality of indivisible or elementary. They consist of a numbers in equal volumes of gas at a fixed nucleus and electrons. The nucleus itself is made temperature and pressure from chemical up of protons and neutrons. The protons and reactions. Kinetic theory justifies this hypothesis. neutrons are again made up of quarks. Even The perfect gas equation can be written as quarks may not be the end of the story. There PV = µ RT (12.3)may be string like elementary entities. Nature always has surprises for us, but the search for where µ is the number of moles and R = NA truth is often enjoyable and the discoveries kB is a universal constant. The temperature T is absolute temperature. Choosing kelvin scale forbeautiful. In this chapter, we shall limit ourselves absolute temperature, R = 8.314 J mol–1K–1.to understanding the behaviour of gases (and a Herelittle bit of solids), as a collection of moving molecules in incessant motion. M N µ = = (12.4) M 0 N A

10.6 SPECIFIC HEAT CAPACITY heat absorbed or given off to change the temperature of unit mass of it by one unit. ThisTake some water in a vessel and start heating it quantity is referred to as the specific heaton a burner. Soon you will notice that bubbles capacity of the substance.begin to move upward. As the temperature is If ∆Q stands for the amount of heat absorbedraised the motion of water particles increases or given off by a substance of mass m when ittill it becomes turbulent as water starts boiling. undergoes a temperature change ∆T, then theWhat are the factors on which the quantity of specific heat capacity, of that substance is givenheat required to raise the temperature of a bysubstance depend? In order to answer this question in the first step, heat a given quantity S 1 ∆ Q s = = (10.11)of water to raise its temperature by, say 20 °C m m ∆ T and note the time taken. Again take the same The specific heat capacity is the property of amount of water and raise its temperature by the substance which determines the change in 40 °C using the same source of heat. Note the the temperature of the substance (undergoing time taken by using a stopwatch. You will find it no phase change) when a given quantity of heat takes about twice the time and therefore, double is absorbed (or given off) by it. It is defined as the the quantity of heat required raising twice the amount of heat per unit mass absorbed or given temperature of same amount of water. off by the substance to change its temperature In the second step, now suppose you take by one unit. It depends on the nature of the double the amount of water and heat it, using substance and its temperature. The SI unit of the same heating arrangement, to raise the specific heat capacity is J kg–1 K–1. temperature by 20 °C, you will find the time taken If the amount of substance is specified in is again twice that required in the first step. terms of moles µ, instead of mass m in kg, we In the third step, in place of water, now heat can define heat capacity per mole of the the same quantity of some oil, say mustard oil, substance by and raise the temperature again by 20 °C. Now note the time by the same stopwatch. You will (10.12) find the time taken will be shorter and therefore, where C is known as molar specific heat the quantity of heat required would be less than capacity of the substance. Like S, C also that required by the same amount of water for depends on the nature of the substance and its the same rise in temperature. temperature. The SI unit of molar specific heat The above observations show that the quantity capacity is J mol–1 K–1. of heat required to warm a given substance However, in connection with specific heat depends on its mass, m, the change in capacity of gases, additional conditions may betemperature, ∆T and the nature of substance. needed to define C. In this case, heat transferThe change in temperature of a substance, when can be achieved by keeping either pressure ora given quantity of heat is absorbed or rejected volume constant. If the gas is held underby it, is characterised by a quantity called the constant pressure during the heat transfer, thenheat capacity of that substance. We define heat it is called the molar specific heat capacity atcapacity, S of a substance as constant pressure and is denoted by Cp. On ∆Q the other hand, if the volume of the gas is S = (10.10) ∆T maintained during the heat transfer, then the where ∆Q is the amount of heat supplied to corresponding molar specific heat capacity is the substance to change its temperature from T called molar specific heat capacity at constant to T + ∆T. volume and is denoted by Cv. For details see You have observed that if equal amount of Chapter 11. Table 10.3 lists measured specific heat is added to equal masses of different heat capacity of some substances at atmospheric substances, the resulting temperature changes pressure and ordinary temperature while Table will not be the same. It implies that every 10.4 lists molar specific heat capacities of some substance has a unique value for the amount of gases. From Table 10.3 you can note that water Reprint 2025-26 THERMAL PROPERTIES OF MATTER 209 Table 10.3 Specific heat capacity of some substances at room temperature and atmospheric pressure Substance Specific heat capacity Substance Specific heat capacity (J kg–1 K–1) (J kg–1 K–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 Tungesten 134.4 Mercury 140 Water 4186.0 has the highest specific heat capacity compared equal to the heat gained by the colder body, to other substances. For this reason water is also provided no heat is allowed to escape to the used as a coolant in automobile radiators, as surroundings. A device in which heat well as, a heater in hot water bags. Owing to its measurement can be done is called a high specific heat capacity, water warms up calorimeter. It consists of a metallic vessel and more slowly than land during summer, and stirrer of the same material, like copper or consequently wind from the sea has a cooling aluminium. The vessel is kept inside a wooden effect. Now, you can tell why in desert