خواص مادة - الفيزياء للعام الأول - الرياضة PDF

‫خواص مادة‪-‬الفرقة األولي‪-‬‬ ‫رياضة مميز‪-‬الئحة موحدة‪-‬‬ ‫د‪.‬حسن عمر ‪ +‬د‪..‬ياسر المحالوى‬ ‫‪0202-0202‬‬ Heat & Properties of Matter (101Phy) For 1st year students (Mathematics group) Prepared by: Physics Department 2021  Heat C...

‫خواص مادة‪-‬الفرقة األولي‪-‬‬ ‫رياضة مميز‪-‬الئحة موحدة‪-‬‬ ‫د‪.‬حسن عمر ‪ +‬د‪..‬ياسر المحالوى‬ ‫‪0202-0202‬‬ Heat & Properties of Matter (101Phy) For 1st year students (Mathematics group) Prepared by: Physics Department 2021  Heat Contents Page number Topic 1 CHAPTER 1 :Heat and Temperature Solved Problems 17 CHAPTER 2: Principle of Calorimetry 24 Solved Problems 36 CHAPTER 3: HEAT TRANSFER 40 Solved Problems 64 CHAPTER 4: Kinetic Theory of Gases 77  CHAPTER 1 Heat and Temperature 1.1 Hot and Cold Bodies When we rub our hands for some time, they become warm. When a block slides on a rough surface, it becomes warm. Press against a rapidly spinning wheel. The wheel slows down and becomes warm. While going on a bicycle, touch the road with your shoe. The bicycle slows down, and the shoe becomes warm. When two vehicles collide with each other during an accident, they become very hot. When an aeroplane crashes, it becomes so hot that it catches fire. In each of these examples, mechanical energy is lost and the bodies in question become hot. Where does the mechanical energy vanish? It goes into the internal energy of the bodies. We conclude that the cold bodies absorb energy to become hot. In other words, a hot body has more internal energy than an otherwise identical cold body. When a hot body is kept in contact with a cold body, the cold body warms up and the hot body cools down. The internal energy of the hot body decreases and the internal energy of the cold body increases. Thus, energy is transferred from the hot body to the cold body when they are placed in contact. Notice that no mechanical work is done during this transfer of energy (neglect any change in volume of the body). This is because there are no displacements involved. This is different from the case when we lift a ball vertically and the energy of the ball- earth system increases or when a compressed spring relaxes, and a block attached to its end speeds up. In the case of lifting the ball, we do some work on the ball and the energy is increased by that amount. In the case of spring-block example, the spring does some work and the kinetic energy of the block increases. 1  The transfer of energy from a hot body to a cold body is nonmechanical process. The energy that is transferred from one body to the other, without any mechanical work involved, is called heat. 1.2.Zeroth Law of the Thermodynamics Two bodies are said to be in thermal equilibrium if no transfer of heat takes place when they are placed in contact. We can now state the Zeroth law of thermodynamics as follow: “If two bodies A and B are in thermal equilibrium and A and C are also in thermal equilibrium , then B and C are also in thermal equilibrium.” The Zeroth law allows us to introduce the concept of temperature to measure the hotness or coldness of a body. All bodies in thermal equilibrium are assigned equal temperature. A hotter body is assigned higher temperature than a colder body. Thus, the temperature of two bodies decide the direction of heat flow when the two bodies are put in contact. Heat flows from the body at higher temperature to the body at lower temperature. 1.3 Thermometers and Temperature Scales We are now in a position to say whether two given bodies are at the same temperature or not. If they are not at the same temperature, we should know which is at higher temperature and which is at lower temperature. Our next task is to define a scale of temperature so that we can give numerical value to the temperature of a body. To do this, we can choose a substance and look for a measurable property of the substance which monotonically changes with temperature. The temperature can then be defined as a chosen function of this property. As an example, take a mass of the mercury in a glass bulb terminating in a long capillary. The length of the mercury column in the capillary, changes with temperature. Each length corresponds to a particular temperature of the mercury. How can we assign a numerical value corresponding to an observed length of the mercury column? 2  The earlier method was to choose two fixed points of temperature which can be easily reproduced in laboratory. The temperature of melting ice at 1 atm (called ice point) and the temperature of boiling water at 1 atm (called steam point) are often chosen as the fixed point. We arbitrarily assign a temperature T1 to the ice point and T2 to the steam point. Suppose the length of the mercury column is when the bulb is kept in melting ice and it is when the bulb is kept in boiling water. Thus, a length of mercury column means that the temperature of the bulb is T1 and a length of the column means that the temperature is T2. The temperature corresponding to any length may be defined by assuming a linear relation between and T. Figure (1.1) (1.1) A change of one degree in temperature will mean a change of in the length of mercury column. Thus, we can graduate the length of the capillary directly in degrees. The centigrade system assumes ice point at 0 oC and the steam point at 100 oC. If the length of the mercury column between its values for 0 oC and 100 oC is divided equally in 100 parts, each part will correspond to a change of 1 oC. Let and denote the lengths of the mercury column at 0 oC and 100 oC respectively. From Eq. (1.1), 3 and giving and Putting these values in Eq. (1.1), the temperature corresponding to a length is given by (1.2) To measure the temperature of the body, the bulb containing mercury is kept in contact with the body and enough time is allowed so that the mercury comes to thermal equilibrium with the body. The temperature of the body is then the same as that of the mercury. The length of the column then gives the temperature according to Eq. (1.2). 1.3.1 Fahrenheit system Another popular system known as Fahrenheit system assumes 32 oF for the ice point and 212 oF for the steam point. A change of 1 oF means 1/180 of the interval between the steam point and the ice point. For example, the average temperature of a normal human body is a round 98 oF. The conversion formula from Centigrade to Fahrenheit scale is (1.3) Eq. (1.3) can also be used to find a relationship between changes in temperature on the Celsius and Fahrenheit scales. The expansion of mercury is just one thermometric property that can be used to define a temperature scale and prepare thermometers. There may be many other thermometric properties. Electric resistance of a metal wire increases monotonically with temperature and may be used to define a temperature scale. If R0 and R100 denote 4 the resistances of a metal wire at ice point and steam point respectively, we can define temperature T corresponding to the resistance RT as (1.4) The temperature of the ice point and the steam point are chosen to be 0 oC and 100 oC as in centigrade scale. A platinum wire is often used to construct a thermometer based on this scale. Such a thermometer is called Platinum resistance thermometer and the temperature scale is called the platinum scale. 1.3.2 Platinum Resistance Thermometer The platinum resistance thermometer works on the principle of Wheatstone bridge used to measure a resistance. In a Wheatstone bridge, four resistances P, Q, R and X are joined in a loop as shown in figure (1.2a). A galvanometer and a battery are also joined as shown. If there is no deflection in the galvanometer and the bridge is called balanced. If the condition is not fulfilled, there is a deflection. Figure (1.2) 5  Figure (1.2b) represents the arrangements for a platinum resistance thermometer. A thin platinum wire is coiled on a mica base and placed in a glass tube. Two connecting wires YY’ going through ebonite lid of the tube are connected to the platinum coil. A similar copper wire XX’, called compensating wire, also goes into the tube as shown in figure. A Wheatstone bridge is arranged as shown in figure (1.2c). Two equal resistances P and Q are connected in two arms of the bridge. A copper coil having resistance roughly equal to that of the platinum coil is connected in the third arm through the compensating wire XX’. The platinum coil is inserted in the fourth arm of the bridge by connecting the points YY’ in that arm. The end C of the wire connected to the galvanometer can slide on a uniform wire AB of length. The end A of this wire is connected to the wire XX’ and the end B is connected to the wire YY’. Thus, the copper coil, the compensating wire and the wire AC are in the third arm and the platinum coil, the connecting wire YY’ and the wire CB are in the fourth arm of the bridge. The test tube containing platinum coil is immersed in the bath of which we can measure the temperature. The end C is slid on AB till the deflection in the galvanometer becomes zero. Let Suppose the resistance of the copper coil connected in the third arm is R, that of the compensating wire is Rc and that of the wire AB is r. The resistance of the connecting wire YY’ is the same as Rc and that of the platinum coil is RT. The resistance of the wire , and that of the wire. The net resistance in the third arm is and that in the fourth arm is. For no deflection in the galvanometer, As , we get 6 or,. If , and are the values of at the ice point, steam point and the temperature T respectively. So, and This gives (1.5) 1.3.3 Absolute Scale and Ideal Gas Scale Eq. (1.4) tells us that the change in resistance when the temperature increases by one degree is and is the same for all temperatures. This means that the resistance increases uniformly as the temperature increases. Note that this conclusion follows from the definition, Eq. (1.4), of temperature and is not the property of platinum or the other metal used to form the resistance. Similarly, it follows that the expansion of mercury is uniform on the mercury scale defined by Eq. (1.2). If we measure the change in resistance of a platinum 7 wire against the temperature measured on mercury scale, we shall find that the resistance varies slowly at lower temperatures and slightly more rapidly at higher temperatures. Similarly, if the expansion of mercury is measured against the temperatures defined on platinum scale Eq. (1.4), we shall find that mercury does not expand at uniform rate as the temperature varies. Thus, the two scales of temperatures do not agree with each other. They are forced to agree at ice point and steam point, but for other physical states the readings of the two thermometers will be different. In general, the scale depends on the properties of thermometric substance used to define the scale. It is possible to define an absolute scale of temperature which does not depend on the thermometric substance and its properties. We now define the ideal gas temperature scale which happens to be identical to the absolute temperature scale. The chosen standard for this purpose is the constant-volume gas thermometer. 