Physics Notes for NEET Chapter 12 PDF

60 554 Thermometry, Thermal Expansion and calorimetry Chapter E3 12 Thermometry, Thermal Expansion and Calorimetry (8) NTP or STP implies 273.15K (0°C = 32°F) ID Temperature Scales of Temperature U Boiling water D YG Freezing water Temperature is defined as the degree of hotness or coldness of a bod...

60 554 Thermometry, Thermal Expansion and calorimetry Chapter E3 12 Thermometry, Thermal Expansion and Calorimetry (8) NTP or STP implies 273.15K (0°C = 32°F) ID Temperature Scales of Temperature U Boiling water D YG Freezing water Temperature is defined as the degree of hotness or coldness of a body. The natural flow of heat is from higher temperature to lower temperature. Two bodies are said to be in thermal equilibrium with each other, when no heat flows from one body to the other. That is when both the bodies are at the same temperature. (1) Temperature is one of the seven fundamental quantities with dimension [ ]. It is a scalar physical quantity with S.I. unit kelvin. ST U (2) When heat is given to a body and its state does not change, the temperature of the body rises and if heat is taken from a body its temperature falls i.e. temperature can be regarded as the effect of cause “heat”. 212°F 100°C 373 K 32°F 0°C 273 K Fahrenheit Celsius Kelvin Fig. 12.1 (°F), Kelvin (K), Reaumer (R), The centigrade (°C), Farenheite Rankine (Ra) are commonly used temperature scales. (1) To construct a scale of temperature, two fixed points are taken. First fixed point is the freezing point (ice point) of water, it is called lower fixed point (LFP). The second fixed point is the boiling point (steam point) of water, it is called upper fixed point (UFP). (2) Celsius scale : In this scale LFP (ice point) is taken 0° and UFP (steam point) is taken 100°. The temperature measured on this scale all in degree Celsius (°C). (3) According to kinetic theory of gases, temperature (macroscopic physical quantity) is a measure of average translational kinetic energy of a molecule (microscopic physical quantity). (3) Farenheite scale : This scale of temperature has LFP as 32° F and UFP as 212°F. The change in temperature of 1° F corresponds to a change of less than 1° on Celsius scale. (4) Although the temperature of a body can to be raised without limit, it cannot be lowered without limit and theoretically limiting low temperature is taken to be zero of the kelvin scale. (4) Kelvin scale : The Kelvin temperature scale is also known as thermodynamic scale. The triple point of water is also selected to be the zero of scale of temperature. The temperature measured on this scale are in Kelvin (K). (5) Highest possible temperature achieved in laboratory is about 10 K while lowest possible temperature attained is 10 K. 8 –8 (6) Temperature of the core of the sun is 10 K while that of its surface is 6000 K. 7 (7) Normal temperature of human body is 310. 15 K (37°C = 98.6°F). The triple point of water is that point on a P-T diagram where the three phases of water, the solid, the liquid and the gas, can coexist in equilibrium. Table 12.1 : Different measuring scales Thermometry, Thermal Expansionand and Calorimetry 555 Scale Symbol for each degree LFP UFP Number of divisions on the scale Celsius °C 0°C 100°C 100 Fahrenheit °F 32°F 212°F 180 Reaumer °R 0°R 80°R 80 Rankine °Ra 460 Ra 672 Ra 212 K 273.15 K 373.15 K 100 Kelvin An instrument used to measure the temperature of a body is called a thermometer It works by absorbing some heat from the body, so the temperature recorded by it is lesser than the actual value unless the body is at constant temperature. Some common types of thermometers are as follows (1) Liquid (mercury) thermometers : In liquid thermometers mercury is preferred over other liquids as its expansion is large and uniform and it has high thermal conductivity and low specific heat. (6) All these temperatures are related to each other by the following relationship (freezing point) 2 (boiling point) (ii) Upper limit of range of mercury thermometer can be raised upto 550°C by filling nitrogen in space over mercury under pressure (which elevates boiling point of mercury). (iv) If alcohol is used instead of mercury then range of temperature measurement becomes – 80°C to 350°C ID (7) The Celsius and Kelvin scales have different zero points but the same size degrees. Therefore any temperature difference is the same on the Celsius and Kelvin scales (T – T)°C = (T – T ) K. 1  50C to 350C (iii) Mercury thermometer with cylindrical bulbs are more sensitive than those with spherical bulbs. C F  32 K  273 R Ra  460 or     5 9 5 4 10.6 2 (i) Range of temperature : E3 C 0 F  32 K  273.15 R 0 Ra  460     100 212  32 373.15  273.15 80  0 672  460 60 (5) Temperature on one scale can be converted into other scale by using the following identity. Reading on any scale  LFP = Constant