areas, jacket, which contains heat insulating material, the earth surface warms up quickly during the like glass wool etc. The outer jacket acts as a day and cools quickly at night. heat shield and reduces the heat loss from the Table 10.4 Molar specific heat capacities of inner vessel. There is an opening in the outer some gases jacket through which a mercury thermometer can be inserted into the calorimeter (Fig. 10.20). Gas Cp (J mol–1K–1) Cv(J mol–1K–1) The following example provides a method by He 20.8 12.5 which the specific heat capacity of a given solid can be determinated by using the principle, heat H2 28.8 20.4 gained is equal to the heat lost. N2 29.1 20.8 ⊳ Example 10.3 A sphere of 0.047 kg O2 29.4 21.1 aluminium is placed for sufficient time in a CO2 37.0 28.5 vessel containing boiling water, so that the sphere is at 100 °C. It is then immediately transfered to 0.14 kg copper calorimeter 10.7 CALORIMETRY containing 0.25 kg water at 20 °C. The A system is said to be isolated if no exchange or temperature of water rises and attains a transfer of heat occurs between the system and steady state at 23 °C. Calculate the specific its surroundings. When different parts of an heat capacity of aluminium. isolated system are at different temperature, a quantity of heat transfers from the part at higher Answer In solving this example, we shall use temperature to the part at lower temperature. the fact that at a steady state, heat given by an The heat lost by the part at higher temperature aluminium sphere will be equal to the heat is equal to the heat gained by the part at lower absorbed by the water and calorimeter. temperature. Mass of aluminium sphere (m1) = 0.047 kg Calorimetry means measurement of heat. Initial temperature of aluminium sphere =100 °C When a body at higher temperature is brought Final temperature = 23 °C in contact with another body at lower Change in temperature (∆T)=(100 °C-23°C)= 77 °C temperature, the heat lost by the hot body is Let specific heat capacity of aluminium be sAl. Reprint 2025-26 210 PHYSICS The amount of heat lost by the aluminium sphere = m 1s Al ∆ T = 0.047kg × s Al × 77 °C Mass of water (m2) = 0.25 kg Mass of calorimeter (m3) = 0.14 kg Initial temperature of water and calorimeter=20 °C Final temperature of the mixture = 23 °C Change in temperature (∆T2) = 23 °C – 20 °C = 3 °C Specific heat capacity of water (sw) = 4.18 × 103 J kg–1 K–1 Specific heat capacity of copper calorimeter = 0.386 × 103 J kg–1 K–1 The amount of heat gained by water and calorimeter = m2 sw ∆T2 + m3scu∆T2 Fig. 10.9 A plot of temperature versus time showing = (m2sw + m3scu) (∆T2) the changes in the state of ice on heating = (0.25 kg × 4.18 × 103 J kg–1 K–1 + 0.14 kg × (not to scale). 0.386 × 103 J kg–1 K–1) (23 °C – 20 °C) The change of state from solid to liquid is called In the steady state heat lost by the aluminium melting or fusion and from liquid to solid is calledsphere = heat gained by water + heat gained by freezing. It is observed that the temperaturecalorimeter. remains constant until the entire amount of the So, 0.047 kg × sAl × 77 °C solid substance melts. That is, both the solid and = (0.25 kg × 4.18 × 103 J kg–1 K–1+ 0.14 kg × the liquid states of the substance coexist in 0.386 × 103 J kg–1 K–1)(3 °C) thermal equilibrium during the change of sAl = 0.911 kJ kg –1 K–1 ⊳ states from solid to liquid. The temperature at which the solid and the liquid states of the substance is in thermal equilibrium with each10.8 CHANGE OF STATE other is called its melting point. It is Matter normally exists in three states: solid, characteristic of the substance. It also depends liquid and gas. A transition from one of these on pressure. The melting point of a substance states to another is called a change of state. Two at standard atomspheric pressure is called its common changes of states are solid to liquid normal melting point. Let us do the following and liquid to gas (and, vice versa). These changes activity to understand the process of melting can occur when the exchange of heat takes place of ice. between the substance and its surroundings. Take a slab of ice. Take a metallic wire and To study the change of state on heating or fix two blocks, say 5 kg each, at its ends. Put cooling, let us perform the following activity. the wire over the slab as shown in Fig. 10.10. Take some cubes of ice in a beaker. Note the You will observe that the wire passes through temperature of ice. Start heating it slowly on a the ice slab. This happens due to the fact that constant heat source. Note the temperature after just below the wire, ice melts at lower every minute. Continuously stir the mixture of temperature due to increase in pressure. When water and ice. Draw a graph between the wire has passed, water above the wire freezes temperature and time (Fig. 10.9). You will observe again. Thus, the wire passes through the slab no change in the temperature as long as there and the slab does not split. This phenomenon is ice in the beaker. In the above process, the of refreezing is called regelation. Skating is temperature of the system does not change even possible on snow due to the formation of water though heat is being continuously supplied. The under the skates. Water is formed due to the heat supplied is being utilised in changing the increase of pressure and it acts as a state from solid (ice) to liquid (water). lubricant. Reprint 2025-26 THERMAL PROPERTIES OF MATTER 211 100 °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 vaporisation. 