1.3.4 Constant volume gas thermometer A gas enclosed in a container has a definite volume and a definite pressure in any given state. If the volume is kept constant, the pressure of a gas increases monotonically with increasing temperature. This property of a gas may be used to construct a thermometer. Figure (1.3) shows a schematic diagram of a constant volume gas thermometer. Figure (1.3) Constant volume gas thermometer 8  A mass of gas is enclosed in a bulb A connected to a capillary BC. The capillary is connected to the manometer CD which contains mercury. The other end of the manometer is open at atmosphere. A vertical meter scale E is fixed in such a way that the height of the mercury column in the tube D can easily be measured. The mercury in the manometer is connected to the mercury reservoir F through a rubber tube. The capillary BC has a fixed mark at C. By raising or lowering the reservoir F, the mercury level in the left part of the manometer is maintained at C. This ensures that the volume of the gas enclosed in the bulb A (and the capillary BC) remains constant. The pressure of the gas is equal to the atmospheric pressure plus the pressure due to the difference of the mercury columns in the manometer. Thus, , where is the atmospheric pressure, is the difference of mercury levels in the manometer tube, is the density of mercury, and is the acceleration due to gravity. If the temperature of the bulb is increased and its volume is kept constant by adjusting the height of the reservoir F, the pressure of the gas P increases. Thus, a temperature scale may be defined by choosing some suitable function of this pressure. Let us assume that the temperature is proportional to the pressure, i.e., (1.6) where c is some constant. In addition, the temperature of triple point of water is assigned a value 273.16 k. (Triple point is a state in which ice, water, and water vapor can stay together in equilibrium.). The unit is called a kelvin and is denoted by the symbol K. To get the value of the constant c in Eq. (1.6), we can put the bulb A in a triple point cell and measure the pressure of the gas. From Eq. (1.6), or,. 9  The temperature of the gas when the pressure is P is obtained by putting this value of c in Eq. (1.6). It is (1.7) To use the thermometer, we must first determine the pressure of the gas at the triple point. This is a fixed value for the thermometer and is used in any measurement. To measure the temperature of a bath of extended volume, the bulb A is dipped in the bath. Sufficient time is allowed so that the gas in the bulb comes to thermal equilibrium with the bath. The reservoir F is adjusted to bring the volume of the gas to its original value and the pressure P of the gas is measured with the manometer. The temperature T on the gas scale is then obtained from Eq. (1.7). One can also define a centigrade scale with gas thermometer. Suppose the pressure of the gas is when the bulb A is placed in melting ice (ice point) and it is when the bulb is placed in the steam bath (steam point). We assign 0 oC to the temperature of the ice point and 100 oC to the steam point. The temperature T corresponding to a pressure P of the gas is defined by O C (1.8) The constant volume gas thermometer allows several errors in the temperature measurements. The main sources of error are the following: (a) The space in the capillary tube BC generally remains out of the heat bath in which the bulb A is placed. The gas in BC is, therefore, not at the same temperature as the gas in A. (b) The volume of the glass bulb changes slightly with temperature allowing the volume of the gas to change. 10 Notice that: it can be shown that the Celsius temperature TC is shifted from the absolute (Kelvin) temperature T by 273.15. Because the size of a Celsius degree is the same as a kelvin, a temperature difference of 5 oC is equal to a temperature difference of 5 K. The two scales differ only in the choice of zero point. The ice point (273.15 K) corresponds to 0.00 oC, and the steam point (373.15 K) is equivalent to 100.00 oC. Figure (1.4) Example (1.1) The pressure of air in the bulb of a constant volume gas thermometer is 73 cm of mercury at 0 oC, 100.3 cm of mercury at 100 oC, and 77.8 cm of mercury at room temperature. Find the room temperature in centigrade. Solution O We have C O O C C 11 1.4 Thermal Expansion of Solids and Liquids Our discussion of the liquid thermometer made use of one of the best-known changes that occur in most substances. As temperature of the substance increases, its volume increases. This phenomenon, known as thermal expansion, plays an important role in numerous applications. Thermal expansion joints, for example, must be included in buildings, concrete highways, and bridges to compensate for changes in dimensions with variations in temperature (Fig.1.5). Figure (1.5) The overall thermal expansion of an object is a consequence of the change in the average separation between its constituent atoms or molecules. To understand this idea, consider how the atoms in a solid substance behave. These atoms are located at fixed equilibrium positions; if an atom is pulled away from its position, a restoring force pulls it back. We can imagine that the atoms are particles connected by springs to their neighboring atoms. (See Fig. (1.6). Figure (1.6) 12  If an atom is pulled away from its equilibrium position, the distortion of the springs provides a restoring force. At ordinary temperatures, the atoms vibrate around their equilibrium positions with an amplitude (maximum distance from the center of vibration) of about 10 -11 m, with an average spacing between the atoms of about 10 -10 m. As the temperature of the solid increases, the atoms vibrate with greater amplitudes and the average separation between them increases. Consequently, the solid as a whole expands. If the thermal expansion of an object is sufficiently small compared with the object’s initial dimensions, then the change in any dimension is, to a good approximation, proportional to the first power of the temperature change. Suppose an object has an initial length L0 along some direction at some temperature T0. Then the length increases by for a change in temperature. So, for small changes in temperature, (1.9) or, where L is the object’s final length T is its final temperature, and the proportionality constant is called the coefficient of linear expansion for a given material and has units of (OC)-1. Table (1.1) lists the coefficients of linear expansion for various materials. 13  Table (1.1) Note that: for these materials is positive, indicating an increase in length with increasing temperature. Example (1.2) A steel railroad track has a length of 30 m when the temperature is 0 oC. What is its length on a hot day when the temperature is 40.0 oC? solution The change in length is The coefficient of linear expansion of steel can be obtained from table (1), so we have 30.000 m 40.0 oC= 0.013 m Add the change to the original length to find the final length 14 Exercise (1.1) What is the length of the same railroad track on a cold winter day when the temperature is 0 oF? [Answer 29.994 m] 1.4.1 Bimetallic Strips and Thermostats How can different coefficients of expansion for metals be used as a temperature gauge and control electronic devices such as air conditioners? When the temperatures of a brass rod and a steel rod of equal length are raised by the same amount from some common initial value, the brass rod expands more than the steel rod because brass has a larger coefficient of expansion than steel. A simple device that uses this principle is a bimetallic strip. Such strips can be found in the thermostats of certain home heating systems. The strip is made by securely bonding two different metals together. As the temperature of the strip increases, the two metals expand by different amounts and the strip bends, as in Figure (1.7). The change in shape can make or break an electrical connection. 15 Figure (1.7): (a) A bimetallic strip bends as the temperature changes because the two metals have different coefficients of expansion. (b) A bimetallic strip used in a thermostat to break or make electrical contact. Application One practical application of thermal expansion is the common technique of using hot water to loosen a metal lid stuck on a glass jar. This works because the circumference of the lid expands more than the rim of the jar. Because the linear dimensions of an object change due to variations in temperature, it follows that surface area and volume of the object also change. Consider a square of material having an initial length on a side and therefore an initial area. As the temperature is increased, the length of each side increases to The new area A is 16. The last term in this expression contains the quantity raised to the second power. Because is much less than one, squaring it makes it even smaller. Consequently, we can neglect this term to get a simpler expression:.. So that, (1.10) where,. The quantity (Greek letter beta) is called the coefficient of area expansion. We can also show that the increase in volume of an object accompanying a change in temperature is (1.11) where. The quantity (Greek letter gamma) is called the coefficient of volume expansion. The proof of Eq. (1.11) is similar to the proof of Eq. (1.10). Note that: and only if the coefficient of linear expansion of the object is the same in all directions. Exercise (1.2) Prove that Solved Problems 1. The pressure of the gas in a constant volume gas thermometer at steam point (273.15 K) is What will be the pressure at the triple point of water? Solution 17  Thus, or, 2. The pressure of air in the bulb of a constant volume gas thermometer at 