for all scales UFP  LFP 1 Thermometry l  l0  100C l100  l0 (2) Gas thermometers : These are more sensitive and accurate than liquid thermometers as expansion of gases is more than that of liquids. The thermometers using a gas as thermoelectric substance are called ideal gas thermometers. These are of two types U A branch of science which deals with the measurement of temperature of a substance is known as thermometry. (v) Formula : t  D YG (1) The linear variation in some physical properties of a substance with change of temperature is the basic principle of thermometry and these properties are defined as thermometric property (x) of the substance. (2) Thermometric properties (x) may be as follows (i) Length of liquid in capillary (ii) Pressure of gas at constant volume. (iii) Volume of gas at constant pressure. (iv) Resistance of a given platinum wire. (3) In old thermometry, freezing point (0°C) and steam point (100°C) are taken to define the temperature scale. So if the thermometric property at temperature 0°C, 100°C and t°C are x , x and x respectively then 100 U 0 x  x0 x  x0 t 0   100C  t C  100  0 x 100  x 0 x100  x 0 ST (4) In modern thermometry instead of two fixed points only one reference point is chosen (triple point of water 273.16 K) the other is itself 0 K where the value of thermometric property is assumed to be zero. So if the value of thermometric property at 0 K, 273.16 K and TK are 0, x and x respectively then Tr T x  273.16 x Tr Thermometers  x   T  273.16  K  x Tr  (i) Constant pressure gas thermometers (a) Principle V  T (b) Formula : t  (if P = constant) V  V0 V  100C or T  273.16 K V100  V0 VTr (ii) Constant volume gas thermometers (a) Principle P  T (b) Formula : t  (if V = constant) P  P0 P  100C or T  273.16 K P100  P0 PTr (c) Range of temperature : Hydrogen gas thermometer : – 200°C to 500°C Nitrogen gas thermometer : – 200°C to 1600°C Helium gas thermometer : – 268°C to 500°C (3) Resistance thermometers : Usually platinum is used in resistance thermometers due to high melting point and large value of temperature coefficient of resistance. Resistance of metals varies with temperature according to relation. R  R0 (1  t) where  is the temperature coefficient of resistance and t is change in temperature. (i) Formula : t  R  R0 R  100C or T  273.16 K R100  R0 RTr (ii) Temperature range : For Platinum resistance thermometer it is – 200°C to 1200°C For Germanium resistance thermometer it is 4 to 77 K. 556 Thermometry, Thermal Expansion and calorimetry (4) Thermoelectric thermometers : These are based on “Seebeck effect” according to which when two distinct metals are joined to form a closed circuit called thermocouple and the difference in temperature is maintained between their junctions, an emf is developed. The emf is called thermo-emf and if one junction is at 0°C, thermoelectric emf varies with temperature of hot junction (t) according to e = at + bt ; where a and b are constants. 2 G When matter is heated without any change in it’s state, it usually expands. According to atomic theory of matter, a symmetry in potential energy curve is responsible for thermal expansion. As with rise in temperature the amplitude of vibration and hence energy of atoms increases, hence the average distance between the atoms increases. So the matter as a whole expands. Fe 60 Cu Fig. 12.2 Thermoelectric thermometers have low thermal capacity and high thermal conductivity, so can be used to measure quickly changing temperature Thermo couple Temperature range Copper-iron thermocouple 0°C to 260°C Iron-constantan thermocouple 0°C to 800°C 2000oC to 3000°C (3) Linear expansion : When a solid is heated and it's length increases, then the expansion is called linear expansion. L0 L0 + L = L (A) Before heating (B) After heating U Tungsten-molybdenum thermocouple (2) Solids can expand in one dimension (linear expansion), two dimension (superficial expansion) and three dimension (volume expansion) while liquids and gases usually suffers change in volume only. ID Table 12.2 : Different temperature range (1) Thermal expansion is minimum in case of solids but maximum in case of gases because intermolecular force is maximum in solids but minimum in gases. E3 Ice Hot D YG (5) Pyrometers : These are the devices used to measure the temperature by measuring the intensity of radiations received from the body. They are based on the fact that the amount of radiations emitted from a body per unit area per second is directly proportional to the fourth power of temperature (Stefan’s law). Telescope Radiations Lamp 0 (L = Original length, T = Temperature change) 0 (ii) Final length L = L (1 + T) 0 (iii) Co-efficient of linear expansion   L L0 T (iv) Unit of  is C 1 or K 1. It's dimension is [ 1 ] (4) Superficial (areal) expansion : When the temperature of a 2D object is changed, it's area changes, then