Thermal Properties of Matter PDF

CHAPTER ELEVEN THERMAL PROPERTIES OF MATTER 11.1 INTRODUCTION We all have common sense notions of heat and temperature. Temperature is a measure of ‘hotness’ of a bo...

CHAPTER ELEVEN THERMAL PROPERTIES OF MATTER 11.1 INTRODUCTION We all have common sense notions of heat and temperature. Temperature is a measure of ‘hotness’ of a body. A kettle 11.1 Introduction with boiling water is hotter than a box containing ice. In 11.2 Temperature and heat physics, we need to define the notion of heat, temperature, 11.3 Measurement of etc., more carefully. In this chapter, you will learn what heat temperature is and how it is measured, and study the various proceses by 11.4 Ideal-gas equation and which heat flows from one body to another. Along the way, absolute temperature you will find out why blacksmiths heat the iron ring before 11.5 Thermal expansion fitting on the rim of a wooden wheel of a horse cart and why 11.6 Specific heat capacity the wind at the beach often reverses direction after the sun 11.7 Calorimetry goes down. You will also learn what happens when water boils 11.8 Change of state or freezes, and its temperature does not change during these 11.9 Heat transfer processes even though a great deal of heat is flowing into or out of it. 11.10 Newton’s law of cooling Summary 11.2 TEMPERATURE AND HEAT Points to ponder Exercises We can begin studying thermal properties of matter with Additional Exercises definitions of temperature and heat. Temperature is a relative measure, or indication of hotness or coldness. A hot utensil is said to have a high temperature, and ice cube to have a low temperature. An object that has a higher temperature than another object is said to be hotter. Note that hot and cold are relative terms, like tall and short. We can perceive temperature by touch. However, this temperature sense is somewhat unreliable and its range is too limited to be useful for scientific purposes. We know from experience that a glass of ice-cold water left on a table on a hot summer day eventually warms up whereas a cup of hot tea on the same table cools down. It means that when the temperature of body, ice-cold water or hot tea in this case, and its surrounding medium are different, heat transfer takes place between the system and the surrounding medium, until the body and the surrounding medium are at the same temperature. We also know that in the case of glass tumbler of ice-cold water, heat flows from the environment to 2020-21 THERMAL PROPERTIES OF MATTER 279 the glass tumbler, whereas in the case of hot tea, it flows from the cup of hot tea to the environment. So, we can say that heat is the form of energy transferred between two (or more) systems or a system and its surroundings by virtue of temperature difference. The SI unit of heat energy transferred is expressed in joule (J) while SI unit of temperature is Kelvin (K), and degree Celsius (oC) is a commonly used unit of temperature. When an object is heated, many changes may take place. Its temperature may rise, it may expand or change state. We will study the effect of heat on different bodies in later sections. Fig. 11.1 A plot of Fahrenheit temperature (tF) versus 11.3 MEASUREMENT OF TEMPERATURE Celsius temperature (tc ). A measure of temperature is obtained using a thermometer. Many physical properties of A relationship for converting between the two materials change sufficiently with temperature. scales may be obtained from a graph of Some such properties are used as the basis for Fahrenheit temperature (t F) versus celsius constructing thermometers. The commonly used temperature (tC) in a straight line (Fig. 11.1), property is variation of the volume of a liquid whose equation is with temperature. For example, in common t F – 32 t C liquid–in–glass thermometers, mercury, alcohol = (11.1) 180 100 etc., are used whose volume varies linearly with temperature over a wide range. 