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 in thermal equilibrium, during the change of state from liquid to vapour. The temperature at which the liquid and the vapour states of the substance coexist is called Fig. 10.10 its boiling point. Let us do the following activity After the whole of ice gets converted into water to understand the process of boiling of water. and as we continue further heating, we shall see Take a round-bottom flask, more than halfthat temperature begins to rise (Fig.10.9). The temperature keeps on rising till it reaches nearly filled with water. Keep it over a burner and fix a 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 into a solid-region, the vapour-region and the liquid-region. The regions are separated by the curves such as sublimation curve (BO), fusion curve (AO) and vaporisation curve (CO). The points on sublimation curve represent states in which solid and vapour phases coexist. The point on the sublimation curve BO represent states in which the solid and vapour phases co-exist. Points on the fusion curve AO represent states in which solid and liquid phase coexist. Points on the vapourisation curve CO represent states in which the liquid and vapour phases coexist. The temperature and pressure at which the fusion curve, the vaporisation 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 pressure 6.11×10–3 Pa. (a) (b) Figure : Pressure-temperature phase diagrams for (a) water and (b) CO2 (not to the scale). Reprint 2025-26 212 PHYSICS thermometer and steam outlet through the cork cork. Keep the flask turned upside down on the of the flask (Fig. 10.11). As water gets heated in stand. Pour ice-cold water on the flask. Water the flask, note first that the air, which was vapours in the flask condense reducing the dissolved in the water, will come out as small pressure on the water surface inside the flask. bubbles. Later, bubbles of steam will form at Water begins to boil again, now at a lower the bottom but as they rise to the cooler water temperature. Thus boiling point decreases with near the top, they condense and disappear. decrease in pressure. Finally, as the temperature of the entire mass This explains why cooking is difficult on hills. of the water reaches 100 °C, bubbles of steam At high altitudes, atmospheric pressure is lower, reach the surface and boiling is said to occur. reducing the boiling point of water as compared The steam in the flask may not be visible but as to that at sea level. On the other hand, boiling it comes out of the flask, it condenses as tiny point is increased inside a pressure cooker by droplets of water, giving a foggy appearance. 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. 10.8.1 Latent Heat In Section 10.8, 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 Fig. 10.11 Boiling process. is added to a given quantity of ice at –10 °C, the temperature of ice increases until it reaches its If now the steam outlet is closed for a few melting point (0 °C). At this temperature, the seconds to increase the pressure in the flask, addition of more heat does not increase the you will notice that boiling stops. More heat temperature but causes the ice to melt, or would be required to raise the temperature changes its state. Once the entire ice melts, (depending on the increase in pressure) before adding more heat will cause the temperature of boiling begins again. Thus boiling point increases the water to rise. A similar situation with increase in pressure. occurs during liquid gas change of state at the Let us now remove the burner. Allow water to boiling point. Adding more heat to boiling water cool to about 80 °C. Remove the thermometer and causes vaporisation, without increase in steam outlet. Close the flask with the airtight temperature. Reprint 2025-26 THERMAL PROPERTIES OF MATTER 213 Table 10.5 Temperatures of the change of state and latent heats for various substances at 1 atm pressure Substance Melting Lf Boiling Lv Point (°C) (105J kg–1) Point (°C) (105J kg–1) Ethanol –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 Note that when heat is added (or removed) depends upon the heat of transformation and during a change of state, the temperature the mass of the substance undergoing a change remains constant. Note in Fig. 10.12 that the of state. Thus, if mass m of a substance slopes of the phase lines are not all the same, undergoes a change from one state to the other, which indicate that specific heats of the various then the quantity of heat required is given by states are not equal. For water, the latent heat of Q = m L fusion and vaporisation are Lf = 3.33 × 105 J kg–1 or L = Q/m (10.13) and Lv = 22.6 × 105 J kg–1, respectively. That is, where L is known as latent heat and is a 3.33 × 105 J of heat is needed to melt 1 kg ice at characteristic of the substance. Its SI unit is 0 °C, and 22.6 × 105 J of heat is needed to convert J kg–1. The value of L also depends on the 1 kg water into steam at 100 °C. So, steam at pressure. Its value is usually quoted at standard 100 °C carries 22.6 × 105 J kg–1 more heat than atmospheric pressure. The latent heat for a solid- water at 100 °C. This is why burns from steam liquid state change is called the latent heat of are usually more serious than those from fusion (Lf), and that for a liquid-gas state change boiling water. is called the latent heat of vaporisation (Lv). ⊳ These are often referred to as the heat of fusion Example 10.4 When 0.15 kg of ice at 0 °C and the heat of vaporisation. A plot of is mixed with 0.30 kg of water at 50 °C in a temperature versus heat for a quantity of water container, the resulting temperature is is shown in Fig. 10.12. The latent heats of some 6.7 °C. Calculate the heat of fusion of ice. substances, their freezing and boiling points, are (swater = 4186 J kg–1 K–1) given in Table 10.5. Answer Heat lost by water = msw (θf–θi)w = (0.30 kg) (4186 J kg–1 K–1) (50.0 °C – 6.7 °C) = 54376.14 J Heat required to melt ice = m2Lf = (0.15 kg) Lf Heat required to raise temperature of ice water to final temperature = mIsw (θf–θi)I = (0.15 kg) (4186 J kg–1 K –1) (6.7 °C – 0 °C) = 4206.93 J Heat lost = heat gained Fig. 10.12 Temperature versus heat for water at 54376.14 J = (0.15 kg) Lf + 4206.93 J 1 atm pressure (not to scale). Lf = 3.34×105 J kg–1. ⊳ Reprint 2025-26 214 PHYSICS ⊳ temperature difference. What are the different Example 10.5 Calculate the heat required ways by which this energy transfer takes to convert 3 kg of ice at –12 °C kept in a place? There are three distinct modes of heat calorimeter to steam at 100 °C at transfer: conduction, convection and radiation atmospheric pressure. Given specific heat (Fig. 