0 oC and 100 oC are 73.00 cm and 100 cm of mercury respectively. Calculate the pressure of the room temperature 20 oC. Solution The room temperature on the scale measured by the thermometer is O C. O Thus, C. or, 3. The pressure of the gas in a constant volume gas thermometer is 80 cm of mercury in melting ice at 1 atm. When the bulb is placed in a liquid, the pressure becomes 160 cm of mercury. Find the temperature of the liquid. Solution For an ideal gas at constant volume, 18  or, The temperature of melting ice at 1 atm is 273.15 K. Thus, the temperature of the liquid is 4. In a constant volume gas thermometer, the pressure of the working gas is measured by the difference in the levels of the mercury in the two arms of a U-tube connected to the gas at one end. When the bulb is placed at the room temperature 27.0 OC, the mercury column in the arm open to atmosphere stands 5.00 cm above the level of the mercury in the other arm. When the bulb is placed in a hot liquid, the difference of mercury levels becomes 45.0 cm. Calculate the temperature of the liquid (Atmospheric pressure = 73.0 cm of mercury). Solution The pressure of the gas= atmospheric pressure + the pressure due to the difference in mercury level. At 27 OC, the pressure is 75 cm + 5 cm = 80 cm of mercury. At liquid temperature, the pressure is 75 cm + 45 cm = 120 cm of mercury. Using the temperature of the liquid is 5. The resistance of a platinum resistance thermometer at ice point, the steam point and the boiling point of Sulphur are 2.50, 3.50 and 6.50 respectively. Find the boiling o point of Sulphur on the platinum scale. The ice point and the steam point measure 0 and 100 o respectively. Solution 19  The temperature on the platinum scale is defined as. The boiling point of Sulphur on this scale is. 6. A platinum resistance thermometer reads 0 o and 100 o at the ice point and the boiling point of water respectively. The resistance of a platinum wire varies with Celsius temperature T as where and. What will be the reading of this thermometer if it is placed a liquid both maintained at 50 oC? Solution The resistance of the wire in the thermometer at 100 oC and 50 oC are and, The temperature T measured on the platinum thermometer is given by 7. An iron of length 50 cm is joined at an end to an aluminium rod of length 100 cm. All measurements refer to 20 oC. Find the length of the composite system at 100 oC and its average coefficient of linear expansion. The coefficient of linear expansion of iron and aluminium are and respectively. Solution The length of the iron rod at 100 oC is 20  [ ] The length of the aluminium rod at 100 oC is [ ] The length of the composite system at 100 oC is 50.048 cm + 100.192 cm = 150.24 cm The length of the composite system at is 150 cm. So, the average coefficient of linear expansion of the composite rod is 8. An iron ring measuring 15.00 cm in diameter is to be shrunk on a pulley which is 15.05 cm in diameter. All measurements refer to the room temperature 20 oC. To what minimum temperature should the ring be heated to make the job possible? Calculate the strain developed in the ring when it comes to the room temperature. Coefficient of linear expansion of iron=. Solution The ring should be heated to increase its diameter from 15.00 cm to 15.05 cm. Using [ ] The temperature = The strain developed. 9. A pendulum clock consists of an iron rod connected to a small, heavy bob. If it is designed to keep correct time at 20 oC, how fast or slow will it go in 24 hours at 40 oC? Coefficient of linear coefficient of iron =. Solution The time period at temperature T is √ 21  √ √ ( ) Thus, ( ) and, ( ) or, [ ][ ] [ ][ ] or, This is the fractional loss of time. As the temperature increases, the time period also increases. Thus, the clock goes slow. The time lost in 24 hours is. 10. A pendulum clock having copper rod keeps correct time at 20 oC. It gains 15 seconds per day if cooled to 0 oC. Calculate the coefficient of linear expansion of copper. Solution The time period at temperature T is √ ( ) Thus, or, 22  is the correct time period. The period at is smaller so that the clock runs fast. The previous equation gives approximately the fractional gain in time. The time gained in 24 hours is or, or, Exercise 1.3 A glass vessel of volume 100 cm-3 is filled with mercury and is heated from 25 oC to 75 oC. What volume of mercury will overflow? Coefficient of linear of glass= and coefficient of volume expansion of mercury is 23  CHAPTER 2 Principle of Calorimetry 2.1 Heat as a Form of Energy When two bodies at different temperatures are placed in contact, the hotter body cools down and the colder body warms up. Energy is thus transferred from a body at higher temperature to a body at lower temperature when they are brought in contact. The energy being transferred between two bodies or between adjacent parts of a body as a result of temperature difference is called heat. Thus, heat is a form of energy. It is energy in transit whenever temperature differences exist. Once it is transferred it becomes the internal energy of the receiving body. It should be clearly understood that the word “heat” is meaningful only as long as the energy is being transferred. Thus, expressions like “heat in a body” or “heat of a body” are meaningless. 2.1.1 Measurement Units of Heat As heat is just energy in transit, its unit in SI is joule. However, another unit of heat “calorie” is in wide use. This unit was formulated much before it was recognized that heat is a form of energy. The old day definition of calorie is as follows: The amount of heat needed to increase the temperature of 1 g of water from 14.5°C to 15.5°C at a pressure of 1 atm is called 1 calorie. The amount of heat needed to raise the temperature of 1 g of water by 1°C depends slightly on the actual temperature of water and the pressure. That is why, the range 14.5°C to 15.5°C and the pressure 1 atm was specified in the definition. We shall ignore this small variation and use one calorie to mean the amount of heat needed to increase the temperature of 1 g of water by 1°C at any region of temperature and pressure. The calorie is now defined in terms of joule as 1 cal = 4.186 joule. We also use the unit kilocalorie which is equal to 1000 calorie as the name indicates. 24 Example (2.1) What is the kinetic energy of a 10 kg mass moving at a speed of 36 km h − 1 in calorie? Solution The kinetic energy is ( ) 2.2 Principle of Calorimetry A simple calorimeter is a vessel generally made of copper with a stirrer of the same material. The vessel is kept in a wooden box to isolate it thermally from the surrounding. A thermometer is used to measure the temperature of the contents of the calorimeter. Objects at different temperatures are made to come in contact with each other in the calorimeter. As a result, heat is exchanged between the objects as well as with the calorimeter. Neglecting any heat exchange with the surrounding, the principle of calorimetry states that the total heat given by the hot objects equals the total heat received by the cold objects. 2.3 Specific Heat Capacity and Molar Heat Capacity When we supply heat to a body, its temperature increases. The amount of heat absorbed depends on the mass of the body, the change in temperature, the material of the body as well as the surrounding conditions, such as pressure. We write the equation (2.1) where is the change in temperature, m is the mass of the body, is the heat supplied, and s is a constant for the given material under the given surrounding conditions. The constant s is called specific heat capacity of the substance. 25  When a solid or a liquid is kept open in the atmosphere and heated, the pressure remains constant. Table (2.1) gives the specific heat capacities of some of the solids and liquids under constant pressure condition. As can be seen from Eq. (1), the SI unit for −1 −1 −1 specific heat capacity is J kg K which is the same as J kg °C −1. The specific heat −1 −1 −1 −1 capacity may also be expressed in cal g K (same as cal g °C ). Specific heat capacity is also called specific heat in short. Table (2.1) Specific heat capacities of some materials The amount of substance in the given body may also be measured in terms of the number of moles. Eq. (2.1) may be rewritten as where n is the number of moles in the sample. The constant C is called molar heat capacity. Example (2.2) A copper block of mass 60 g is heated till its temperature is increased by 20°C. Find the heat supplied to the block. [Specific heat capacity of copper = 0.09 cal g −1 °C −1]. Solution The heat supplied is = (60 g) (0.09 cal g −1 °C −1) (20°C) = 108 cal. 26 The quantity ms is called the heat capacity of the body. Its unit is J K −1. The mass of water having the same heat capacity as a given body, is called the water equivalent of the body. 2.3.1 Determination of Specific Heat in Laboratory Figure (2.1) shows Regnault’s apparatus to determine the specific heat capacity of a solid heavier than water, and insoluble in it. A wooden partition P separates a steam chamber O and a calorimeter C. The steam chamber O is a double-walled cylindrical vessel. Steam can be passed in the space between the two walls through an inlet A and it can escape through an outlet B. The upper part of the vessel is closed by a cork. The given solid may be suspended in the vessel by a thread passing through the cork. A thermometer T1 is also inserted into the vessel to record the temperature of the solid. The steam chamber is kept on a wooden platform with a removable wooden disc D closing the bottom hole of the chamber. Figure (2.1) Regnault’s apparatus To start with, the experimental solid (in the form of a ball or a block) is weighed and then suspended in the steam chamber. Steam is prepared by boiling water in a separate boiler and is passed through the steam chamber. A calorimeter with a stirrer is weighed and sufficient amount of water is kept in it so that the solid may be completely 27 immersed in it. The calorimeter is again weighed with water to get the mass of the water. The initial temperature of the water is noted. When the temperature of the solid becomes constant (say for 15 minutes), the partition P is removed, the calorimeter is taken below the steam chamber, the wooden disc D is removed and the thread is cut to drop the solid in the calorimeter. The calorimeter is taken to its original place and is stirred. The maximum temperature