11.4 IDEAL-GAS EQUATION AND Thermometers are calibrated so that a ABSOLUTE TEMPERATURE numerical value may be assigned to a given temperature in an appropriate scale. For the Liquid-in-glass thermometers show different definition of any standard scale, two fixed readings for temperatures other than the fixed reference points are needed. Since all points because of differing expansion properties. substances change dimensions with A thermometer that uses a gas, however, gives temperature, an absolute reference for the same readings regardless of which gas is expansion is not available. However, the used. Experiments show that all gases at low necessary fixed points may be correlated to the densities exhibit same expansion behaviour. The physical phenomena that always occur at the variables that describe the behaviour of a given same temperature. The ice point and the steam quantity (mass) of gas are pressure, volume, and point of water are two convenient fixed points temperature (P, V, and T )(where T = t + 273.15; and are known as the freezing and boiling t is the temperature in °C). When temperature points, respectively. These two points are the is held constant, the pressure and volume of a temperatures at which pure water freezes and quantity of gas are related as PV = constant. boils under standard pressure. The two familiar This relationship is known as Boyle’s law, after temperature scales are the Fahrenheit Robert Boyle (1627–1691), the English Chemist temperature scale and the Celsius temperature who discovered it. When the pressure is held scale. The ice and steam point have values constant, the volume of a quantity of the gas is 32 °F and 212 °F, respectively, on the Fahrenheit related to the temperature as V/T = constant. scale and 0 °C and 100 °C on the Celsius scale. This relationship is known as Charles’ law, On the Fahrenheit scale, there are 180 equal after French scientist Jacques Charles (1747– intervals between two reference points, and on 1823). Low-density gases obey these the Celsius scale, there are 100. laws, which may be combined into a single 2020-21 280 PHYSICS Fig. 11.3 A plot of pressure versus temperature and Fig. 11.2 Pressure versus temperature of a low extrapolation of lines for low density gases density gas kept at constant volume. indicates the same absolute zero temperature. relationship. Notice that since PV = constant named after the British scientist Lord Kelvin. On and V/T = constant for a given quantity of gas, this scale, – 273.15 °C is taken as the zero point, then PV/T should also be a constant. This that is 0 K (Fig. 11.4). relationship is known as ideal gas law. It can be written in a more general form that applies not just to a given quantity of a single gas but to any quantity of any low-density gas and is known as ideal-gas equation: PV = µR T or PV = µRT (11.2) where, µ is the number of moles in the sample of gas and R is called universal gas constant: R = 8.31 J mol–1 K–1 In Eq. 11.2, we have learnt that the pressure and volume are directly proportional to temperature : PV ∝ T. This relationship allows a gas to be used to measure temperature in a constant volume gas thermometer. Holding the volume of a gas constant, it gives P ∝T. Thus, Fig. 11.4 Comparision of the Kelvin, Celsius and with a constant-volume gas thermometer, Fahrenheit temperature scales. temperature is read in terms of pressure. A plot The size of unit in Kelvin and Celsius of pressure versus temperature gives a straight temperature scales is the same. So, temperature line in this case, as shown in Fig. 11.2. on these scales are related by However, measurements on real gases deviate T = tC + 273.15 (11.3) from the values predicted by the ideal gas law at low temperature. But the relationship is linear 11.5 THERMAL EXPANSION over a large temperature range, and it looks as though the pressure might reach zero with You may have observed that sometimes sealed decreasing temperature if the gas continued to bottles with metallic lids are so tightly screwed be a gas. The absolute minimum temperature that one has to put the lid in hot water for some for an ideal gas, therefore, inferred by time to open it. This would allow the metallic lid extrapolating the straight line to the axis, as in to expand, thereby loosening it to unscrew Fig. 11.3. This temperature is found to be easily. In case of liquids, you may have observed – 273.15 °C and is designated as absolute zero. that mercury in a thermometer rises, when the Absolute zero is the foundation of the Kelvin thermometer is put in slightly warm water. If temperature scale or absolute scale temperature we take out the thermometer from the warm 2020-21 THERMAL PROPERTIES OF MATTER 281 water the level of mercury falls again. Similarly, Table 11.1 Values of coef ficient of linear in case of gases, a balloon partially inflated in a expansion for some material cool room may expand to full size when placed in warm water. On the other hand, a fully Material α l (10–5 K–1) inflated balloon when immersed in cold water Aluminium 2.5 would start shrinking due to contraction of the Brass 1.8 air inside. Iron 1.2 It is our common experience that most Copper 1.7 substances expand on heating and contract on Silver 1.9 cooling. A change