10.13). capacity of ice = 2100 J kg–1 K–1, specific heat capacity of water = 4186 J kg– 1 K–1, latent heat of fusion of ice = 3.35 × 105 J kg–1 and latent heat of steam = 2.256 ×106 J kg–1. Answer We have Mass of the ice, m = 3 kg specific heat capacity of ice, sice = 2100 J kg–1 K–1 specific heat capacity of water, swater = 4186 J kg–1 K–1 latent heat of fusion of ice, Lf ice = 3.35 × 105 J kg–1 latent heat of steam, Lsteam Fig. 10.13 Heating by conduction, convection and = 2.256 × 106 J kg–1 radiation. Now, Q = heat required to convert 3 kg of 10.9.1 Conduction ice at –12 °C to steam at 100 °C, Conduction is the mechanism of transfer of heat Q1 = heat required to convert ice at between two adjacent parts of a body because –12 °C to ice at 0 °C. of their temperature difference. Suppose, one end = m sice ∆T1 = (3 kg) (2100 J kg–1. of a metallic rod is put in a flame, the other end K–1) [0–(–12)]°C = 75600 J of the rod will soon be so hot that you cannot Q2 = heat required to melt ice at hold it by your bare hands. Here, heat transfer 0 °C to water at 0 °C takes place by conduction from the hot end of = m Lf ice = (3 kg) (3.35 × 105 J kg–1) the rod through its different parts to the other = 1005000 J end. Gases are poor thermal conductors, while Q3 = heat required to convert water liquids have conductivities intermediate between at 0 °C to water at 100 °C. solids and gases. = msw ∆T2 = (3kg) (4186J kg–1 K–1) Heat conduction may be described (100 °C) quantitatively as the time rate of heat flow in a = 1255800 J material for a given temperature difference. Q4 = heat required to convert water Consider a metallic bar of length L and uniform cross-section A with its two ends maintained at at 100 °C to steam at 100 °C. different temperatures. This can be done, for = m Lsteam = (3 kg) (2.256 ×106 example, by putting the ends in thermal contact J kg–1) with large reservoirs at temperatures, say, TC and = 6768000 J TD, respectively (Fig. 10.14). Let us assume theSo, Q = Q1 + Q2 + Q3 + Q4 ideal condition that the sides of the bar are fully = 75600J + 1005000 J insulated so that no heat is exchanged between + 1255800 J + 6768000 J the sides and the surroundings. = 9.1×106 J ⊳ After sometime, a steady state is reached; the temperature of the bar decreases uniformly with 10.9 HEAT TRANSFER distance from TC to TD; (TC>TD). The reservoir at C supplies heat at a constant rate, whichWe have seen that heat is energy transfer transfers through the bar and is given out atfrom one system to another or from one part the same rate to the reservoir at D. It is foundof a system to another part, arising due to Reprint 2025-26 THERMAL PROPERTIES OF MATTER 215 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 from overheating. Table 10.6 Thermal conductivities of some material Fig. 10.14 Steady state heat flow by conduction in a bar with its two ends maintained at Material Thermal conductivity temperatures TC and TD; (TC > TD). (J s–1 m–1 K–1 ) Metals experimentally that in this steady state, the rate of flow of heat (or heat current) H is proportional Silver 406 to the temperature difference (TC – TD) and the Copper 385 area of cross-section A and is inversely Aluminium 205 proportional to the length L : Brass 109 Steel 50.2 TC – TD Lead 34.7 H = KA (10.14) L Mercury 8.3 The constant of proportionality K is called the thermal conductivity of the material. The Non-metals greater the value of K for a material, the more rapidly will it conduct heat. The SI unit of K is Insulating brick 0.15 J s–1 m –1 K–1 or W m –1 K–1. The thermal Concrete 0.8 conductivities of various substances are listed Body fat 0.20 in Table 10.6. These values vary slightly with Felt 0.04 temperature, but can be considered to be Glass 0.8 constant over a normal temperature range. Ice 1.6 Compare the relatively large thermal Glass wool 0.04 conductivities of good thermal conductors and, Wood 0.12 metals, with the relatively small thermal Water 0.8 conductivities of some good thermal insulators, such as wood and glass wool. You may have Gases noticed that some cooking pots have copper coating on the bottom. Being a good conductor Air 0.024 of heat, copper promotes the distribution of heat Argon 0.016 over the bottom of a pot for uniform cooking. Hydrogen 0.14 Plastic foams, on the other hand, are good ⊳ insulators, mainly because they contain pockets Example 10.6 What is the temperature of of air. Recall that gases are poor conductors, the steel-copper junction in the steady and note the low thermal conductivity of air in state of the system shown in Fig. 10.15. the Table 10.5. Heat retention and transfer are Length of the steel rod = 15.0 cm, length important in many other applications. Houses of the copper rod = 10.0 cm, temperature made of concrete roofs get very hot during of the furnace = 300 °C, temperature of the summer days because thermal conductivity of other end = 0 °C. The area of cross section concrete (though much smaller than that of a of the steel rod is twice that of the copper metal) is still not small enough. Therefore, people, rod. (Thermal conductivity of steel usually, prefer to give a layer of earth or foam = 50.2 J s –1 m –1 K –1; and of copper insulation on the ceiling so that heat transfer is = 385 J s–1m–1 K–1). Reprint 2025-26 216 PHYSICS Answer Given, L1 = L2= L = 0.1 m, A1 = A2= A= 0.02 m2 K1 = 79 W m–1 K –1, K2 = 109 W m–1 K–1, T1 = 373 K, and T2 = 273 K. Under steady state condition, the heat current (H1) through iron bar is equal to the Fig. 10.15 heat current (H2) through brass bar. Answer The insulating material around the rods So, H = H1 = H2 reduces heat loss from the sides of the rods. T1 – T 0 ) K 2 A 2 ( T 0 – T 2 )Therefore, heat flows only along the length of K 1 A1 ( = = the rods. Consider any cross section of the rod. L 1 L 2 In the steady state, heat flowing into the element For A1 = A2 = A and L1 = L2 = L, this equation must equal the heat flowing out of it; otherwise leads to there would be a net gain or loss of heat by the K1 (T1 – T0) = K2 (T0 – T2) element and its temperature would not be Thus, the junction temperature T0 