the expansion is called superficial expansion. U A Red glass filter Fig. 12.4 (i) Change in length  L = L T L0 Fig. 12.3 ST (i) These can be used to measure temperatures ranging from 800°C to 6000°C. L L0 (ii) They cannot measure temperature below 800° C because the amount of radiations is too small to be measured. (6) Vapour pressure thermometer : These are used to measure very low temperatures. They are based on the fact that saturated vapour pressure P of a liquid depends on the temperature according to the relation c log P  a  bT  T The range of these thermometers varies from 0.71 K to 120 K for different liquid vapours. L Fig. 12.5 (i) Change in area is A = A T 0 (A = Original area, T = Temperature change) 0 (ii) Final area A = A (1 + T) 0 (iii) Co-efficient of superficial expansion   Thermal Expansion (iv) Unit of  is °C or K. –1 –1 A A0 T Thermometry, Thermal Expansionand and Calorimetry 557 (5) Volume or cubical expansion : When a solid is heated and it's volume increases, then the expansion is called volume or cubical expansion. heating due to unequal linear expansion of the two metal. The strip will bend with metal of greater  on outer side i.e. convex side. Steel L0 Brass L0 L0 V Room temperature Higher temperature (A) (B) Bimetallic strip (i) Change in volume is V  V0T Fig. 12.6 (V = Original volume, T = change in temperature) 0 V V0 T At room temperature –1  2  3 (2) Effect of temperature on the time period of a simple pendulum : A pendulum clock keeps proper time at temperature . If temperature is increased to  (  ) then due to linear expansion, length of pendulum and hence its time period will increase. ID    :  : 1: 2 : 3 Fractional change in time period (i) Hence for the same rise in temperature T 1    T 2 (i) Due to increment in its time period, a pendulum clock becomes slow in summer and will lose time. U Percentage change in area = 2  percentage change in length. Percentage change in volume = 3  percentage change in length. Loss of time in a time period T  D YG (ii) The three coefficients of expansion are not constant for a given solid. Their values depend on the temperature range in which they are measured. (iii) The values of , ,  are independent of the units of length, area and volume respectively. (iv) For anisotropic solids    x   y   z where  ,  , and  x y z represent the mean coefficients of linear expansion along three mutually perpendicular directions. U (7) Contraction on heating : Some rubber like substances contract with rising temperature, because transverse vibration of atoms of substance dominate over longitudinal vibration which is responsible for expansion. Table 12.3 :  and  for some materials [K–1 or (°C)–1] ST Material (D) Fig. 12.7 (6) More about ,  and  : The co-efficient ,  and  for a solid are related to each other as follows  At high temperature (C) (iv) Unit of  is °C or K. –1 E3 (ii) Final volume V  V0 (1  T ) (iii) Volume co-efficient of expansion   OFF 60 ON [K–1 or (°C)–1] Steel 1.2  10–5 3.6  10–5 Copper 1.7  10–5 5.1  10–5 Brass 2.0  10–5 6.0  10–5 Aluminium 2.4  10–5 7.2  10–5 Application of Thermal Expansion in Solids (1) Bi-metallic strip : Two strips of equal lengths but of different materials (different coefficient of linear expansion) when join together, it is called “bi-metallic strip”, and can be used in thermostat to break or make electrical contact. This strip has the characteristic property of bending on 1   T 2 (ii) Time lost by the clock in a day (t = 86400 sec) t  1 1   t    (86400)  43200   sec 2 2 (iii) The clock will lose time i.e. will become slow if     (in summer) and will gain time i.e. will become fast if     (in winter). (iv) The gain or loss in time is independent of time period T and depends on the time interval t. (v) Since coefficient of linear expansion () is very small for invar, hence pendulums are made of invar to show the correct time in all seasons. (3) Thermal stress in a rigidly fixed rod : When a rod whose ends are rigidly fixed such as to prevent expansion or contraction, undergoes a change in temperature, due to thermal expansion or contraction, a compressive or tensile stress is developed in it. Due to this thermal stress the rod will exert a large force on the supports. If the change in temperature of a rod of length L is  then (A) Heating (B) Cooling Fig. 12.8 Thermal strain  L   L L 1    As  L      558 Thermometry, Thermal Expansion and calorimetry expands less) but later on, it starts rising due to faster expansion of the liquid. stress    AsY  strain    So Thermal stress  Y  R P Q or Force on the supports F  YA   PQ represents expansion of vessel QR represents the real (4) Error in scale reading due to expansion or contraction : If a scale gives correct reading at temperature , at temperature  (  ) due to linear expansion of scale, the scale will expand and scale reading will be