in the temperature of a body Gold 1.4 causes change in its dimensions. The increase Glass (pyrex) 0.32 in the dimensions of a body due to the increase Lead 0.29 in its temperature is called thermal expansion. The expansion in length is called linear Similarly, we consider the fractional change expansion. The expansion in area is called area expansion. The expansion in volume is called ∆V in volume, , of a substance for temperature volume expansion (Fig. 11.5). V change ∆T and define the coefficient of volume expansion (or volume expansivity), α V as  ∆V  1 αV =  (11.5)  V  ∆T Here α V is also a characteristic of the substance but is not strictly a constant. It ∆l ∆A ∆V depends in general on temperature (Fig 11.6). It = a l ∆T = 2a l ∆T = 3a l ∆T l A V is seen that αV becomes constant only at a high temperature. (a) Linear expansion (b) Area expansion (c) Volume expansion Fig. 11.5 Thermal Expansion. If the substance is in the form of a long rod, then for small change in temperature, ∆T, the fractional change in length, ∆l/l, is directly proportional to ∆T. ∆l = α1 ∆T (11.4) l where α1 is known as the coefficient of linear expansion (or linear expansivity) and is characteristic of the material of the rod. In Table 11.1, typical average values of the coefficient of Fig. 11.6 Coefficient of volume expansion of copper linear expansion for some material in the as a function of temperature. temperature range 0 °C to 100 °C are given. From this Table, compare the value of αl for glass and Table 11.2 gives the values of coefficient of copper. We find that copper expands about five volume expansion of some common substances times more than glass for the same rise in in the temperature range 0–100 °C. You can see temperature. Normally, metals expand more and that thermal expansion of these substances have relatively high values of αl. (solids and liquids) is rather small, with material, 2020-21 282 PHYSICS like pyrex glass and invar (a special iron-nickel lakes and ponds, freeze at the top first. As a lake alloy) having particularly low values of αV. From cools toward 4 °C, water near the surface loses this Table we find that the value of αv for energy to the atmosphere, becomes denser, and alcohol (ethanol) is more than mercury and sinks; the warmer, less dense water near the expands more than mercury for the same rise bottom rises. However, once the colder water on in temperature. top reaches temperature below 4 °C, it becomes less dense and remains at the surface, where it Table 11.2 Values of coefficient of volume freezes. If water did not have this property, lakes expansion for some substances and ponds would freeze from the bottom up, which would destroy much of their animal and Material αv ( K–1) plant life. Aluminium 7 × 10–5 Gases, at ordinary temperature, expand more Brass 6 × 10–5 than solids and liquids. For liquids, the Iron 3.55 × 10–5 coefficient of volume expansion is relatively Paraffin 58.8 × 10–5 independent of the temperature. However, for Glass (ordinary) 2.5 × 10–5 gases it is dependent on temperature. For an Glass (pyrex) 1 × 10–5 ideal gas, the coefficient of volume expansion at Hard rubber 2.4 × 10–4 constant pressure can be found from the ideal Invar 2 × 10–6 gas equation: Mercury 18.2 × 10–5 PV = µRT Water 20.7 × 10–5 At constant pressure Alcohol (ethanol) 110 × 10–5 P∆V = µR ∆T ∆V ∆T Water exhibits an anomalous behaviour; it = V T contracts on heating between 0 °C and 4 °C. The volume of a given amount of water decreases 1 i.e., αv = for ideal gas (11.6) as it is cooled from room temperature, until its T temperature reaches 4 °C, [Fig. 11.7(a)]. Below At 0 °C, αv = 3.7 × 10–3 K–1, which is much 4 °C, the volume increases, and therefore, the larger than that for solids and liquids. density decreases [Fig. 11.7(b)]. Equation (11.6) shows the temperature This means that water has the maximum dependence of αv; it decreases with increasing density at 4 °C. This property has an important temperature. For a gas at room temperature and environmental effect: bodies of water, such as constant pressure, αv is about 3300 × 10–6 K–1, as Temperature (°C) Temperature (°C) (a) (b) Fig. 11.7 Thermal expansion of water. 2020-21 THERMAL PROPERTIES OF MATTER 283 much as order(s) of magnitude larger than the Answer coefficient of volume expansion of typical liquids. ∆A3 = (∆a) (∆b) There is a simple relation between the ∆Al = a (∆b) coefficient of volume expansion ( α v ) and coefficient of linear expansion (αl). Imagine a ∆b cube of length, l, that expands equally in all directions, when its temperature increases by b ∆T. We have ∆a a ∆l = αl l ∆T so, ∆V = (l+∆l)3 – l3
