11.3 ZEROTH LAW OF THERMODYNAMICS (a) Imagine two systems A and B, separated by an adiabatic wall, while each is in contact with a third system C, via a conducting wall [Fig. 11.2(a)]. The states of the systems (i.e., their macroscopic variables) will change until both A and B come to thermal equilibrium with C. After this is achieved, suppose that the adiabatic wall between A and B is replaced by a conducting wall and C is insulated from A and B by an adiabatic wall [Fig.11.2(b)]. It is found that the states of A and B change no (b) further i.e. they are found to be in thermal Fig. 11.2 (a) Systems A and B are separated by an equilibrium with each other. This observation adiabatic wall, while each is in contact forms the basis of the Zeroth Law of with a third system C via a conducting Thermodynamics, which states that ‘two wall. (b) The adiabatic wall between A systems in thermal equilibrium with a third and B is replaced by a conducting wall, system separately are in thermal equilibrium while C is insulated from A and B by an adiabatic wall.with each other’. R.H. Fowler formulated this * Both the variables need not change. It depends on the constraints. For instance, if the gases are in containers of fixed volume, only the pressures of the gases would change to achieve thermal equilibrium. Reprint 2025-26 THERMODYNAMICS 229

5.5 WORK DONE BY A VARIABLE FORCE A constant force is rare. It is the variable force, which is more commonly encountered. Fig. 5.3 is a plot of a varying force in one dimension. If the displacement ∆x is small, we can take the force F (x) as approximately constant and the work done is then ∆W =F (x) ∆x Fig. 5.3(a) Reprint 2025-26 76 PHYSICS The work done by the frictional force is Wf → area of the rectangle AGHI Wf = (−50) × 20 = − 1000 J The area on the negative side of the force axis has a negative sign. ⊳ 5.6 THE WORK-ENERGY THEOREM FOR A VARIABLE FORCE We are now familiar with the concepts of workFig. 5.3 (a) The shaded rectangle represents the work done by the varying force F(x), over and kinetic energy to prove the work-energy the small displacement ∆x, ∆W = F(x) ∆x. theorem for a variable force. We confine (b) adding the areas of all the rectangles we ourselves to one dimension. The time rate of find that for ∆x →0, the area under the curve change of kinetic energy is is exactly equal to the work done by F(x). d K d  1 2  =⊳ d t  2 m v  Example 5.5 A woman pushes a trunk on d t a railway platform which has a rough d v surface. She applies a force of 100 N over a = m v d t distance of 10 m. Thereafter, she gets progressively tired and her applied force = F v (from Newton’s Second Law) reduces linearly with distance to 50 N. The d x total distance through which the trunk has = F d t been moved is 20 m. Plot the force applied Thus by the woman and the frictional force, which dK = Fdx is 50 N versus displacement. Calculate the Integrating from the initial position (x i ) to final work done by the two forces over 20 m. position ( x f ), we have Answer K f x f F dx ∫ d K = ∫ K i x i where, Ki and K f are the initial and final kinetic energies corresponding to x i and x f. x f F d x or (5.8a) K f − K i = ∫ Fig. 5.4 Plot of the force F applied by the woman and x i the opposing frictional force f versus From Eq. (5.7), it follows that displacement. Kf − Ki = W (5.8b) The plot of the applied force is shown in Fig. 5.4. At x = 20 m, F = 50 N (≠ 0). We are given Thus, the WE theorem is proved for a variable that the frictional force f is |f|= 50 N. It opposes force. motion and acts in a direction opposite to F. It While the WE theorem is useful in a variety of is therefore, shown on the negative side of the problems, it does not, in general, incorporate the force axis. complete dynamical information of Newton’s The work done by the woman is second law. It is an integral form of Newton’s WF → area of the rectangle ABCD + area of second law. Newton’s second law is a relation the trapezium CEID between acceleration and force at any instant of 1 time. Work-energy theorem involves an integral WF = 100 × 10 + (100 + 50) × 10 over an interval of time. In this sense, the temporal 2 = 1000 + 750 (time) information contained in the statement of = 1750 J Newton’s second law is ‘integrated over’ and is Reprint 2025-26 WORK, ENERGY AND POWER 77 not available explicitly. Another observation is that are like ‘compressed springs’. They possess a Newton’s second law for two or three dimensions large amount of potential energy. An earthquake