of the two steady. Thus in the steady state, rate of heat bars is flowing across a cross section of the rod is the same at every point along the length of the ( K1T1 + K 2 T2 ) combined steel-copper rod. Let T be the T0 = ( K1 + K 2 ) temperature of the steel-copper junction in the Using this equation, the heat current H through steady state. Then, either bar is K1 A1 (300 − T ) K 2 A 2 ( T – 0 ) K1 A ( T1 – T0 ) K 2 A ( T0 – T2 ) = H = = L 1 L 2 L L 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 J s–1 m–1 K –1, K2 = 385 J s–1 m–1 K –1, we have Using these equations, the heat current H′ 50.2 × 2 ( 300 − T ) 385 T = through the compound bar of length L1 + L2 = 2L 15 10 and the equivalent thermal conductivity K′, of which gives T = 44.4 °C ⊳ the compound bar are given by ⊳ K ′ A ( T1 – T2 ) Example 10.7 An iron bar (L1 = 0.1 m, A1 = H ′ = = H 0.02 m 2, K 1 = 79 W m –1 K –1) and a 2 L brass bar (L2 = 0.1 m, A2 = 0.02 m2, 2 K1 K 2 K2 = 109 W m–1K–1) are soldered end to end K ′ = K1 + K 2 as shown in Fig. 10.16. The free ends of the iron bar and brass bar are maintained ( K1T1 + K 2 T2 ) at 373 K and 273 K respectively. Obtain (i) T0 = ( K 1 + K 2 ) expressions for and hence compute (i) the temperature of the junction of the two bars, 79 W m –1 K –1 109 W m –1 K –1 273 K ) ( )( 373 K ) + ( )( (ii) the equivalent thermal conductivity of = –1 –1 –1 –1 79 W m K + 109 W m K the compound bar, and (iii) the heat current through the compound bar. = 315 K 2 K 1 K 2 (ii) K ′ = K 1 + K 2 2×(79 W m –1 K –1 ) ×(109 W m –1 K –1 ) = –1 –1 –1 –1 79 W m K +109 W m K Fig 10.16 = 91.6 W m–1 K–1 Reprint 2025-26 THERMAL PROPERTIES OF MATTER 217 of water do. This occurs both because water has K ′ A ( T1 – T2 )(iii) H ′ = H = a greater specific heat capacity and because 2 L mixing currents disperse the absorbed heat 91.6 W m K × 0.02 m ( –1 –1 2 throughout the great volume of water. The air ) ( ) × ( 373 K–273 K ) in contact with the warm ground is heated by = 2× (0.1 m ) conduction. It expands, becoming less dense = 916.1 W ⊳ than the surrounding cooler air. As a result, the warm air rises (air currents) and the other air 10.9.2 Convection moves (winds) to fill the space-creating a sea breeze near a large body of water. Cooler air Convection is a mode of heat transfer by actual descends, and a thermal convection cycle is set motion of matter. It is possible only in fluids. up, which transfers heat away from the land.Convection can be natural or forced. In natural At night, the ground loses its heat more quickly,convection, gravity plays an important part. and the water surface is warmer than the land.When a fluid is heated from below, the hot part As a result, the cycle is reveresed (Fig. 10.17).expands and, therefore, becomes less dense. The other example of natural convection isBecause of buoyancy, it rises and the upper the steady surface wind on the earth blowingcolder part replaces it. This again gets heated, in from north-east towards the equator, therises up and is replaced by the relatively colder so-called trade wind. A resonable explanationpart of the fluid. The process goes on. This mode is as follows: the equatorial and polar regions ofof heat transfer is evidently different from the earth receive unequal solar heat. Air at theconduction. Convection involves bulk transport of different parts of the fluid. earth’s surface near the equator is hot, while In forced convection, material is forced to move the air in the upper atmosphere of the poles is by a pump or by some other physical means. The cool. In the absence of any other factor, a common examples of forced convection systems convection current would be set up, with the are forced-air heating systems in home, the air at the equatorial surface rising and moving human circulatory system, and the cooling out towards the poles, descending and system of an automobile engine. In the human streaming in towards the equator. The rotation body, the heart acts as the pump that circulates of the earth, however, modifies this convection blood through different parts of the body, current. Because of this, air close to the equator transferring heat by forced convection and has an eastward speed of 1600 km/h, while it maintaining it at a uniform temperature. is zero close to the poles. As a result, the air Natural convection is responsible for many descends not at the poles but at 30° N (North) familiar phenomena. During the day, the latitude and returns to the equator. This is ground heats up more quickly than large bodies called trade wind. Fig. 10.17 Convection cycles. Reprint 2025-26 218 PHYSICS 10.9.3 Radiation contents of the bottle. The outer wall similarly reflects back any incoming radiation. The spaceConduction and convection require some between the walls is evacuted to reduce material as a transport medium. These modes conduction and convection losses and the flask of heat transfer cannot operate between bodies is supported on an insulator, like cork. The separated by a distance in vacuum. But the device is, therefore, useful for preventing hot earth does receive heat from the Sun across a contents (like, milk) from getting cold, or huge distance. Similarly, we quickly feel the alternatively, to store cold contents (like, ice). warmth of the fire nearby even though air conducts poorly and before convection takes 10.9.4 Blackbody Radiation some time to set in. The third mechanism for We have so far not mentioned the wavelength heat transfer needs no medium; it is called content of thermal radiation. The important radiation and the energy so transferred by thing about thermal radiation at any electromagnetic waves is called radiant energy. temperature is that it is not of one (or a few) In an electromagnetic wave, electric and wavelength(s) but has a continuous spectrum magnetic fields oscillate in space and time. Like from the small to the long wavelengths. The any wave, electromagnetic waves can have energy content of radiation, however, varies for different wavelengths and can travel in vacuum different wavelengths. Figure 10.18 gives the with the same speed, namely