3l2 ∆l (11.7) In Equation (11.7), terms in (∆l)2 and (∆l)3 have been neglected since ∆l is small compared to l. So ∆A2 = b (∆a) Fig. 11.8 3V ∆l ∆V = = 3V αl ∆T (11.8) l Consider a rectangular sheet of the solid which gives material of length a and breadth b (Fig. 11.8 ). αv = 3αl (11.9) When the temperature increases by ∆T, a increases by ∆a = αl a∆T and b increases by ∆b What happens by preventing the thermal = αlb ∆T. From Fig. 11.8, the increase in area expansion of a rod by fixing its ends rigidly? ∆A = ∆A1 +∆A2 + ∆A3 Clearly, the rod acquires a compressive strain ∆A = a ∆b + b ∆a + (∆a) (∆b) due to the external forces provided by the rigid = a αlb ∆T + b αl a ∆T + (αl)2 ab (∆T)2 support at the ends. The corresponding stress = αl ab ∆T (2 + αl ∆T) = αl A ∆T (2 + αl ∆T) set up in the rod is called thermal stress. For Since αl
10 –5 K–1, from Table 11.1, the example, consider a steel rail of length 5 m and product αl ∆T for fractional temperature is small area of cross-section 40 cm2 that is prevented in comparision with 2 and may be neglected. from expanding while the temperature rises by Hence, 10 °C. The coefficient of linear expansion of steel is αl(steel) = 1.2 × 10–5 K–1. Thus, the compressive t ∆l strain is = αl(steel) ∆T = 1.2 × 10–5 × 10=1.2 × 10–4. l Example 11.2 A blacksmith fixes iron ring t Youngs modulus of steel is Y (steel) = 2 × 1011 N m–2. Therefore, the thermal stress developed is on the rim of the wooden wheel of a horse cart. The diameter of the rim and the iron ∆F  ∆l  ring are 5.243 m and 5.231 m, respectively = Y steel   = 2.4 × 10 7 N m –2 , which A  l  at 27 °C. To what temperature should the corresponds to an external force of ring be heated so as to fit the rim of the wheel?  ∆l  ∆F = AYsteel   = 2.4 × 107 × 40 × 10–4 j 105N.  l  Answer If two such steel rails, fixed at their outer ends, are in contact at their inner ends, a force of this Given, T1 = 27 °C magnitude can easily bend the rails. LT1 = 5.231 m LT2 = 5.243 m Example 11.1 Show that the coefficient t So, of area expansion, (∆A/A)/∆T, of a LT2 =LT1 [1+αl (T2–T1)] rectangular sheet of the solid is twice its linear expansivity, αl. 5.243 m = 5.231 m [1 + 1.20×10–5 K–1 (T2–27 °C)] or T2 = 218 °C. t 2020-21 284 PHYSICS 11.6 SPECIFIC HEAT CAPACITY heat absorbed or given off to change the Take some water in a vessel and start heating it temperature of unit mass of it by one unit. This on a burner. Soon you will notice that bubbles quantity is referred to as the specific heat begin to move upward. As the temperature is capacity of the substance. raised the motion of water particles increases If ∆Q stands for the amount of heat absorbed till it becomes turbulent as water starts boiling. or given off by a substance of mass m when it What are the factors on which the quantity of undergoes a temperature change ∆T, then the heat required to raise the temperature of a specific heat capacity, of that substance is given substance depend? In order to answer this by question in the first step, heat a given quantity S 1 ∆Q of water to raise its temperature by, say 20 °C s= = (11.11) 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 no phase change) when a given quantity of heat it takes about twice the time and therefore, is absorbed (or given off) by it. It is defined as the double the quantity of heat required raising twice amount of heat per unit mass absorbed or given the 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 If the amount of substance is specified in taken is again twice that required in the first terms of moles µ, instead of mass m in kg, we step. can define heat capacity per mole of the In the third step, in place of water, now heat substance by the same quantity of some oil, say mustard oil, and raise the temperature again by 20 °C. Now S 1 ∆Q C= = (11.12) note the time by the same stopwatch. You will µ µ ∆T 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 be temperature, ∆T and the nature of substance. needed to define C. In this case, heat transfer The change in temperature of a substance, when can be achieved by keeping either pressure or a given quantity of heat is absorbed or rejected volume constant. If the gas is held under by it, is characterised by a quantity called the constant pressure during the heat transfer, then heat capacity of that substance. We define heat it is called the molar specific heat capacity at capacity, 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= (11.10) ∆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 12. Table 11.3 