of the mixture is noted. Calculation: Let the mass of the solid = m1 mass of the calorimeter and the stirrer = m2 mass of the water = m3 specific heat capacity of the solid = s1 specific heat capacity of the material of the calorimeter (and stirrer) = s2 specific heat capacity of water = s3 initial temperature of the solid = T1 initial temperature of the calorimeter, stirrer and water = T2 final temperature of the mixture = T. We have, heat lost by the solid = m1 s1 (T1 − T) heat gained by the calorimeter (and the stirrer) = m2 s2 (T − T2) and heat gained by the water = m3 s3 (T − T2). Assuming no loss of heat to the surrounding, the heat lost by the solid goes into the calorimeter, stirrer and water. Thus, m1 s1 (T1 − T)= m2 s2 (T − T2)+ m3 s3 (T − T2) (2.2) or, 28  − 1 − 1 Knowing the specific heat capacity of water (s3 = 4186 J kg K ) and that of the − 1 − 1 material of the calorimeter and the stirrer (s2 = 389 J kg K if the material be copper), one can calculate s1. Note that: specific heat capacity of a liquid can also be measured with the Renault's apparatus. Here a solid of known specific heat capacity s1 is used and the experimental liquid is taken in the calorimeter in place of water. The solid should be denser than the liquid. Using the same procedure and with the same symbols, we get an equation identical to Eq. (2) above, that is, m1 s1 (T1 − T)= m2 s2 (T − T2)+ m3 s3 (T − T2) in which s3 is the specific heat capacity of the liquid. We get, 2.4 Latent Heat of Fusion and Vaporization Apart from raising the temperature, heat supplied to a body may cause a phase change such as solid to liquid or liquid to vapor. During this process of melting or vaporization, the temperature remains constant. The amount of heat needed to melt a solid of mass m may be written as (2.3) where L is a constant for the given material (and surrounding conditions). This constant L is called specific latent heat of fusion. The term latent heat of fusion is also used to mean the same thing. Eq. (2.3) is also valid when a liquid change its phase to vapor. The constant L in this case is called the specific latent heat of vaporization or simply latent heat of vaporization. When a vapor condenses or a liquid solidifies, heat is released to the surrounding. 29  In solids, the forces between the molecules are large and the molecules are almost fixed in their positions inside the solid. In a liquid, the forces between the molecules are weaker and the molecules may move freely inside the volume of the liquid. However, they are not able to come out of the surface. In vapors or gases, the intermolecular forces are almost negligible, and the molecules may move freely anywhere in the container. When a solid melt, its molecules move apart against the strong molecular attraction. This needs energy which must be supplied from outside. Thus, the internal energy of a given body is larger in liquid phase than in solid phase. Similarly, the internal energy of a given body in vapor phase is larger than that in liquid phase. Example (2.3) A piece of ice of mass 100 g and at temperature 0°C is put in 200 g of water at 25°C. How much ice will melt as the temperature of the water reaches 0°C ? The specific heat capacity of water = 4200 J kg − 1 K − 1 and the specific latent heat of fusion of ice = 3.4 × 10 5 J kg − 1. Solution The heat released as the water cools down from 25°C to 0°C is = (0.2 kg) (4200 J kg − 1 K − 1) (25 K) = 21000 J. The amount of ice melted by this much heat is given by = 62 g. 30 2.4.1 Measurement of Latent Heat of Fusion of Ice An empty calorimeter (together with a stirrer) is weighed. About two third of the calorimeter is filled with water and is weighed again. Thus, one gets the mass of the water. The initial temperature of the water is measured with the help of a thermometer. A piece of ice is taken and as it starts melting it is dried with a blotting paper and put into the calorimeter. The water is stirred keeping the ice always fully immersed in it. The minimum temperature reached is recorded. This represents the temperature when all the ice has melted. The calorimeter with its contents is weighed again. Thus, one can get the mass of the ice that has melted in the calorimeter. Calculation: Let the mass of the calorimeter (with stirrer) = m1 mass of water = m2 mass of ice = m3 initial temperature of the calorimeter and the water (in Celsius) = T1 final temperature of the calorimeter and the water (in Celsius) = T2 temperature of the melting ice = 0 °C specific latent heat of fusion of ice = L specific heat capacity of the material of the calorimeter (and stirrer) = s1 specific heat capacity of water = s2. We have, heat lost by the calorimeter (and the stirrer) = m1s1(T1 − T2) heat lost by the water kept initially in the calorimeter = m2s2(T1 – T2) heat gained by the ice during its fusion to water = m3L heat gained by this water in coming from 0°C to T2 = m3s2 T2. Assuming no loss of heat to the surrounding, m1s1(T1 − T2) + m2s2(T1 – T2) = m3L + m3s2 T2 31 Knowing the specific heat capacity of water and that of the material of the calorimeter and the stirrer, one can calculate the specific latent heat of fusion of ice L. 2.4.2 Measurement of Latent Heat of Vaporization of Water The specific latent heat of vaporization of water can be measured by the arrangement showed in Figure (2.2). Steam is prepared by boiling water in a boiler A. The cork of the boiler has two holes. A thermometer T1 is inserted into one to measure the temperature of the steam and the other contains a bent glass tube to carry the steam to a steam trap B. A tube C with one end bent and the other end terminated in a jet is fitted in the steam trap. Another tube D is fitted in the trap which is used to drain out the extra steam and water condensed at the bottom. Figure (2.2) To start the experiment, an empty calorimeter (together with the stirrer) is weighed. About half of it is filled with water and is weighed again. Thus, one gets the mass of the water. The initial temperature of the water and the calorimeter is measured 32 by a thermometer T2. Water is kept in the boiler A and is heated. As it boils, the steam passes to the steam trap and then comes out through the tubes C and D. After some steam has gone out (say for 5 minutes), the temperature of the steam is noted. The calorimeter with the water is kept below the tube C so that steam goes into the calorimeter. The water in the calorimeter is continuously stirred and the calorimeter is removed after the temperature in it increases by about 5°C. The final temperature of the water in the calorimeter is noted. The calorimeter together with the water (including the water condensed) is weighed. From this, one gets the mass of the steam that condensed in the calorimeter. Let the mass of the calorimeter (with the stirrer) = m1 mass of the water = m2 temperature of the steam = T1 initial temperature of the water in the calorimeter = T2 final temperature of the water in the calorimeter = T3 specific latent heat of vaporization of water = L specific heat capacity of the material of the calorimeter (and the stirrer) = s1 specific heat capacity of water = s2. We have, heat gained by the calorimeter (and the stirrer) = m1s1(T3 − T2) heat gained by the water kept initially in the calorimeter = m2s2(T3 − T2) heat lost by the steam in condensing = m3L heat lost by the condensed water in cooling from temperature T1 to T3 = m3s2(T1 − T3). Assuming no loss of heat to the surrounding, m1 s1 (T3 − T2) + m2 s2 (T3 − T2)= m3 L + m3 s2 (T1 − T3) or, Knowing the specific heat capacity of water and that of the material of the calorimeter, one can calculate the specific latent heat of vaporization of water L. 33 Example (2.4) A calorimeter of water equivalent 15 g contains 165 g of water at 25°C. Steam at 100°C is passed through the water for some time. The temperature is increased to 30°C and the mass of the calorimeter and its contents is increased by 1⋅5 g. Calculate the specific latent heat of vaporization of water. Specific heat capacity of water is 1 cal g − 1 °C − 1. Solution Let L be the specific latent heat of vaporization of water. The mass of the steam condensed is 1.5 g. Heat lost in condensation of steam is The condensed water cools from 100°C to 30°C. Heat lost in this process is Heat supplied to the calorimeter and to the cold water during the rise in temperature from 25°C to 30°C is If no heat is lost to the surrounding, or, 34 2.5 Mechanical Equivalent of Heat In early days heat was not recognized as a form of energy. Heat was supposed to be something needed to raise the temperature of a body or to change its phase. Calorie was defined as the unit of heat. Several experiments were performed to show that the temperature may also be increased by doing mechanical work on the system. These experiments established that heat is equivalent to mechanical energy and measured how much mechanical energy is equivalent to a calorie. If mechanical work W produces the same temperature change as heat H, we write, W=JH (2.3) where J is called mechanical equivalent of heat. It is clear that if W and H are both measured in the same unit then J = 1. If W is measured in joule (work done by a force of 1 N in displacing an object by 1 m in its direction) and H in calorie (heat required to raise the temperature of 1 g of water by 1°C) then J is expressed in joule per calorie. The value of J gives how many joules of mechanical work is needed to raise the temperature of 1 g of water by 1°C. 35  Solved Problems 1. Calculate the amount of heat required to convert 1.00 kg of ice at −10°C into steam at 100°C at normal pressure. Specific heat capacity of ice = 2100 J kg − 1 K − 1, latent heat of fusion of ice = 3.36 × 10 5 J kg − 1, specific heat capacity of water = 4200 J kg − 1 K − 1 and latent heat of vaporization of water = 2.25 × 10 6 J kg − 1. Solution Heat required to take the ice from −10°C to 0°C = (1 kg) (2100 J kg − 1 K − 1) (10 K) = 21000 J. Heat required to melt the ice at 0°C to water = (1 kg) (3.36 × 10 5 J kg − 1) = 336000 J. Heat required to take 1 kg of water from 0°C to 100°C = (1 kg) (4200 J kg − 1 K − 1) (100 K) = 420000 J. Heat required to convert 1 kg of water at 100°C into steam = (1 kg) (2.25 × 10 6 J kg − 1) = 2.25 × 10 6 J. Total heat required = 3.03 × 10 6 J. 2. A 5 g piece of ice at −20°C is put into 10 g of water at 30°C. Assuming that heat is exchanged only between the ice and the water, find the final temperature of the mixture. Specific heat capacity of ice = 2100 J kg − 1 °C − 1, specific heat capacity of water = 4200 J kg − 1 °C − 1 and latent heat of fusion of ice = 3.36 × 10 5 J kg − 1. Solution The heat given by the water when it cools down from 30°C to 0°C is (0.01 kg) (4200 J kg − 1 °C − 1) (30°C) = 1260 J. The heat required to bring the ice to 0°C is (0.005 kg) (2100 J kg − 1 °C − 1) (20°C) = 210 J. The heat required to melt 5 g of ice is (0.005 kg) (3.36 × 10 5 J kg − 1) = 1680 J. We see that whole of the ice cannot be melted as the required amount of heat is not provided by the water. Also, the heat is enough to bring the ice to 0°C. Thus, the final temperature of the mixture is 0°C with some of the ice melted. 