the speed of light experimental curves for radiation energy per unit i.e., 3 × 108 m s–1 . You will learn these matters in area per unit wavelength emitted by a blackbody more detail later, but you now know why heat versus wavelength for different temperatures. transfer by radiation does not need any medium and why it 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 gas. The electromagnetic radiation emitted by a body by virtue of its temperature, like 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 colour of the body. We find that black bodies absorb and emit radiant energy better than bodies of lighter Fig. 10.18: Energy emitted versus wavelength colours. This fact finds many applications in our for a blackbody at different daily life. We wear white or light coloured clothes temperatures in summer, so that they absorb the least heat Notice that the wavelength λm for which energy from the Sun. However, during winter, we use is the maximum decreases with increasing dark coloured clothes, which absorb heat from temperature. The relation between λm and T is the sun and keep our body warm. The bottoms of given by what is known as Wien’s Displacement utensils for cooking food are blackened so that Law: they absorb maximum heat from fire and transfer λm T = constant (10.15) it to the vegetables to be cooked. Similarly, a Dewar flask or thermos bottle is The value of the constant (Wien’s constant) a device to minimise heat transfer between the is 2.9 × 10–3 m K. This law explains why the contents of the bottle and outside. It consists colour of a piece of iron heated in a hot flame of a double-walled glass vessel with the inner first becomes dull red, then reddish yellow, and and outer walls coated with silver. Radiation finally white hot. Wien’s law is useful for from the inner wall is reflected back to the estimating the surface temperatures of celestial Reprint 2025-26 THERMAL PROPERTIES OF MATTER 219 bodies like, the moon, Sun and other stars. Light For a body with emissivity e, the relation from the moon is found to have a maximum modifies to intensity near the wavelength 14 µm. By Wien’s H = eσ A (T4 – Ts 4) (10.18)law, the surface of the moon is estimated to have a temperature of 200 K. Solar radiation has a As an example, let us estimate the heat maximum at λm = 4753 Å. This corresponds to radiated by our bodies. Suppose the surface area T = 6060 K. Remember, this is the temperature of a person’s body is about 1.9 m2 and the room of the surface of the sun, not its interior. temperature is 22 °C. The internal body The most significant feature of the temperature, as we know, is about 37 °°C. The blackbody radiation curves in Fig. 10.18 is that skin temperature may be 28°°C (say). Thethey are universal. They depend only on the emissivity of the skin is about 0.97 for thetemperature and not on the size, shape or relevant region of electromagnetic radiation. Thematerial of the blackbody. Attempts to explain rate of heat loss is:blackbody radiation theoretically, at the beginning of the twentieth century, spurred the H = 5.67 × 10–8 × 1.9 × 0.97 × {(301)4 – (295)4} quantum revolution in physics, as you will learn in later courses. = 66.4 W Energy can be transferred by radiation over which is more than half the rate of energy large distances, without a medium (i.e., in production by the body at rest (120 W). To vacuum). The total electromagnetic energy prevent this heat loss effectively (better than radiated by a body at absolute temperature T ordinary clothing), modern arctic clothing has is proportional to its size, its ability to radiate an additional thin shiny metallic layer next to (called emissivity) and most importantly to its the skin, which reflects the body’s radiation. temperature. For a body, which is a perfect radiator, the energy emitted per unit time (H) 10.10 NEWTON’S LAW OF COOLING is given by We all know that hot water or milk when left on H = AσT 4 (10.16) a table begins to cool, gradually. Ultimately it attains the temperature of the surroundings. To where A is the area and T is the absolute study how slow or fast a given body can cool on temperature of the body. This relation obtained exchanging heat with its surroundings, let us experimentally by Stefan and later proved perform the following activity. theoretically by Boltzmann is known as Stefan- Take some water, say 300 mL, in a Boltzmann law and the constant σ is called calorimeter with a stirrer and cover it with a Stefan-Boltzmann constant. Its value in SI units two-holed lid. Fix the stirrer through one hole is 5.67 × 10–8 W m–2 K–4. Most bodies emit only a and fix a thermometer through another hole fraction of the rate given by Eq. 10.16. A substance in the lid and make sure that the bulb of like lamp black comes close to the limit. One, thermometer is immersed in the water. Note therefore, defines a dimensionless fraction e the reading of the thermometer. This reading called emissivity and writes, T1 is the temperature of the surroundings. H = AeσT 4 (10.17) Heat the water kept in the calorimeter till it Here, e = 1 for a perfect radiator. For a tungsten attains a temperature, say 40 °C above room lamp, for example, e is about 0.4. Thus, a tungsten temperature (i.e., temperature of the lamp at a temperature of 3000 K and a surface surroundings). Then, stop heating the water area of 0.3 cm2 radiates at the rate H = 0.3 × by removing the heat source. Start the 10–4 × 0.4 × 5.67 × 10–8 × (3000)4 = 60 W. stop-watch and note the reading of the A body at temperature T, with surroundings thermometer after a fixed interval of time, say at temperatures Ts, emits, as well as, receives after every one minute of stirring gently with energy. For a perfect radiator, the net rate of the stirrer. Continue to note the temperature loss of radiant energy is (T2) of water till it attains a temperature about 5 °C above that of the surroundings. Then, plot H = σA (T 4 – Ts4) Reprint 2025-26 220 PHYSICS a graph by taking each value of temperature From Eqs. (10.15) and (10.16) we have ∆T = T2 – T1 along y-axis and the coresponding dT 2value of t along