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 11.4 lists molar specific heat capacities of some substance has a unique value for the amount of gases. From Table 11.3 you can note that water 2020-21 THERMAL PROPERTIES OF MATTER 285 Table 11.3 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 11.4 Molar specific heat capacities of inner vessel. There is an opening in the outer some gases jacket through which a mercury thermometer Gas Cp (J mol–1K–1) Cv(J mol–1K–1) can be inserted into the calorimeter (Fig. 11.20). The following example provides a method by He 20.8 12.5 which the specific heat capacity of a given solid H2 28.8 20.4 can be determinated by using the principle, heat gained is equal to the heat lost. N2 29.1 20.8 Example 11.3 A sphere of 0.047 kg t 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 11.7 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. 2020-21 286 PHYSICS The amount of heat lost by the aluminium sphere = m1s 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. 11.9 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) In the steady state heat lost by the aluminium The change of state from solid to liquid is sphere = heat gained by water + heat gained by called melting and from liquid to solid is called calorimeter. fusion. It is observed that the temperature remains constant until the entire amount of the So, 0.047 kg × sAl × 77 °C solid substance melts. That is, both the solid = (0.25 kg × 4.18 × 103 J kg–1 K–1+ 0.14 kg × and the liquid states of the substance coexist 0.386 × 103 J kg–1 K–1)(3 °C) in thermal equilibrium during the change of sAl = 0.911 kJ kg –1 K–1 t states from solid to liquid. The temperature at which the solid and the liquid states of the 11.8 CHANGE OF STATE substance is in thermal equilibrium with each 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. 11.10. 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. 11.9). You will again. Thus, the wire passes through the slab observe no change in the temperature as long and the slab does not split. This phenomenon as there is ice in the beaker. In the above process, of refreezing is called regelation. Skating is the temperature of the system does not change possible on snow due to the formation of water even though heat is being continuously supplied. under the skates. Water is formed due to the The heat supplied is being utilised in changing increase of pressure and it acts as a the state from solid (ice) to liquid (water). lubricant. 2020-21 THERMAL PROPERTIES OF MATTER 287 100 °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. 11.10 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 that temperature begins to rise (Fig.11.9). The Take a round-bottom flask, more than half 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) Fig. 11.11: Pressure-temperature phase diagrams for (a) water and (b) CO2 (not to the scale). 2020-21 288 PHYSICS thermometer and steam outlet through the cork cork. Keep the f lask turned upside down on the of the flask (Fig. 11.11). 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. 11.8.1 Latent Heat In Section 11.8, 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. 11.11 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. 2020-21 THERMAL PROPERTIES OF MATTER 289 Table 11.5 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. 11.12 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=mL fusion and vaporisation are Lf = 3.33 × 105 J kg–1 or L = Q/m (11.13) 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 11.4 When 0.15 kg of ice at 0 °C t 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. 11.12. 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 11.5. 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. 11.12 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. t 2020-21 290 PHYSICS temperature difference. What are the different Example 11.5 Calculate the heat required t 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. 11.13). 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. 