is in vector form whereas the work-energy results when these fault lines readjust. Thus, theorem is in scalar form. In the scalar form, potential energy is the ‘stored energy’ by virtue information with respect to directions contained of the position or configuration of a body. The in Newton’s second law is not present. body left to itself releases this stored energy in ⊳ the form of kinetic energy. Let us make our notion Example 5.6 A block of mass m = 1 kg, of potential energy more concrete. moving on a horizontal surface with speed The gravitational force on a ball of mass m is vi = 2 m s–1 enters a rough patch ranging mg . g may be treated as a constant near the earth from x = 0.10 m to x = 2.01 m. The retarding surface. By ‘near’ we imply that the height h of force Fr on the block in this range is inversely the ball above the earth’s surface is very small proportional to x over this range, compared to the earth’s radius RE (h <<RE) so that −k we can ignore the variation of g near the earth’s Fr = for 0.1 < x < 2.01 m surface*. In what follows we have taken the x upward direction to be positive. Let us raise the = 0 for x < 0.1m and x > 2.01 m ball up to a height h. The work done by the external where k = 0.5 J. What is the final kinetic agency against the gravitational force is mgh. This energy and speed vf of the block as it work gets stored as potential energy. crosses this patch ? Gravitational potential energy of an object, as a function of the height h, is denoted by V(h) and it Answer From Eq. (5.8a) is the negative of work done by the gravitational 2.01 ( −k ) force in raising the object to that height. d x V (h) = mgh K f = K i + ∫ x 0.1 If h is taken as a variable, it is easily seen that the gravitational force F equals the negative of 1 2 2.01 = mv i − k ln ( x ) 0.1 the derivative of V(h) with respect to h. Thus, 2 d F = − V(h) = −m g 1 2 d h = mv i − k ln (2.01/0.1) 2 The negative sign indicates that the = 2 − 0.5 ln (20.1) gravitational force is downward. When released, the ball comes down with an increasing speed. = 2 − 1.5 = 0.5 J Just before it hits the ground, its speed is given v f = 2K f / m = 1 m s−1 by the kinematic relation, v2 = 2gh This equation can be written as Here, note that ln is a symbol for the natural 1logarithm to the base e and not the logarithm to the base 10 [ln X = loge X = 2.303 log10 X]. ⊳ 2 m v2 = m g h which shows that the gravitational potential5.7 THE CONCEPT OF POTENTIAL ENERGY energy of the object at height h, when the object The word potential suggests possibility or is released, manifests itself as kinetic energy of capacity for action. The term potential energy the object on reaching the ground. brings to one’s mind ‘stored’ energy. A stretched Physically, the notion of potential energy is bow-string possesses potential energy. When it applicable only to the class of forces where work is released, the arrow flies off at a great speed. done against the force gets ‘stored up’ as energy. The earth’s crust is not uniform, but has When external constraints are removed, it discontinuities and dislocations that are called manifests itself as kinetic energy. Mathematically, fault lines. These fault lines in the earth’s crust (for simplicity, in one dimension) the potential * The variation of g with height is discussed in Chapter 7 on Gravitation. Reprint 2025-26 78 PHYSICS energy V(x) is defined if the force F(x) can be which means that K + V, the sum of the kinetic written as and potential energies of the body is a constant. Over the whole path, xi to xf, this means that d V F ( x ) = − d x Ki + V(xi ) = Kf + V(xf) (5.11) The quantity K +V(x), is called the totalThis implies that mechanical energy of the system. Individually xf Vf the kinetic energy K and the potential energy ∫ F(x) d x = − ∫ d V = Vi − V f V(x) may vary from point to point, but the sum x i Vi is a constant. The aptness of the term The work done by a conservative force such as ‘conservative force’ is now clear. gravity depends on the initial and final positions Let us consider some of the definitions of a only. In the previous chapter we have worked conservative force. on examples dealing with inclined planes. If an l A force F(x) is conservative if it can be derived object of mass m is released from rest, from the from a scalar quantity V(x) by the relation top of a smooth (frictionless) inclined plane of given by Eq. (5.9). The three-dimensional height h, its speed at the bottom generalisation requires the use of a vector is 