36 3. An aluminium container of mass 100 g contains 200 g of ice at −20°C. Heat is added to − 1 the system at a rate of 100 cal s. What is the temperature of the system after 4 minutes ? Draw a rough sketch showing the variation in the temperature of the system − 1 − 1 as a function of time. Specific heat capacity of ice = 0.5 cal g °C , specific heat capacity of aluminium = 0.2 cal g − 1 °C − 1, specific heat capacity of water = 1 cal g − 1 °C − 1 and latent heat of fusion of ice = 80 cal g − 1. Solution Total heat supplied to the system in 4 minutes is Q = 100 cal s − 1 × 240 s = 2.4 × 104 cal. The heat required to take the system from − 20°C to 0°C = (100 g) × (0.2 cal g − 1 °C − 1) × (20°C) + (200 g) × (0.5 cal g − 1 °C − 1) × (20°C) = 400 cal + 2000 cal = 2400 cal. The time taken in this process = s = 24 s. The heat required to melt the ice at 0°C= (200 g) (80 cal g − 1) = 16000 cal. The time taken in this process = s = 160 s. If the final temperature is T, the heat required to take the system to the final temperature is = (100 g) (0.2 cal g − 1 °C − 1) T + (200 g) (1 cal g − 1 °C − 1) T. Thus, 2.4 × 10 4 cal = 2400 cal + 16000 cal + (220 cal °C − 1) T or, The variation in the temperature as a function of time can be sketched as 37 4. A thermally isolated vessel contains 100 g of water at 0°C. When air above the water is pumped out, some of the water freezes and some evaporates at 0°C itself. Calculate the mass of the ice formed if no water is left in the vessel. Latent heat of vaporization of water at 0°C = 2.10 × 10 6 J kg − 1 and latent heat of fusion of ice = 3.36 × 10 5 J kg − 1. Solution Total mass of the water = M = 100 g. Latent heat of vaporization of water at 0°C = L 1 = 21.0 × 105 J kg − 1. Latent heat of fusion of ice = L 2 = 3.36 × 105 J kg −1. Suppose, the mass of the ice formed = m. Then, the mass of water evaporated = M − m. Heat taken by the water to evaporate = (M − m) L1 and heat given by the water in freezing = mL2. Thus, mL2 = (M − m)L1 or,. 5. A lead bullet penetrates a solid object and melts. Assuming that 50% of its kinetic energy was used to heat it, calculate the initial speed of the bullet. The initial temperature of the bullet is 27°C and its melting point is 327°C. Latent heat of fusion of lead = 2.5 × 104 J kg−1 and specific heat capacity of lead = 125 J kg 1 K 1. Solution Let the mass of the bullet = m. Heat required to take the bullet from 27°C to 327°C 38  = m × (125 J kg − 1 K − 1) (300 K) = m × (3.75 × 10 4 J kg − 1). Heat required to melt the bullet = m × (2.5 × 10 4 J kg − 1). If the initial speed be v, the kinetic energy is and hence the heat developed is ( ). Thus, or, 6. A lead ball at 30°C is dropped from a height of 6.2 km. The ball is heated due to the air resistance and it completely melts just before reaching the ground. The molten substance falls slowly on the ground. Calculate the latent heat of fusion of lead. −1 − 1 Specific heat capacity of lead = 126 J kg °C and melting point of lead = 330°C. Assume that any mechanical energy lost is used to heat the ball. Use g = 10 m s − 2. Solution The initial gravitational potential energy of the ball = mgh = m × (10 m s –2) × (6.2 × 10 3 m) = m × (6.2 × 10 4 m 2 s –2) = m × (6.2 × 104 J kg− 1). All this energy is used to heat the ball as it reaches the ground with a small velocity. Energy required to take the ball from 30°C to 330°C is m × (126 J kg − 1 °C − 1) × (300°C) = m × 37800 J kg − 1 and energy required to melt the ball at 330°C = mL where L = latent heat of fusion of lead. Thus, m × (6.2 × 10 4 J kg− 1) = m × 37800 J kg − 1 + mL or, L = 2.4 × 10 4 J kg− 1. 39  CHAPTER 3 HEAT TRANSFER Heat can be transferred from one place to another by three different methods, namely, conduction, convection and radiation. Conduction usually takes place in solids, convection in liquids and gases, and no medium is required for radiation. 3.1 Thermal Conduction If one end of a metal rod is placed in a heater, the temperature of the other end gradually increases. Heat is transferred from one end of the rod to the other end. This transfer takes place due to molecular collisions and the process is called heat conduction. The molecules at one end of the rod gain heat from the stove and their average kinetic energy increases. As these molecules collide with the neighboring molecules having less kinetic energy, the energy is shared between these two groups. The kinetic energy of these neighboring molecules increases. As they collide with their neighbors on the colder side, they transfer energy to them. This way, heat is passed along the rod from molecule to molecule. The average position of a molecule does not change and hence, there is no mass movement of matter. 3.2 Thermal Conductivity The ability of a material to conduct heat is measured by thermal conductivity (defined below) of the material. Figure (3.1) 40  Consider a slab of uniform cross section A and length x. Let one face of the slab be maintained at temperature T1 and the other at T2. Also, suppose the remaining surface is covered with a nonconducting material so that no heat is transferred through the sides. After sufficient time, steady state is reached and the temperature at any point remains unchanged as time passes. In such a case, the amount of heat crossing per unit time through any cross section of the slab is equal. If ∆Q amount of heat crosses through any cross section in time ∆t, ∆Q/∆t is called the heat current. It is found that in steady state the heat current is proportional to the area of cross section A, proportional to the temperature difference (T1 − T2) between the ends and inversely proportional to the length x. Thus, (3.1) where is a constant for the material of the slab and is called the thermal conductivity of the material. If the area of cross section is not uniform or if the steady-state conditions are not reached, the equation can only be applied to a thin layer of material perpendicular to the heat flow. If A be the area of cross section at a place, dx be a small thickness along the direction of heat flow, and dT be the temperature difference across the layer of thickness dx, the heat current through this cross section is (3.2) The quantity dT/dx is called the temperature gradient. The minus sign indicates that dT/dx is negative along the direction of the heat flow. The measurement unit of thermal conductivity can be easily worked out using equation (3.2). The SI unit is Js−1m−1K−1 or Wm−1K−1. As a change of 1 K and a change of 1°C are the same, the unit may also be written as Wm−1°C−1. 41 Example (3.1) One face of a copper cube of edge 10 cm is maintained at 100°C and the opposite face is maintained at 0°C. All other surfaces are covered with an insulating material. Find the amount of heat flowing per second through the cube. Thermal conductivity of copper is 385 Wm −1°C −1. Solution The heat flows from the hotter face towards the colder face. The area of cross section perpendicular to the heat flow is A = (10 cm) 2. The amount of heat flowing per second is = (385 Wm 1°C 1) × (0.1 m) 2 / 0.1 m 3850 W. In general, solids are better conductors than liquids and liquids are better conductors than gases. Metals are much better conductors than nonmetals. This is because, in metals we have a large number of “free electrons” which can move freely anywhere in the body of the metal. These free electrons help in carrying the thermal energy from one place to another in a metal. We can now understand why cooking utensils are made of metals whereas their handles are made of plastic or wood. When a rug is placed in bright sun on a tiled floor, both the rug and the floor acquire the same temperature. But it is much more difficult to stay bare foot on the tiles than to stay on the rug. This is because the conductivity of the rug is lesser than the tiles, and hence, the heat current going into the foot is smaller. This can be shown in figure (3.2). Thermal conductivities of some materials can be shown in Table (3.1). 42  Figure (3.2) Table (3.1): gives thermal conductivities of some materials. 43  3.2.1 Thermal Resistance The quantity in Eq. (1) is called the thermal resistance R. Writing the heat current ∆Q/∆t as we have (3.3) This is mathematically equivalent to Ohm’s law to be introduced in a later chapter. Many results derived from Ohm’s law are also valid for thermal conduction. Example (3.2) Find the thermal resistance of an aluminum rod of length 20 cm and area of cross section 1 cm2. The heat current is along the length of the rod. Thermal conductivity of aluminum = 200 Wm−1K−1. Solution The thermal resistance is 3.2.2 SERIES AND PARALLEL CONNECTION OF RODS (a) Series Connection Consider two rods of thermal resistances R1 and R2 joined one after the other as shown in figure (3.3). Figure (3.3) The free ends are kept at temperatures T1 and T2 with T1> T2. In steady state, any heat that goes through the first rod also goes through the second rod. Thus, the same heat current passes through the two rods. Such a connection of rods is called a series 44 connection. Suppose the temperature of the junction is T. From Eq. (3.3), the heat current through the first rod is or, (i) and that through the second rod is or, (ii) Adding (i) and (ii), or, Thus, the two rods together are equivalent to a single rod of thermal resistance R1 + R2. If more than two rods are joined in series, the equivalent thermal resistance is given by R = R1 + R 2 + R3 + … (b) Parallel Connection Now, suppose the two rods are joined at their ends as shown in figure (3.4). Figure (3.4) The left ends of both the rods are kept at temperature T1 and the right ends are kept at temperature T2. So, the same temperature