11.13 Heating by conduction, convection and = 2.256 × 106 J kg–1 radiation. Now, Q = heat required to convert 3 kg of 11.9.1 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 at 100 °C to steam at 100 °C. cross-section A with its two ends maintained at = m L steam = (3 kg) (2.256×10 6 different temperatures. This can be done, for J kg–1) example, by putting the ends in thermal contact with large reservoirs at temperatures, say, TC and = 6768000 J TD, respectively (Fig. 11.14). Let us assume the So, 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 t After sometime, a steady state is reached; the temperature of the bar decreases uniformly with 11.9 HEAT TRANSFER distance from TC to TD; (TC>TD). The reservoir at We have seen that heat is energy transfer C supplies heat at a constant rate, which from one system to another or from one part transfers through the bar and is given out at of a system to another part, arising due to the same rate to the reservoir at D. It is found 2020-21 THERMAL PROPERTIES OF MATTER 291 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 11.6 Thermal conductivities of some material Fig. 11.14 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 (11.14) 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 11.6. 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 11.6 What is the temperature of t 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. 11.15. the Table 11.5. 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 summer days because thermal conductivity of the other end = 0 °C. The area of cross concrete (though much smaller than that of a section of the steel rod is twice that of the metal) is still not small enough. Therefore, people, copper 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–1K–1). 2020-21 292 PHYSICS Answer Given, L1 = L2= L = 0.1 m, A1 = A2= A= 0.02 m2 K 1 = 79 W m –1 K –1 , K 2 = 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. 11.15 heat current (H2) through brass bar. Answer The insulating material around the So, H = H1 = H2 rods reduces heat loss from the sides of the rods. Therefore, heat flows only along the length of K1 A1 ( T1 – T0 ) K 2 A 2 ( T 0 – T2 ) = = the rods. Consider any cross section of the rod. L1 L2 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 2T2 ) T0 = combined steel-copper rod. Let T be the ( 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 A2 (T – 0) K1 A (T1 – T0 ) = K 2 A(T0 – T2 ) L1 L2 H= = 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 50.2 × 2 ( 300 − T ) Using these equations, the heat current H′ 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 t the compound bar are given by Example 11.7 An iron bar (L1 = 0.1 m, A1 K ′ A (T1 – T2 ) H′= = H t = 0.02 m 2 , K 1 = 79 W m –1 K –1) and a 2L brass bar (L 2 = 0.1 m, A 2 = 0.02 m 2 , 2 K1 K 2 K2 = 109 W m–1K–1) are soldered end to end K′ = as shown in Fig. 11.16. The free ends of K1 + K 2 the iron bar and brass bar are maintained ( K1T1 + K 2T2 ) at 373 K and 273 K respectively. Obtain (i) T0 = expressions for and hence compute (i) the ( K1 + K 2 ) temperature of the junction of the two bars, = (79 W m –1 ) ( K –1 (373 K ) + 109 W m –1 K –1 (273 K ) ) (ii) the equivalent thermal conductivity of 79 W m K –1 –1 + 109 W m K –1 –1 the compound bar, and (iii) the heat current through the compound bar. = 315 K 2K 1 K 2 (ii) K ′ = K + K 1 2 2 × (79 W m –1 K –1 ) × (109 W m –1 K –1 ) = 79 W m –1 K –1 +109 W m –1 K –1 Fig 11.16 = 91.6 W m–1 K–1 2020-21 THERMAL PROPERTIES OF MATTER 293 of water do. This occurs both because water has K ′ A (T1 – T2 ) (iii) H ′ = H = a greater specific heat capacity and because 2L mixing currents disperse the absorbed heat throughout the great volume of water. The air = (91.6 W m –1 ) ( ) K –1 × 0.02 m 2 × (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 t than the surrounding cooler air. As a result, the warm air rises (air currents) and the other air 11.9.2 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. 11.17). expands and, therefore, becomes less dense. Because of buoyancy, it rises and the upper The other example of natural convection is colder part replaces it. This again gets heated, the steady surface wind on the earth blowing rises up and is replaced by the relatively colder in from north-east towards the equator, the part of the fluid. The process goes on. This mode so-called trade wind. A resonable explanation of heat transfer is evidently different from is as follows: the equatorial and polar regions of conduction. Convection involves bulk transport the earth receive unequal solar heat. Air at the 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. 11.17 Convection cycles. 2020-21 294 PHYSICS 11.9.3 Radiation contents of the bottle. The outer wall similarly reflects back any incoming radiation. The space Conduction 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 11.9.4 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 11.18 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 area per unit wavelength emitted by a blackbody in 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. 