lesser than true value so that, expansion of liquid PR Represent the apparent expansion of liquid TV  SR [1    ] with   (    ) i.e. a 0 SR 0 Fig. 12.11 (3) The actual increase in the volume of the liquid = The apparent increase in the volume of liquid + the increase in the volume of the vessel. (4) Liquids have two coefficients of volume expansion. a SR 0 a 60 True value = Scale reading [1   (    )] (i) Co-efficient of apparent expansion ( ) : It is due to apparent (that appears to be, but is not) increase in the volume of liquid if expansion of vessel containing the liquid is not taken into account. at  >  TV > SR TV = SR at  <  TV < SR Fig. 12.9 Apparent expansion in volume (V )a  Initial volume   V   a  (ii) Co-efficient of real expansion ( ) : It is due to the actual increase in volume of liquid due to heating. r ID However, if     , due to contraction of scale, scale reading will be more than true value, so true value will be lesser than scale reading and will still be given by above equation with   (    ) negative. E3 a at  r  Real increase in volume (V )r  Initial volume   V   (iii) Also coefficient of expansion of flask  Vessel  (V )Vessel V   (iv)  Real   Apparent   Vessel a A r D YG U (5) Expansion of cavity : Thermal expansion of an isotropic object may be imagined as a photographic enlargement. So if there is a hole A in a plate C (or cavity A inside a body C), the area of hole (or volume of cavity) will increase when body expands on heating, just as if the hole (or cavity) were solid B of the same material. Also the expansion of area (or volume) of the body C will be independent of shape and size of hole (or cavity), i.e., will be equal to that of D. r a B (v) Change (apparent change) in volume in liquid relative to vessel is Vapp  V  app  = V( Real   Vessel )  V ( r  3 )  = Coefficient of linear expansion of the vessel. Table 12.4 : Different level of liquid in vessel D b Expansion of A = Expansion of B Expansion of C = Expansion of D V  C b Level  Real   Vessel (=3)  app  0 Vapp is positive Level of liquid in vessel will rise on heating.  Real   Vessel (=3)  app  0 Vapp is negative Level of liquid in vessel will fall on heating.  Real   Vessel (=3)  app  0 Vapp  0 level of liquid in vessel will remain same. Fig. 12.10 (6) Some other application U (i) When rails are laid down on the ground, space is left between the ends of two rails. (ii) The transmission cable are not tightly fixed to the poles. Thermal Expansion in Liquids (1) Liquids do not have linear and superficial expansion but these only have volume expansion. (2) Since liquids are always to be heated along with a vessel which contains them so initially on heating the system (liquid + vessel), the level of liquid in vessel falls (as vessel expands more since it absorbs heat and liquid This anomalous behaviour of water causes ice to form first at the surface of a lake in cold weather. As winter approaches, the water temperature increases initially at the surface. The water there sinks because of its increased density. Consequently, the surface reaches 0°C first and the lake becomes covered with ice. Aquatic life is able to survive the cold winter as the lake bottom remains unfrozen at a temperature of about 4°C. At 4°C, density of water is maximum while its specific volume is minimum. max min 0°C 4°C Temperature Density Anomalous behaviour (v) A glass stopper jammed in the neck of a glass bottle can be taken out by warming the neck of the bottle Anomalous behaviour (iv) The iron rim to be put on a cart wheel is always of slightly smaller diameter than that of wheel. (5) Anomalous expansion of water : Generally matter expands on heating and contracts on cooling. In case of water, it expands on heating if its temperature is greater than 4°C. In the range 0°C to 4°C, water contracts on heating and expands on cooling, i.e.  is negative. This behaviour of water in the range from 0°C to 4°C is called anomalous expansion. vol/mass ST (iii) Test tubes, beakers and crucibles are made of pyrex-glass or silica because they have very low value of coefficient of linear expansion. 0°C 4°C (A) Temperature (B) Fig. 12.12 Thermometry, Thermal Expansionand and Calorimetry 559 (3) The amount of heat (Q) is given to a body depends upon it's mass (m), change in it's temperature ( ° = ) and nature of material i.e. Q  m.c.  ; where c = specific heat of material. (4) Heat is a scalar quantity. It's units are joule, erg, cal, kcal etc. Now with rise in temperature by  °C, due to expansion, volume of the body will increase while density of liquid will decrease according to the relations V   V(1   S  ) and     /(1   L  ) T h  V  g (1   S  )   (1   L  ) Th Vg app As  S   L  Th  Th with rise in temperature thrust also to 64°F (7) In solids thermal energy is present in the form of kinetic energy, in liquids, in the form of translatory energy of molecules. In gas it is due to the random motion of molecules. (8) Heat always flows from a body of higher temperature to lower temperature till their temperature becomes equal (Thermal equilibrium). decreases and apparent weight of body increases. Variation of Density with Temperature For a given mass  1 V (9) The heat required for a given temperature increase depends only on how many atoms the sample contains, not on the mass of an individual atom. ID Most substances (solid and liquid) expand when they are heated, i.e., volume of a given mass of a substance increases on heating, so the density  Specific Heat When a body is heated it's temperature rises (except during a change in phase).  V V V 1       V V  V V  V 1    D YG       (1    )1 =  (1    ) 1    U 1   as   . V   (6) British Thermal Unit (BTU) : One BTU is the quantity of heat required to raise the temperature of one pound ( 1 lb ) of water from 63°F 1 BTU = 778 ft. lb = 252 cal = 1055 J and apparent weight of the body W = Actual weight – Thrust should decrease Also 1 kcal = 1000 cal = 4186 J and 1 cal = 4.18 J E3 So the thrust Th  V  g  (5) The calorie (cal) is defined as the amount of heat required to raise the temperature of 1 gm of water from 14.5°C to 15.5°C. 60 (6) Effect of temperature on upthrust : The thrust on V volume of a body in a liquid of density  is given by Th  Vg (1) Gram specific heat : The amount of heat energy required to raise the temperature of unit mass of a body through 1° C (or K) is called specific heat of the material of the body. If Q heat changes the temperature of mass m by  then specific heat Expansion of Gases Q m  Gases have no definite shape, therefore gases have only volume expansion. Since the expansion of container is negligible in comparison to the gases, therefore gases have only real expansion. c (1) Coefficient of volume expansion : At constant pressure, the unit volume of a given mass of a gas, increases with 1°C rise of temperature, is called coefficient of volume expansion. [L T  V 1  V0   Final volume V   V(1   ) U  2 1 P 1  P   Final pressure P  P(1    ) For an ideal gas, coefficient of volume expansion is equal to the 1 coefficient of pressure expansion i.e.     C 1 273 Heat Dimension : ] (ii) For an infinitesimal temperature change d and corresponding quantity of heat dQ. Specific heat c  ST (2) Coefficient of pressure expansion :   (i) Units : Calorie/gm  °C (practical), J/kg  K (S.I.) 2 1 dQ. m d (2) Molar specific heat : Molar specific heat of a substance is defined as the amount of heat required to raise the temperature of one gram mole of the substance through a unit degree it is represented by (capital) C. Molar specific heat (C)  M Gram specific heat (c) (M = Molecular mass of substance) Q 1 Q  m    m   where, Number of moles    M  (1) The form of energy which is exchanged among various bodies or system on account of temperature difference is defined as heat. CM (2) We can change the temperature of a body by giving heat (temperature rises) or by removing heat (temperature falls) from body. Units : calorie/mole  °C (practical); J/mole  kelvin (S.I.) Dimension : [ML2T 2 1 ] 560 Thermometry, Thermal Expansion and calorimetry Specific Heat of Solids When a solid is heated through a small range of temperature, its volume remains more or less constant. Therefore specific heat of a solid may be called its specific heat at constant volume C. (3) The variation of specific heat with temperature for water is shown in the figure. Usually this temperature dependence of specific heat is neglected. Y Sp. heat cal/g C° V 3R CV 1.004 1.000 0.996 X V (2) With rise in temperature, C increases and at a particular temperature (called Debey's temperature) it becomes constant = 3R = 6 cal/mole  kelvin = 25 J/mole  kelvin (4) Dulong and Petit law : Average molar specific heat of all metals at room temperature is constant, being nearly equal to 3R = 6 cal. mole K = 25 J mole K , where R is gas constant for one mole of the gas. This statement is known as Dulong and Petit law. D YG cal gm  C cal Joule.  2100 kg  C kg  C Table 12.5 : Specific heat of some solids at room temperature and atmospheric pressure Specific heat Molar specific heat (J-kg–1 K–1) (J-g mole–1 K–1) 900.0 24.4 386.4 24.5 236.1 25.5 Lead 127.7 26.5 Tungsten 134.4 24.9 Copper ST U Silver (3) If the gas is compressed suddenly and no heat is supplied from outside i.e. Q = 0, but the temperature of the gas raises on the account of compression.  