difference is maintained between the ends 45 of each rod. Such a connection of rods is called a parallel connection. The heat current going through the first rod is and that through the second rod is The total heat current going through the left end is ( ) or, where (iii) Thus, the system of the two rods is equivalent to a single rod of thermal resistance R given by (i). If more than two rods are joined in parallel, the equivalent thermal resistance R is given by 46 3.2.3 Measurement of Thermal Conductivity of a solid Figure (3.5) Figure (3.5) shows Searle’s apparatus to measure the thermal conductivity of a solid. The solid is taken in the form of a cylindrical rod. One end of the rod goes into a steam chamber. A copper tube is coiled around the rod near the other end of the rod. A steady flow of water is maintained in the copper tube. Water enters the tube at the end away from the steam chamber and it leaves at the end nearer to it. Thermometers T3 and T4 are provided to measure the temperatures of the outgoing and incoming water. Two holes are drilled in the rod and mercury is filled in these holes to measure the temperature of the rod at these places with the help of thermometers T1 and T2. The whole apparatus is covered properly with layers of an insulating material like wool or felt to prevent any loss of heat from the sides. Steam is passed in the steam chamber and a steady flow of water is maintained. The temperatures of all the four thermometers rise initially and ultimately become constant in time as the steady state is reached. The readings T1, T2, T3 and T4 are noted in steady state. A beaker is weighed and the water coming out of the copper tube is collected in it for a fixed time t measured by a stop clock. The beaker together with the water is weighed. 47 The mass m of the water collected is then calculated. The area of cross section of the rod is calculated by measuring its radius with a slide caliper. The distance between the holes in the rod is measured with the help of a divider and a meter scale. Let the length of the rod between the holes be x and the area of cross section of the rod be A. If the thermal conductivity of the material of the rod is K, the rate of heat flow (heat current) from the steam chamber to the rod is In a time, t, the chamber supplies a heat (i) As the mass of the water collected in time t is m, the heat taken by the water is (ii) where s is the specific heat capacity of water. As the entire rod is covered with an insulating material and the temperature of the rod does not change with time at any point, any heat given by the steam chamber must go into the flowing water. Hence, the same Q is used in (i) and (ii). Thus, or, (3.4) 48 3.3 CONVECTION In convection, heat is transferred from one place to the other by the actual motion of heated material. For example, in a hot air blower, air is heated by a heating element and is blown by a fan. The air carries the heat wherever it goes. When water is kept in a vessel and heated on a heater, the water at the bottom gets heat due to conduction through the vessel’s bottom. Its density decreases and consequently it rises. Thus, the heat is carried from the bottom to the top by the actual movement of the parts of the water. If the heated material is forced to move, say by a blower or a pump, the process of heat transfer is called forced convection. If the material moves due to difference in density, it is called natural or free convection. Natural convection and the anomalous expansion of water play important roles in saving the lives of aquatic animals like fishes when the atmospheric temperature goes below 0°C. As the water at the surface is cooled, it becomes denser and goes down. The less cold water from the bottom rises up to the surface and gets cooled. This way the entire water is cooled to 4°C. As the water at the surface is further cooled, it expands and the density decreases. Thus, it remains at the surface and gets further cooled. Finally, it starts freezing. Heat is now lost to the atmosphere by the water only due to conduction through the ice. As ice is a poor conductor of heat, the further freezing is very slow. The temperature of the water at the bottom remains constant at 4°C for a large period of time. The atmospheric temperature ultimately improves, and the animals are saved. The main mechanism for heat transfer inside a human body is forced convection. Heart serves as the pump and blood as the circulating fluid. Heat is lost to the atmosphere through all the three processes conduction, convection and radiation. The rate of loss depends on the clothing, the tiredness and perspiration, atmospheric temperature, air current, humidity and several other factors. The system, however, transports the just required amount of heat and hence maintains a remarkably constant body temperature. 49 3.4 RADIATION The process of radiation does not need any material medium for heat transfer. Energy is emitted by a body and this energy travels in the space just like light. When it falls on a material body, a part is absorbed, and the thermal energy of the receiving body is increased. The energy emitted by a body in this way is called radiant energy, thermal radiation or simply radiation. Thus, the word “radiation” is used in two meanings. It refers to the process by which the energy is emitted by a body, is transmitted in space and falls on another body. It also refers to the energy itself which is being transmitted in space. The heat from the sun reaches the earth by this process, travelling millions of kilometers of empty space. 3.5 Theory of Exchange Way back in 1792, Pierre Prévost put forward the theory of radiation in a systematic way now known as the theory of exchange. According to this theory, all bodies radiate thermal radiation at all temperatures. The amount of thermal radiation radiated per unit time depends on the nature of the emitting surface, its area and its temperature. The rate is faster at higher temperatures. Besides, a body also absorbs part of the thermal radiation emitted by the surrounding bodies when this radiation falls on it. If a body radiates more than what it absorbs, its temperature falls. If a body radiates less than what it absorbs, its temperature rises. Now, consider a body kept in a room for a long time. One finds that the temperature of the body remains constant and is equal to the room temperature. The body is still radiating thermal radiation. But it is also absorbing part of the radiation emitted by the surrounding objects, walls, etc. We thus conclude that when the temperature of a body is equal to the temperature of its surroundings, it radiates at the same rate as it absorbs. If we place a hotter body in the room, it radiates at a faster rate than the rate at which it absorbs. Thus, the body suffers a net loss of thermal energy in any given time and its temperature decreases. Similarly, if a colder body is kept in a warm surrounding, it radiates less to the surrounding than what it absorbs from the surrounding. 50 Consequently, there is a net increase in the thermal energy of the body and the temperature rises. 3.6 Blackbody Radiation Consider two bodies of equal surface areas suspended in a room. One of the bodies has polished surface and the other is painted black. After sufficient time, the temperature of both the bodies will be equal to the room temperature. As the surface areas of the bodies are the same, equal amount of radiation falls on the two surfaces. The polished surface reflects a large part of it and absorbs a little, while the black-painted surface reflects a little and absorbs a large part of it. As the temperature of each body remains constant, we conclude that the polished surface radiates at a slower rate and the black- painted surface radiates at a faster rate. So, good absorbers of radiation are also good emitters. A body that absorbs all the radiation falling on it is called a blackbody. Such a body will emit radiation at the fastest rate. The radiation emitted by a blackbody is called blackbody radiation. The radiation inside an enclosure with its inner walls maintained at a constant temperature has the same properties as the blackbody radiation and is also called blackbody radiation. A blackbody is also called an ideal radiator. A perfect blackbody, absorbing 100% of the radiation falling on it, is only an ideal concept. Among the materials, lampblack is close to a blackbody. It reflects only about 1% of the radiation falling on it. If an enclosure is painted black from inside and a small hole is made in the wall (figure (3.6)) the hole acts as a very good blackbody. Figure (3.6) 51 Any radiation that falls on the hole goes inside. This radiation has little chance to come out of the hole again and it gets absorbed after multiple reflections. The cone directly opposite to the hole (figure 6) ensures that the incoming radiation is not directly reflected back to the hole. 3.6.1 Kirchhoff’s Law We have learnt that good absorbers of radiation are also good radiators. This aspect is described quantitatively by Kirchhoff’s law of radiation. Before stating the law let us define certain terms. 