11.18: 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 (11.15) 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 2020-21 THERMAL PROPERTIES OF MATTER 295 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 (T 4 – Ts4) (11.18) 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. 11.18 is that skin temperature may be 28°C (say). The they are universal. They depend only on the emissivity of the skin is about 0.97 for the temperature and not on the size, shape or relevant region of electromagnetic radiation. The material 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 = 66.4 W learn in later courses. 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) 11.9.5 Greenhouse Effect is given by The earth’s surface is a source of thermal H = AσT 4 (11.16) radiation as it absorbs energy received from the Sun. The wavelength of this radiation lies in the where A is the area and T is the absolute long wavelength (infrared) region. But a large temperature of the body. This relation obtained portion of this radiation is absorbed by experimentally by Stefan and later proved greenhouse gases, namely, carbon dioxide theoretically by Boltzmann is known as Stefan- (CO2); methane (CH 4); nitrous oxide (N 2O); Boltzmann law and the constant σ is called chlorofluorocarbon (CFxClx); and tropospheric Stefan-Boltzmann constant. Its value in SI units ozone (O3). This heats up the atmosphere which, is 5.67 × 10–8 W m–2 K–4. Most bodies emit only a in turn, gives more energy to earth, resulting in fraction of the rate given by Eq. 11.16. A substance warmer surface. This increases the intensity of like lamp black comes close to the limit. One, radiation from the surface. The cycle of therefore, defines a dimensionless fraction e processes described above is repeated until no called emissivity and writes, radiation is available for absorption. The net H = AeσT 4 (11.17) result is heating up of earth’s surface and Here, e = 1 for a perfect radiator. For a tungsten atmosphere. This is known as Greenhouse lamp, for example, e is about 0.4. Thus, a tungsten Effect. Without the Greenhouse Effect, the lamp at a temperature of 3000 K and a surface temperature of the earth would have been –18°C. area of 0.3 cm2 radiates at the rate H = 0.3 × Concentration of greenhouse gases has 10–4 × 0.4 × 5.67 × 10–8 × (3000)4 = 60 W. enhanced due to human activities, making the A body at temperature T, with surroundings earth warmer. According to an estimate, average temperature of earth has increased by 0.3 to at temperatures Ts, emits, as well as, receives 0.6°C, since the beginning of this century energy. For a perfect radiator, the net rate of because of this enhancement. By the middle of loss of radiant energy is the next century, the earth’s global temperature H = σA (T 4 – Ts4) may be 1 to 3°C higher than today. This global 2020-21 296 PHYSICS warming may cause problem for human life, From the graph you can infer how the cooling plants and animals. Because of global warming, of hot water depends on the difference of its ice caps are melting faster, sea level is rising, temperature from that of the surroundings. You and weather pattern is changing. Many coastal will also notice that initially the rate of cooling cities are at the risk of getting submerged. The is higher and decreases as the temperature of enhanced Greenhouse Effect may also result in the body falls. expansion of deserts. All over the world, efforts The above activity shows that a hot body loses are being made to minimise the effect of global heat to its surroundings in the form of heat warming. radiation. The rate of loss of heat depends on the difference in temperature between the body 11.10 NEWTON’S LAW OF COOLING and its surroundings. Newton was the first to study, in a systematic manner, the relation We all know that hot water or milk when left on between the heat lost by a body in a given a table begins to cool, gradually. Ultimately it enclosure and its temperature. attains the temperature of the surroundings. To According to Newton’s law of cooling, the rate study how slow or fast a given body can cool on of loss of heat, – dQ/dt of the body is directly exchanging heat with its surroundings, let us proportional to the difference of temperature perform the following activity. ∆T = (T2–T1) of the body and the surroundings. T a k e s o m e w a t e r, s a y 3 0 0 m L , i n a The law holds good only for small difference of calorimeter with a stirrer and cover it with a temperature. Also, the loss of heat by radiation two-holed lid. Fix the stirrer through one hole depends upon the nature of the surface of the and fix a thermometer through another hole body and the