0 Q  0 m( ) m  c U specific heat  T 3 (exception are Sn, Pb and Pt) Aluminium (2) Hence specific heat of a gas, which is the amount of heat energy required to raise the temperature of one gram of gas through a unit degree shall not have a single or unique value. –1 (5) Debey's law : It was observed that at very low temperature molar Substance (1) In case of gases, heat energy supplied to a gas is spent not only in raising the temperature of the gas but also in expansion of gas against atmospheric pressure. ID –1 –1 ice 100 E3 (3) For most of the solids, Debye temperature is close to room temperature. In S.I. c  500 80 Specific Heat of Gases V (6) Specific heat of ice : In C.G.S. cice  0.5 60 (4) As specific heat of water is Fig. very12.14 large; by absorbing or releasing large amount of heat its temperature changes by small amount. This is why, it is used in hot water bottles or as coolant in radiators. Fig. at 12.13 (1) From the graph it is clear that T = 0, C tends to zero –1 20 40 Temp. in °C 60 Debye temp. T 1.0089 Specific Heat of Liquid (Water) (1) Among all known solids and liquids specific heat of water is maximum i.e. water takes more time to heat and more time to cool w.r.t. other solids and liquids. (2) It is observed that by increasing temperature, initially specific heat of water goes on decreasing, becomes minimum at 37° C and then it start increasing. Specific heat of water is – 1 cal cal J  1000  4200 gm  C kg  C kg  C (This value is obtained between the temperature 14.5°C to 15.5°C) (4) If the gas is heated and allowed to expand at such a rate that rise in temperature due to heat supplied is exactly equal to fall in temperature due to expansion of the gas. i.e.  = 0  c Q Q   m( ) 0 (5) If rate of expansion of the gas were slow, the fall in temperature of the gas due to expansion would be smaller than the rise in temperature of the gas due to heat supplied. Therefore, there will be some net rise in temperature of the gas i.e. T will be positive.  c Q  Positive m ( ) (6) If the gas were to expand very fast, fall of temperature of gas due to expansion would be greater than rise in temperature due to heat supplied. Therefore, there will be some net fall in temperature of the gas i.e.  will be negative.  c Q  Negative m( ) Hence the specific heat of gas can have any positive value ranging from zero to infinity. Further it can even be negative. The exact value depends upon the mode of heating the gas. Out of many values of specific heat of a gas, two are of special significance, namely C and C , in the chapter “Kinetic theory of gases” we will discussed this topic in detail. P Specific heat of steam : c steam  0.47 cal / gm  C Phase Change and Latent Heat V Thermometry, Thermal Expansionand and Calorimetry 561 (1) Phase : We use the term phase to describe a specific state of matter, such as solid, liquid or gas. A transition from one phase to another is called a phase change. But when a solid gets converted to a liquid, then the increase in volume is negligible. Hence very less amount of heat is required. So, latent heat of vaporisation is more than the latent heat of fusion. (i) For any given pressure a phase change takes place at a definite temperature, usually accompanied by absorption or emission of heat and a change of volume and density. Thermal Capacity and Water Equivalent (ii) In phase change ice at 0°C melts into water at 0°C. Water at 100°C boils to form steam at 100°C. Heat taken (–  Q) (1) Thermal capacity : It is defined as the amount of heat required to raise the temperature of the whole body (mass m) through 0°C or 1K. Thermal capacity  mc  C  The value of thermal capacity of a body depends upon the nature of the body and its mass. 60 Solid at 0°C Dimension : [ML2 T 2 1 ] , Unit : cal/°C (practical) Joule/k (S.I.) Liquid at 0°C (2) Water Equivalent : Water equivalent of a body is defined as the mass of water which would absorb or evolve the same amount of heat as is done by the body in rising or falling through the same range of temperature. It is represented by W. Heat given (+  Q) E3 Heat given (+  Q) (A) Heat taken (–  Q) Liquid at 100°C Q  If m = Mass of the body, c = Specific heat of body,  = Rise in temperature. Vapours at 100°C Then heat given to body Q  mc (B) If same amount of heat is given to W gm of water and its temperature also rises by . Then ID Fig. 12.15 heat given to water Q  W  1   … (ii) D YG (iv) In case of change of state if the molecules come closer, energy is released and if the molecules move apart, energy is absorbed. (2) Latent heat : The amount of heat required to change the state of the mass m of the substance is written as : Q = mL, where L is the latent heat. Latent heat is also called as Heat of Transformation. It's unit is cal/gm or J/kg and Dimension: [L2T 2 ] ST U (i) Latent heat of fusion : The latent heat of fusion is the heat energy required to change 1 kg of the material in its solid state at its melting point to 1 kg of the material in its liquid state. It is also the amount of heat energy released when at melting point 1 kg of liquid changes to 1 kg of solid. For water at its normal freezing temperature or melting point (0° C), the latent heat of fusion (or latent heat of ice) is LF  Lice  