3.6.2 Emissive Power Figure (3.7) Consider a small area ∆A of a body emitting thermal radiation. Consider a small solid angle ∆ω about the normal to the radiating surface. Let the energy radiated by the area ∆A of the surface in the solid angle ∆ω in time ∆t be ∆U. We define emissive power of the body as Thus, emissive power denotes the energy radiated per unit area per unit time per unit solid angle along the normal to the area. 52 3.6.3 Absorptive Power Absorptive power of a body is defined as the fraction of the incident radiation that is absorbed by the body. If we denote the absorptive power by , As all the radiation incident on a blackbody is absorbed, the absorptive power of a blackbody is unity. Note that: the absorptive power is a dimensionless quantity, but the emissive power is not. Kirchhoff’s Law The ratio of emissive power to absorptive power is the same for all bodies at a given temperature and is equal to the emissive power of a blackbody at that temperature. Thus, Kirchhoff’s law tells that if a body has high emissive power, it should also have high absorptive power to have the ratio E/a same. Similarly, a body having low emissive power should have low absorptive power. Kirchhoff’s law may be easily proved by a simple argument as described below. Figure (3.8) 53  Consider two bodies A and B of identical geometrical shapes placed in an enclosure. Suppose A is an arbitrary body and B is a blackbody. In thermal equilibrium, both the bodies will have the same temperature as the temperature of the enclosure. Suppose an amount ∆U of radiation falls on the body A in a given time ∆t. As A and B have the same geometrical shapes, the radiation falling on the blackbody B is also ∆U. The blackbody absorbs all of this ∆U. As the temperature of the blackbody remains constant, it also emits an amount ∆U of radiation in that time. If the emissive power of the blackbody is E0, we have (i) where k is a constant. Let the absorptive power of A be a. Thus, it absorbs an amount a∆U of the radiation falling on it in time ∆t. As its temperature remains constant, it must also emit the same amount a ∆U in that time. If the emissive power of the body A is E, we have (ii) The same proportionality constant k is used in (i) and (ii) because the two bodies have identical geometrical shapes and radiation emitted in the same time ∆t is considered. From (i) and (ii), or, or, This proves Kirchhoff’s law. 54 3.7 Nature of Thermal Radiation Thermal radiation, once emitted, is an electromagnetic wave like light. It, therefore, obeys all the laws of wave theory. The wavelengths are still small compared to the dimensions of usual obstacles encountered, so the rules of geometrical optics are valid, i.e., it travels in a straight line, casts shadow, is reflected and refracted at the change of medium, etc. The radiation emitted by a body is a mixture of waves of different wavelengths. However, only a small range of wavelength has significant contribution in the total radiation. The radiation emitted by a body at a temperature of 300 K (room temperature) has significant contribution from wavelengths around 9550 nm which is in long infrared region (visible light has a range of about 380–780 nm). As the temperature of the emitter increases, this dominant wavelength decreases. At around 1100 K, the radiation has a good contribution from red region of wavelengths and the object appears red. At temperatures around 3000 K, the radiation contains enough shorter wavelengths and the object appears white. Even at such a high temperature most significant contributions come from wavelengths around 950 nm. Figure (3.9) The relative importance of different wavelengths in a thermal radiation can be studied qualitatively from figure (3.9). Here the intensity of radiation near a given 55 wavelength is plotted against the wavelength for different temperatures. We see that as the temperature is increased, the wavelength corresponding to the highest intensity decreases. In fact, this wavelength λm is inversely proportional to the absolute temperature of the emitter. So, λmT = b (3.5) where b is a constant. This equation is known as the Wien’s displacement law. For a blackbody, the constant b appearing in equation (3.5) is measured to be 0.288 cm K and is known as the Wien constant. Example (3.3) The light from the sun is found to have a maximum intensity near the wavelength of 470 nm. Assuming that the surface of the sun emits as a blackbody, calculate the temperature of the surface of the sun. Solution For a blackbody, λmT = 0.288 cm K. Thus, T = 0.288 cmK /470 nm = 6130 K. Distribution of radiant energy among different wavelengths has played a very significant role in the development of quantum mechanics and in our understanding of nature in a new way. Classical physics had predicted a very different and unrealistic wavelength distribution. Planck put forward a bold hypothesis that radiation can be emitted or absorbed only in discrete steps, each step involving an amount of energy given by E = nhν where ν is the frequency of the radiation and n is an integer. A new fundamental constant h named as Planck constant was introduced in physics. This opened the gateway of modern physics through which we look into the atomic and subatomic world. 56 3.7.1 Stefan–Boltzmann Law The energy of thermal radiation emitted per unit time by a blackbody of surface area A is given by (3.6) where σ is a universal constant known as Stefan–Boltzmann constant and T is its – 8 –2 –4 temperature on absolute scale. The measured value of σ is 5.67 × 10 Wm K. Equation (6) itself is called the Stefan–Boltzmann law. Stefan had suggested this law from experimental data available on radiation and Boltzmann derived it from thermodynamical considerations. The law is also quoted as Stefan’s law and the constant σ as Stefan constant. A body, which is not a blackbody, emits less radiation than given by equation (3.6). It is, however, proportional to T4. The energy emitted by such a body per unit time is written as (3.7) where e is a constant for the given surface having a value between 0 and 1. This constant is called the emissivity of the surface. It is zero for completely reflecting surface and is unity for a blackbody. Using Kirchhoff’s law, (i) where a is the absorptive power of the body. The emissive power E is proportional to the energy radiated per unit time, that is, proportional to u. Using equations (6) and (7) in (i), or e= a 57 Thus, emissivity and absorptive power have the same value. Consider a body of emissivity e kept in thermal equilibrium in a room at temperature T0. The energy of radiation absorbed by it per unit time should be equal to the energy emitted by it per unit time. This is because the temperature remains constant. Thus, the energy of the radiation absorbed per unit time is Now, suppose the temperature of the body is changed to T but the room temperature remains T0. The energy of the thermal radiation emitted by the body per unit time is The energy absorbed per unit time by the body is Thus, the net loss of thermal energy per unit time is (3.8) Example (3.4) A blackbody of surface area 10 cm2 is heated to 127°C and is suspended in a room at temperature 27°C. Calculate the initial rate of loss of heat from the body to the room. Solution For a blackbody at temperature T, the rate of emission is. When it is kept in a room at temperature T0, the rate of absorption is. The net rate of loss of heat is Here A = 10 × 10 – 4 m 2, T = 400 K and T0 = 300 K. Thus, = (5.67 × 10 – 8 Wm –2K –4) (= 10 × 10 – 4 m 2) (400 4-300 4) K 4 = 0.99 W 58 3.7.2 Newton’s Law of Cooling Suppose a body of surface area A at an absolute temperature T is kept in a surrounding having a lower temperature T0. The net rate of loss of thermal energy from the body due to radiation is If the temperature difference is small, we can write or, ( ) ( ) Thus, The body may also lose thermal energy due to convection in the surrounding air. For small temperature difference, the rate of loss of heat due to convection is also proportional to the temperature difference and the area of the surface. This rate may, therefore, be written as The net rate of loss of thermal energy due to convection and radiation is If s be the specific heat capacity of the body and m its mass, the rate of fall of temperature is 59 Thus, for small temperature difference between a body and its surrounding, the rate of cooling of the body is directly proportional to the temperature difference and the surface area exposed. We can write (3.9) This is known as Newton’s law of cooling. The constant b depends on the nature of the surface involved and the surrounding conditions. The minus sign indicates that if T > T0, dT/dt is negative, that is, the temperature decreases with time. As the difference in temperature is the same for absolute and Celsius scale. Example (3.5) A liquid cool from 70°C to 60°C in 5 minutes. Calculate the time taken by the liquid to cool from 60°C to 50°C, if the temperature of the surrounding is constant at 30°C. Solution The average temperature of the liquid in the first case is The average temperature difference from the surrounding is The rate of fall of temperature is From Newton’s law of cooling, or, … (i) 60 In the second case, the average temperature of the liquid is so that, If it takes a time t to cool down from 60°C to 50°C, the rate of fall of temperature is From Newton’s law of cooling and (i), or, t = 7 min. 3.8 Detection and Measurement of Radiation Several instruments are used to detect and measure the amount of thermal radiation. We describe two of them here, a bolometer and a thermopile. 3.8.1 Bolometer The bolometer is based on the theory of Wheatstone bridge which was introduced while discussing resistance thermometer. A thin (a small fraction of a millimetre) foil of platinum is taken and strips are cut from it to leave a grid-type structure as shown in figure (3.10a). Four such grids, identical in all respect, are prepared and joined with a battery and a galvanometer to form a Wheatstone bridge (figure 3.10b). Grid 1 faces the radiation to be detected or measured. The particular arrangement of the four grids ensures that the radiation passing through the empty spaces in grid 1 falls on the strips of grid 4. Grids 2 and 3 are protected from the radiation. 61  Figure (3.10) Bolometer When no radiation falls on the bolometer, all the grids have the same resistance so that R1 = R2 = R3 = R4. Thus, , and the bridge is balanced. There is no deflection in the galvanometer. When radiation falls on the bolometer, grids 1 and 4 get heated. As the temperature increases, the resistances R1 and R4 increase and hence the product R1 R4 increases. On the other hand, R2 R3 remains unchanged. Thus, or, The bridge becomes unbalanced and there is a deflection in the galvanometer which indicates the presence of radiation. The magnitude of deflection is related to the amount of radiation falling on the bolometer. The bolometer is usually enclosed in a glass bulb evacuated to low pressures. This increases the sensitivity. 62 3.8.2 Thermopile A thermopile is based on the principle of Seebeck effect. Figure (3.11) illustrates the principle. Figure (3.11) Thermopile Two dissimilar metals A and B are joined to form two junctions J1 and J2. A galvanometer is connected between the junctions through the metal B. If the junctions are at the same temperature, there is no deflection in the galvanometer. But if the temperatures of the junctions are different, the galvanometer deflects. Such an instrument is called a thermocouple. 