x-axis (Fig. 10.19). – m s = k ( T2 – T1 ) dt dT2 k = – dt = – K dt (10.21) T 2 – T1 ms where K = k/m s On integrating, ∆ loge (T2 – T1) = – K t + c (10.22) or T2 = T1 + C′ e–Kt; where C′ = ec (10.23) Equation (10.23) enables you to calculate the time of cooling of a body through a particular Fig. 10.19 Curve showing cooling of hot water range of temperature. with time. For small temperature differences, the rate From the graph you can infer how the cooling of cooling, due to conduction, convection, and of hot water depends on the difference of its radiation combined, is proportional to the temperature from that of the surroundings. You difference in temperature. It is a valid will also notice that initially the rate of cooling approximation in the transfer of heat from a is higher and decreases as the temperature of radiator to a room, the loss of heat through the the body falls. wall of a room, or the cooling of a cup of tea on The above activity shows that a hot body loses the table. 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’s law of cooling, the rate of loss of heat, – dQ/dt of the body is directly proportional to the difference of temperature ∆T = (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 Fig. 10.20 Verification of Newton’s Law of cooling. can write Newton’s law of cooling can be verified with the help of the experimental set-up shown in – (10.19) Fig. 10.20(a). The set-up consists of a double- walled vessel (V) containing water between where k is a positive constant depending upon the two walls. A copper calorimeter (C)the area and nature of the surface of the body. containing hot water is placed inside theSuppose a body of mass m and specific heat double-walled vessel. Two thermometerscapacity s is at temperature T2. Let T1 be the through the corks are used to note thetemperature of the surroundings. If the temperatures T2 of water in calorimeter andtemperature falls by a small amount dT2 in time dt, then the amount of heat lost is T1 of hot water in between the double walls, respectively. Temperature of hot water in the dQ = ms dT2 calorimeter is noted after equal intervals of ∴ Rate of loss of heat is given by time. A graph is plotted between log e (T2–T1) [or ln(T2–T1)] and time (t). The nature of the dQ dT2 = ms (10.20) dt dt Reprint 2025-26 THERMAL PROPERTIES OF MATTER 221 graph is observed to be a straight line having 8 °C a negative slope as shown in Fig. 10.20(b). This = K ( 70 °C ) is in support of Eq. 10.22. 2 min ⊳ The average of 69 °C and 71 °C is 70 °C, which Example 10.8 A pan filled with hot food is 50 °C above room temperature. K is the same cools from 94 °C to 86 °C in 2 minutes when for this situation as for the original. the room temperature is at 20 °C. How long will it take to cool from 71 °C to 69 °C? 2 °C = K (50 °C) Time Answer The average temperature of 94 °C and When we divide above two equations, we 86 °C is 90 °C, which is 70 °C above the room have temperature. Under these conditions the pan cools 8 °C in 2 minutes. 8 °C/2 min K (70 °C) Using Eq. (10.21), we have = 2 °C/time K (50 °C) Change in temperature = K ∆ T Time = 0.7 min Time = 42 s ⊳ SUMMARY 1. Heat is a form of energy that flows between a body and its surrounding medium by virtue of temperature difference between them. The degree of hotness of the body is quantitatively represented by temperature. 2. A temperature-measuring device (thermometer) makes use of some measurable property (called thermometric property) that changes with temperature. Different thermometers lead to different temperature scales. To construct a temperature scale, two fixed points are chosen and assigned some arbitrary values of temperature. The two numbers fix the origin of the scale and the size of its unit. 3. The Celsius temperature (tC) and the Farenheit temperare (tF)are related by tF = (9/5) tC + 32 4. The ideal gas equation connecting pressure (P), volume (V) and absolute temperature (T) is : PV = µRT where µ is the number of moles and R is the universal gas constant. 5. In the absolute temperature scale, the zero of the scale corresponds to the temperature where every substance in nature has the least possible molecular activity. The Kelvin absolute temperature scale (T ) has the same unit size as the Celsius scale (Tc ), but differs in the origin : TC = T – 273.15 6. The coefficient of linear expansion (αl ) and volume expansion (αv ) are defined by the relations : ∆=l αl ∆ T l ∆V = αV ∆ T V Reprint 2025-26 222 PHYSICS where ∆l and ∆V denote the change in length l and volume V for a change of temperature ∆T. The relation between them is : αv = 3 αl 7. The specific heat capacity of a substance is defined by 1 ∆Q s = m ∆T where m is the mass of the substance and ∆Q is the heat required to change its temperature by ∆T. The molar specific heat capacity of a substance is defined by 1 ∆ Q C = µ ∆ T where µ is the number of moles of the substance. 8. The latent heat of fusion (Lf) is the heat per unit mass required to change a substance from solid into liquid at the same temperature and pressure. The latent heat of vaporisation (Lv) is the heat per unit mass required to change a substance from liquid to the vapour state without change in the temperature and pressure. 9. The three modes of heat transfer are conduction, convection and radiation. 10. In conduction, heat is transferred between neighbouring parts of a body through molecular collisions, without any flow of matter. For a bar of length L and uniform cross section A with its ends maintained at temperatures TC and TD, the rate of flow of heat H is : T − T H = K A C D L where K is the thermal conductivity of the material of the bar. 11. Newton’s Law of Cooling says that the rate of cooling of a body is proportional to the excess temperature of the body over the surroundings : d Q = – k ( T2 – T1 ) d t Where T1 is the temperature of the surrounding medium and T2 is the temperature of the body. Reprint 2025-26 THERMAL PROPERTIES OF MATTER 223 POINTS TO PONDER 1. The relation connecting Kelvin temperature (T ) and the Celsius temperature tc T = tc + 273.15 and the assignment T = 273.16 K for the triple point of water are exact relations (by choice). With this choice, the Celsius temperature of the melting point of water and boiling point of water (both at 1 atm pressure) are very close to, but not exactly equal to 0 °C and 100 °C respectively. In the original Celsius scale, these latter fixed points were exactly at 0 °C and 100 °C (by choice), but now the triple point of water is the preferred choice for the fixed point, because it has a unique temperature. 