80 cal / gm  60 kJ / mol  336 kilo joule / kg (ii) Latent heat of vaporisation : The latent heat of vaporisation is the heat energy required to change 1 kg of the material in its liquid state at its boiling point to 1 kg of the material in its gaseous state. It is also the amount of heat energy released when 1 kg of vapour changes into 1 kg of liquid. For water at its normal boiling point or condensation temperature (100°C), the latent heat of vaporisation (latent heat of steam) is LV  Lsteam  540 cal / gm  40.8 kJ / mol  2260 kilo joule / kg (iii) Latent heat of vaporisation is more than the latent heat of fusion. This is because when a substance gets converted from liquid to vapour, there is a large increase in volume. Hence more amount of heat is required. [As c water  1 ] From equation (i) and (ii) Q  mc  W  1   U (iii) 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 vapours or gases, the intermolecular forces are almost negligible and the molecules may move freely anywhere in the container. When a solid melts, 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 vapour phase is larger than that in liquid phase. ….. (i)  Water equivalent (W) = mc gm (i) Unit : Kg (S.I.) Dimension : [ML0 T 0 ] (ii) Unit of thermal capacity is J/kg while unit of water equivalent is kg. (iii) Thermal capacity of the body and its water equivalent are numerically equal. (iv) If thermal capacity of a body is expressed in terms of mass of water it is called water-equivalent of the body. Some Important Terms (1) Evaporation : Vaporisation occurring from the free surface of a liquid is called evaporation. Evaporation is the escape of molecules from the surface of a liquid. This process takes place at all temperatures and increases with the increase of temperature. Evaporation leads to cooling because the faster molecules escape and, therefore, the average kinetic energy of the molecules of the liquid (and hence the temperature) decreases. (2) Melting (or fusion)/freezing (or solidification) : The phase change of solid to liquid is called melting or fusion. The reverse phenomenon is called freezing or solidification. Evaporation cools hot water produced by power plants Fig. 12.16 When pressure is applied on ice, it melts. As soon as the pressure is Fig. 12.17 562 Thermometry, Thermal Expansion and calorimetry removed, it freezes again. This phenomenon is called regelation. R.H.(%)  An increase of pressure opposes the melting of such substances and their melting point is raised. (11) Variation of latent heat with temperature and pressure : The latent heat of vapourization of a substance varies with temperature and hence pressure because the boiling point depends on pressure. It increases as the temperature is decreased. For example, water at 1 atm boils at 100°C and has latent heat 2259 Jg but at 0.5 atm it boils at 82°C and has latent heat 2310 Jg –1 –1 P L Critical point Vaporization S Triple point V Critical point P S L Vaporization Triple point V ID (6) Vapour pressure : When the Fig. 12.18 space above a liquid is closed, it soon becomes saturated with vapour and a dynamic equilibrium is established. The pressure exerted by this vapour is called Saturated Vapour Pressure (S.V.P.) whose value depends only on the temperature – it is independent of any external pressure. If the volume of the space is reduced, some vapour liquefies, but the pressure is unchanged. 60 (5) Hoar frost : Direct conversion of vapours into solid is called hoar frost. This process is just reverse of the process of sublimation, e.g., formation of snow by freezing of clouds. (10) Variation of melting point with pressure : For those substances with contract on melting (e.g. water and rubber), the melting point decreases with pressure. The reason is the pressure helps shrinking and hence melting. Most substances expand on melting. (e.g. max, sulpher etc.) Fusion (4) Sublimation : Sublimation is the conversion of a solid directly into vapours. Sublimation takes place when boiling point is less than the melting point. A block of ice sublimates into vapours on the surface of moon because of very very low pressure on its surface. Heat required to change unit mass of solid directly into vapours at a given temperature is called heat of sublimation at that temperature. S.V.P. at dew point  100 S.V. P. at given temperatu re E3 (3) Vaporisation/liquefication (condensation) : The phase change from liquid to vapour is called vaporisation. The reverse transition is called liquefication or condensation. Fig. 12.20for (a) a substance (e.g., water) The figures show the P-T graphs which contracts on melting an (b) a substance (e.g. wax) which expands on melting. The P-T graph consists of three curves. U A saturated vapour does not obey the gas law whereas the unsaturated vapour obeys them fairly well. However, a vapour differs from a gas in