2 gh irrespective of the angle of inclination. derivative, which is outside the scope of this book.Thus, at the bottom of the inclined plane it l The work done by the conservative forceacquires a kinetic energy, mgh. If the work done depends only on the end points. This can be or the kinetic energy did depend on other factors seen from the relation, such as the velocity or the particular path taken W = Kf – Ki = V (xi) – V(xf)by the object, the force would be called non- which depends on the end points. conservative. l A third definition states that the work done The dimensions of potential energy are by this force in a closed path is zero. This is [ML2T –2] and the unit is joule (J), the same as once again apparent from Eq. (5.11) since kinetic energy or work. To reiterate, the change xi = xf .in potential energy, for a conservative force, ∆V is equal to the negative of the work done by Thus, the principle of conservation of total mechanical energy can be stated asthe force ∆V = − F(x) ∆x (5.9) The total mechanical energy of a system is In the example of the falling ball considered in conserved if the forces, doing work on it, are this section we saw how potential energy was conservative. The above discussion can be made moreconverted to kinetic energy. This hints at an concrete by considering the example of theimportant principle of conservation in mechanics, gravitational force once again and that of thewhich we now proceed to examine. spring force in the next section. Fig. 5.5 depicts

11.2 THERMAL EQUILIBRIUM independent variables. Let the pressure and Equilibrium in mechanics means that the net volume of the gases be (PA, VA) and (PB, VB) external force and torque on a system are zero. respectively. Suppose first that the two systems The term ‘equilibrium’ in thermodynamics appears are put in proximity but are separated by an * Thermodynamics may also involve other variables that are not so obvious to our senses e.g. entropy, enthalpy, etc., and they are all macroscopic variables. However, a thermodynamic state is specified by five state variables viz., pressure, volume, temperature, internal energy and entropy. Entropy is a measure of disorderness in the system. Enthalpy is a measure of total heat content of the system. Reprint 2025-26 228 PHYSICS adiabatic wall – an insulating wall (can be law in 1931 long after the first and second Laws movable) that does not allow flow of energy (heat) of thermodynamics were stated and so numbered. from one to another. The systems are insulated The Zeroth Law clearly suggests that when two from the rest of the surroundings also by similar systems A and B, are in thermal equilibrium, adiabatic walls. The situation is shown there must be a physical quantity that has the schematically in Fig. 11.1 (a). In this case, it is same value for both. This thermodynamic found that any possible pair of values (PA, VA) will variable whose value is equal for two systems in be in equilibrium with any possible pair of values thermal equilibrium is called temperature (T ). (PB, VB). Next, suppose that the adiabatic wall is Thus, if A and B are separately in equilibrium replaced by a diathermic wall – a conducting wall with C, TA = TC and TB = TC. This implies that that allows energy flow (heat) from one to another. TA = TB i.e. the systems A and B are also in It is then found that the macroscopic variables of thermal equilibrium. the systems A and B change spontaneously until We have arrived at the concept of temperature both the systems attain equilibrium states. After formally via the Zeroth Law. The next question that there is no change in their states. The is : how to assign numerical values to situation is shown in Fig. 11.1(b). The pressure temperatures of different bodies ? In other words, and volume variables of the two gases change to how do we construct a scale of temperature ? (PB ′, VB ′) and (PA ′, VA ′) such that the new states Thermometry deals with this basic question to of A and B are in equilibrium with each other*. which we turn in the next section. There is no more energy flow from one to another. We then say that the system A is in thermal equilibrium with the system B. What characterises the situation of thermal equilibrium between two systems ? You can guess the answer from your experience. In thermal equilibrium, the temperatures of the two systems are equal. We shall see how does one arrive at the concept of temperature in thermodynamics? The Zeroth law of thermodynamics provides the clue.