2. A liquid in equilibrium with vapour has the same pressure and temperature throughout the system; the two phases in equilibrium differ in their molar volume (i.e. density). This is true for a system with any number of phases in equilibrium. 3. Heat transfer always involves temperature difference between two systems or two parts of the same system. Any energy transfer that does not involve temperature difference in some way is not heat. 4. Convection involves flow of matter within a fluid due to unequal temperatures of its parts. A hot bar placed under a running tap loses heat by conduction between the surface of the bar and water and not by convection within water. EXERCISES 10.1 The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales. 10.2 Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between TA and TB ? 10.3 The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law : R = Ro [1 + α (T – To )] The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω ? 10.4 Answer the following : (a) The triple-point of water is a standard fixed point in modern thermometry. Why ? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale) ? (b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0 °C and 100 °C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale ? (c) The absolute temperature (Kelvin scale) T is related to the temperature tc on the Celsius scale by tc = T – 273.15 Why do we have 273.15 in this relation, and not 273.16 ? (d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale ? 10.5 Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made : Reprint 2025-26 224 PHYSICS Temperature Pressure Pressure thermometer A thermometer B Triple-point of water 1.250 × 105 Pa 0.200 × 105 Pa Normal melting point 1.797 × 105 Pa 0.287 × 105 Pa of sulphur (a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B ? (b) What do you think is the reason behind the slight difference in answers of thermometers A and B ? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings ? 10.6 A steel tape 1m long is correctly calibrated for a temperature of 27.0 °C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 °C. What is the actual length of the steel rod on that day ? What is the length of the same steel rod on a day when the temperature is 27.0 °C ? Coefficient of linear expansion of steel = 1.20 × 10–5 K–1 . 10.7 A large steel wheel is to be fitted on to a shaft of the same material. At 27 °C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range : αsteel = 1.20 × 10–5 K–1. 10.8 A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C? Coefficient of linear expansion of copper = 1.70 × 10–5 K–1. 10.9 A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of –39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm ? Co-efficient of linear expansion of brass = 2.0 × 10–5 K–1; Young’s modulus of brass = 0.91 × 1011 Pa. 10.10 A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °C? Is there a ‘thermal stress’ developed at the junction ? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2.0 × 10–5 K–1, steel = 1.2 × 10–5 K–1 ). 10.11 The coefficient of volume expansion of glycerine is 49 × 10–5 K–1. What is the fractional change in its density for a 30 °C rise in temperature ? 10.12 A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = 0.91 J g–1 K–1. 10.13 A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 J g–1 K–1; heat of fusion of water = 335 J g–1 ). 10.14 In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150 °C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27 °C. The final temperature is 40 °C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal ? 10.15 Given below are observations on molar specific heats at room temperature of some common gases. Reprint 2025-26 THERMAL PROPERTIES OF MATTER 225 Gas Molar specific heat (Cv ) (cal mo1–1 K–1) Hydrogen 4.87 Nitrogen 4.97 Oxygen 5.02 Nitric oxide 4.99 Carbon monoxide 5.01 Chlorine 6.17 The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine ? 10.16 A child running a temperature of 101°F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98 °F in 20 minutes, what is the average rate of extra evaporation caused, by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g–1. 10.17 A ‘thermacole’ icebox is a cheap and an efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45 °C, and co-efficient of thermal conductivity of thermacole is 0.01 J s–1 m–1 K–1. [Heat of fusion of water = 335 × 103 J kg–1] 10.18 A brass boiler has a base area of 0.15 m2 and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s–1 m–1 K–1 ; Heat of vaporisation of water = 2256 × 103 J kg–1. 10.19 Explain why : (a) a body with large reflectivity is a poor emitter (b) a brass tumbler feels much colder than a wooden tray on a chilly day (c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace (d) the earth without its atmosphere would be inhospitably cold (e) heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water 10.20 A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30 °C. The temperature of the surroundings is 20 °C. Reprint 2025-26 CHAPTER ELEVEN THERMODYNAMICS 11.1 INTRODUCTION In previous chapter we have studied thermal properties of matter. In this chapter we shall study laws that govern thermal energy. We shall study the processes where work is 11.1 Introduction converted into heat and vice versa. In winter, when we rub 11.2 Thermal equilibrium our palms together, we feel warmer; here work done in rubbing 11.3 Zeroth law of produces the ‘heat’. Conversely, in a steam engine, the ‘heat’ Thermodynamics of the steam is used to do useful work in moving the pistons,