63  Solved Problems 1. The lower surface of a slab of stone of face-area 3600 cm2 and thickness 10 cm is exposed to steam at 100°C. A block of ice at 0°C rests on the upper surface of the slab. 4.8 g of ice melts in one hour. Calculate the thermal conductivity of the stone. Latent heat of fusion of ice = 3.36 × 105 J kg−1. Solution The amount of heat transferred through the slab to the ice in one hour is Q = (4.8 × 10 − 3 kg) × (3.36 × 105 J kg−1) = 4.8 × 336 J. Using the equation or, = 1.24 × 10− 3 W m−1°C−1. 2. An icebox made of 1.5 cm thick Styrofoam has dimensions 60 cm × 60 cm × 30 cm. It contains ice at 0°C and is kept in a room at 40°C. Find the rate at which the ice is melting. Latent heat of fusion of ice = 3.36 × 105 J kg−1 and thermal conductivity of Styrofoam = 0.04 W m−1°C−1. Solution The total surface area of the walls = 2(60 cm × 60 cm + 60 cm × 30 cm + 60 cm × 30 cm) = 1.44 m2. The thickness of the walls = 1.5 cm = 0.015 m. The rate of heat flow into the box is The rate at which the ice melts is 64 3. A closed cubical box is made of perfectly insulating material and the only way for heat to enter or leave the box is through two solid cylindrical metal plugs, each of cross sectional area 12 cm2 and length 8 cm fixed in the opposite walls of the box. The outer surface of one plug is kept at a temperature of 100°C while the outer surface of the other plug is maintained at a temperature of 4°C. The thermal conductivity of the material of the plug is 2.0 Wm−1°C−1. A source of energy generating 13 W is enclosed inside the box. Find the equilibrium temperature of the inner surface of the box assuming that it is the same at all points on the inner surface. Solution Figure (3i) The situation is shown in figure (3i). Let the temperature inside the box be T. The rate at which heat enters the box through the left plug is The rate of heat generation in the box = 13 W. The rate at which heat flows out of the box through the right plug is In the steady state or, or, or, = 52°C + 216.67°C ≈ 269°C. 65 4. A bar of copper of length 75 cm and a bar of steel of length 125 cm are joined together end to end. Both are of circular cross section with diameter 2 cm. The free ends of the copper and the steel bars are maintained at 100°C and 0°C respectively. The curved surfaces of the bars are thermally insulated. What is the temperature of the copper– steel junction? What is the amount of heat transmitted per unit time across the junction? Thermal conductivity of copper is 386 J s−1 m−1°C−1 and that of steel is 46 J s−1 m−1°C−1. Solution Figure (3ii) The situation is shown in figure (3ii). Let the temperature at the junction be T (on Celsius scale). The same heat current passes through the copper and the steel rods. Thus, or, or, The rate of heat flow is. 66 5. Two parallel plates A and B are joined together to form a compound plate (figure 3.iii). The thicknesses of the plates are 4.0 cm and 2.5 cm respectively and the area of cross section is 100 cm2 for each plate. The thermal conductivities are KA = 200 W m−1°C−1 for the plate A and KB = 400 W m−1°C−1 for the plate B. The outer surface of the plate A is maintained at 100°C and the outer surface of the plate B is maintained at 0°C. Find (a) the rate of heat flow through any cross section, (b) the temperature at the interface and (c) the equivalent thermal conductivity of the compound plate. Figure(3.iii) Solution (a) Let the temperature of the interface be θ.The area of cross section of each plate is A = 100 cm2= 0.01 m2. The thicknesses are xA = 0.04 m and xB = 0.025 m. The thermal resistance of the plate A is and that of the plate B is The equivalent thermal resistance is [ ⁄ ⁄ ] (i) Thus, [ ⁄ ⁄ ] * + 67 (b) We have ⁄ or, ⁄ or, (c) If K is the equivalent thermal conductivity of the compound plate, its thermal resistance is Comparing with (i), or, ⁄ ⁄. 68 6. A room has a 4 m × 4 m × 10 cm concrete roof (K = 1.26 Wm −1°C−1). At some instant, the temperature outside is 46°C and that inside is 32°C. (a) Neglecting convection, calculate the amount of heat flowing per second into the room through the roof. (b) Bricks (K = 0.65 Wm−1°C−1) of thickness 7.5 cm are laid down on the roof. Calculate the new rate of heat flow under the same temperature conditions. Solution The area of the roof = 4 m × 4 m = 16 m2. The thickness x = 10 cm = 0.10 m. (a) The thermal resistance of the roof is The heat current is (b) The thermal resistance of the brick layer is The equivalent thermal resistance is The heat current is 69 7. An electric heater is used in a room of total wall area 137 m2 to maintain a temperature of 20°C inside it, when the outside temperature is − 10°C. The walls have three different layers of materials. The innermost layer is of wood of thickness 2.5 cm, the middle layer is of cement of thickness 1.0 cm and the outermost layer is of brick of thickness 25.0 cm. Find the power of the electric heater. Assume that there is no heat loss through the floor and the ceiling. The thermal conductivities of wood, cement and brick are 0.125 Wm−1°C−1, 1.5 Wm−1°C−1 and 1.0 Wm−1°C−1 respectively. Solution Figure (3.vi) The situation is shown in figure (3.vi). The thermal resistances of the wood, the cement and the brick layers are and As the layers are connected in series, the equivalent thermal resistance is The heat current is The heater must supply 9000 W to compensate the outflow of heat. 70 8. Three rods of material x and three of material y are connected as shown in figure (3.v). All the rods are identical in length and cross-sectional area. If the end A is maintained at 60°C and the junction E at 10°C, calculate the temperature of the junction B. The thermal conductivity of x is 800 W m−1°C−1 and that of y is 400 W m−1°C −1. Figure (3.v) Solution It is clear from the symmetry of the figure that the points C and D are equivalent in all respect and hence, they are at the same temperature, say T. No heat will flow through the rod CD. We can, therefore, neglect this rod in further analysis. Let and A be the length and the area of cross section of each rod. The thermal resistances of AB, BC and BD are equal. Each has a value As the rod CD has no effect, we can say that the rods BC and CE are joined in series. Their equivalent thermal resistance is Also, the rods BD and DE together have an equivalent thermal resistance The resistances R3 and R4 are joined in parallel and hence their equivalent thermal resistance is given by 71 or, This resistance R5 is connected in series with AB. Thus, the total arrangement is equivalent to a thermal resistance Figure (3vi) shows the successive steps in this reduction. Figure (3vi) The heat current through A is This current passes through the rod AB. We have or, Putting from ( ) and (3ii), So, 72 9. A rod CD of thermal resistance 5.0 K W −1 is joined at the middle of an identical rod AB as shown in figure (3vii). The ends A, B and D are maintained at 100°C, 0°C and 25°C respectively. Find the heat current in CD. Figure (3vii) Solution −1 The thermal resistance of AC is equal to that of CB and is equal to 2.5 K W. Suppose the temperature at C is T. The heat currents through AC, CB and CD are and We also have i.e. Thus, 73 10. A blackbody of surface area 1 cm2 is placed inside an enclosure. The enclosure has a constant temperature 27°C and the blackbody is maintained at 327°C by heating it electrically. What electric power is needed to maintain the temperature? [σ = 6.00 × 10 – 8 W m–2 K–4.] Solution The area of the blackbody is A = 10 − 4 m2, its temperature is T1 = 327°C = 600 K and the temperature of the enclosure is T2 = 27°C = 300 K. The blackbody emits radiation at the rate of. The radiation falls on it (and gets absorbed) at the rate of. The net rate of loss of energy is –. The heater must supply this much of power. Thus, the power needed is – ( – ). 11. An electric heater emits 1000 W of thermal radiation. The coil has a surface area of 0.020 m2. Assuming that the coil radiates like a blackbody, find its temperature? [σ = 6.00 × 10 – 8 W m–2 K–4.] Solution Let the temperature of the coil be T. The coil will emit radiation at a rate. Thus, or, Thus,. 74 12. The earth receives solar radiation at a rate of 8.2 J cm−2min−1. Assuming that the sun radiates like a blackbody, calculate the surface temperature of the sun. The angle subtended by the sun on the earth is 0.53° and the Stefan constant σ = 5.67×10 – 8 W m– 2 K–4. Solution Let the diameter of the sun be D and its distance from the earth be R. From the question, The radiation emitted by the surface of the sun per unit time is ( ) At distance R, this radiation falls on an area 4πR2 in unit time. The radiation received at the earth’s surface per unit time per unit area is, therefore, ( ) Thus, ( ) or, 75 13. The temperature of a body falls from 40°C to 36°C in 5 minutes when placed in a surrounding of constant temperature 16°C. Find the time taken for the temperature of the body to become 32°C. Solution The mean temperature of the body as it cools from 40°C to 36°C is. The rate of decrease of temperature is. Newton’s law of cooling is or, So, Let the time taken for the temperature to become 32°C be t. During this period, The mean temperature is. Now, Thus, 76  CHAPTER 4 Kinetic Theory of Gases 4.1 Introduction We have seen that the pressure p, the volume V and the temperature T of any gas at low densities obey the equation PV= n RT, where n is the number of moles in the gas and R is the gas constant having value 8.314 JK−1mol – 1. The temperature T is defined on the absolute scale. We define the term ideal gas to mean a gas which always obeys this equation. The real gases available to us are good approximation of an ideal gas at low density but deviate from it when the density is increased. Any sample of a gas is made of molecules. A molecule is the smallest unit having all the chemical properties of the sample. The observed behavior of a gas results from the detailed behavior of its large number of molecules. The kinetic theory of gases attempts to develop a model of the molecular behavior which should result in the observed behavior of an ideal gas. 4.2 Assumptions of Kinetic Theory of Gases 1. All gases are made of molecules moving randomly in all directions. 2. The size of a molecule is much smaller than the average separation between the molecules. 77  3. The molecules exert no force on each other or on the walls of the container except during collision. 4. All collisions between two molecules or between a molecule and a wall are perfectly elastic. Also, the time spent d