area of the exposed surface. We in the lid and make sure that the bulb of can write thermometer is immersed in the water. Note the reading of the thermometer. This reading – (11.19) T1 is the temperature of the surroundings. Heat the water kept in the calorimeter till it where k is a positive constant depending upon attains a temperature, say 40 °C above room the area and nature of the surface of the body. temperature (i.e., temperature of the Suppose a body of mass m and specific heat surroundings). Then, stop heating the water capacity s is at temperature T2. Let T1 be the by removing the heat source. Start the temperature of the surroundings. If the stop-watch and note the reading of the temperature falls by a small amount dT2 in time thermometer after a fixed interval of time, say dt, then the amount of heat lost is after every one minute of stirring gently with dQ = ms dT2 the stirrer. Continue to note the temperature ∴ Rate of loss of heat is given by (T2) of water till it attains a temperature about 5 °C above that of the surroundings. Then, plot dQ dT = ms 2 (11.20) a graph by taking each value of temperature dt dt ∆T = T2 – T1 along y-axis and the coresponding From Eqs. (11.15) and (11.16) we have value of t along x-axis (Fig. 11.19). dT2 –m s = k (T2 – T1 ) dt dT2 k =– dt = – K dt (11.21) T2 – T1 ms where K = k/m s ∆ On integrating, log e (T2 – T1) = – K t + c (11.22) or T2 = T1 + C′ e –Kt; where C′ = e c (11.23) Equation 11.23 enables you to calculate the Fig. 11.19 Curve showing cooling of hot water time of cooling of a body through a particular with time. range of temperature. 2020-21 THERMAL PROPERTIES OF MATTER 297 For small temperature differences, the rate time. A graph is plotted between log e (T2–T1) of cooling, due to conduction, convection, and [or ln(T2–T1)] and time (t). The nature of the radiation combined, is proportional to the graph is observed to be a straight line having difference in temperature. It is a valid a negative slope as shown in Fig. 11.20(b). This approximation in the transfer of heat from a is in support of Eq. 11.22. radiator to a room, the loss of heat through the Example 11.8 A pan filled with hot food t wall of a room, or the cooling of a cup of tea on the table. cools from 94 °C to 86 °C in 2 minutes when the room temperature is at 20 °C. How long will it take to cool from 71 °C to 69 °C? Answer The average temperature of 94 °C and 86 °C is 90 °C, which is 70 °C above the room temperature. Under these conditions the pan cools 8 °C in 2 minutes. Using Eq. (11.21), we have Change in temperature = K ∆T Time 8 °C = K ( 70 °C) 2 min The average of 69 °C and 71 °C is 70 °C, which Fig. 11.20 Verification of Newton’s Law of cooling. is 50 °C above room temperature. K is the same Newton’s law of cooling can be verified with for this situation as for the original. the help of the experimental set-up shown in 2 °C Fig. 11.20(a). The set-up consists of a double- = K (50 °C) walled vessel (V) containing water between Time the two walls. A copper calorimeter (C) When we divide above two equations, we containing hot water is placed inside the have double-walled vessel. Two thermometers through the corks are used to note the 8 °C/2 min K (70 °C) = temperatures T2 of water in calorimeter and 2 °C/time K (50 °C) T1 of hot water in between the double walls, respectively. Temperature of hot water in the Time = 0.7 min calorimeter is noted after equal intervals of = 42 s t 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. 2020-21 298 PHYSICS 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 = α l ∆T l ∆V = αV ∆T V 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 C D H=K A 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 : dQ = – k (T2 – T1 ) dt Where T1 is the temperature of the surrounding medium and T2 is the temperature of the body. 2020-21 THERMAL PROPERTIES OF MATTER 299 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 11.1 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. 11.2 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 ? 11.3 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 ? 11.4 Answer the following : (a) The triple-point of water is a standard fixed point in modern thermometry. 2020-21 300 PHYSICS 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 ? 11.5 Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made : 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 ? 11.6 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. 11.7 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. 11.8 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.