that the former can be liquefied by pressure alone, whereas the latter cannot be liquefied unless it is first cooled. The latent heat of fusion shows similar but less pronounced pressure T T dependence. D YG (7) Boiling : As the temperature of a liquid is increased, the rate of evaporation also increases. A stage is reached when bubbles of vapour start forming in the body of the liquid which rise to the surface and escape. A liquid boils at a temperature at which the S.V.P. is equal to the external pressure. It is a fast process. The boiling point changes on mixing impurities. Fig. 12.19 Joule's Law (Heat and Mechanical Work) ST U (8) Dew point : It is that temperature at which the mass of water vapour present in a given volume of air is just sufficient to saturate it, i.e. the temperature at which the actual vapour pressure becomes equal to the saturated vapuor pressure. (i) Sublimation curve which connects points at which vapour (V) and solid (S) exist in equilibrium. (ii) Vapourization curve which shows vapour and liquid (L) existing in equilibrium. (iii) Fusion curve which shows liquid and solid existing in equilibrium. The three curves meet at a single point which is called the triple point. It is that unique temperature-pressure point for a substance at which all the three phases exist in equilibrium. (12) Freezing mixture : If salt is added to ice, then the temperature of mixture drops down to less than 0°C. This is so because, some ice melts down to cool the salt to 0°C. As a result, salt gets dissolved in the water formed and saturated solution of salt is obtained; but the ice point (freeing point) of the solution formed is always less than that of pure water. So, ice cannot be in the solid state with the salt solution at 0°C. The ice which is in contact with the solution, starts melting and it absorbs the required latent heat from the mixture, so the temperature of mixture falls down. (9) Humidity : Atmospheric air always contains some water vapour. The mass of water vapour per unit volume is called absolute humidity. The ratio of the mass of water vapour (m) actually present in a given volume of air to the mass of water vapour (M) required to saturate the same volume at the same temperature is called the relative humidity (R.H.). m Generally, it is expressed as a percentage, i.e., R.H.(%)   100(%) M R.H. May also be defined as the ratio of the actual vapour pressure (p) p of water at the same temperature, i.e. R.H.(%)   100(%) P Thus R.H. may also be defined as Whenever heat is converted into mechanical work or mechanical work is converted into heat, then the ratio of work done to heat produced always W remains constant. i.e. W  Q or J Q Thermometry, Thermal Expansionand and Calorimetry 563 Joule. kcal  4.2  10 3 (4) When water in a stream falls from height h, then its potential energy is converted into heat and temperature of water rises slightly. From W = JQ  mgh = J (mc  ) [where m = Mass of water, c = Specific heat of water,  = temperature rise] gh C Jc  Rise in temperature    1 mv 2  J ( m s  ) 2 c = Specific D YG L = Latent heat of bullet  v2   L     2 J    C      c       U  Rise in temperature ST (6) If m kg ice-block falls down through some height (h) and melts partially (m' kg) then its potential energy gets converted into heat of melting. From W = JQ  mgh  J m ' L  h  m1 c11  m 2 c 2 2 m1 c1  m 2 c 2 Table 12.6 : Temperature of mixture in different cases Condition Temperature of mixture  mix  If bodies are of same mass m '  JL    m  g  If ice-block melts completely then m' = m  h  JL meter g Principle of Calorimetry Calorimetry means 'measuring heat'. When two bodies (one being solid and other liquid or both being liquid) at different temperatures are mixed, heat will be transferred from body at higher temperature to a body at lower temperature till both acquire same temperature. The body at higher temperature releases heat while body at lower temperature absorbs it, so that m11  m 2 2 m1  m 2  mix  m1 = m2 1c1   2 c 2  mix  If m1 = m2 and c1 = c2 Phase change mix  m1c1 (1   mix )  m 2 c2 ( mix   2 )  i.e. c1 = c2 If the temperature of bullet rises upto the melting point of the bullet and bullet melts then. W = J(Q +Q ) From Temperature change m 2 , gram specific heat c 1 and c 2 , temperatures  1 and  2 (1   2 ) are mixed together such that temperature of mixture at equilibrium is  Hence, Heat lost = Heat gained U v2 C 2 Jc 1  mv 2  J (mc   mL ) ; 2 It means the temperature of mixture can never be lesser than lower temperatures (as a body cannot be cooled below the temperature of cooling body) and greater than higher temperature (as a body cannot be heated above the temperature of heating body). Furthermore usually rise in temperature of one body is not equal to the fall in temperature of the other body though heat gained by one body is equal to the heat lost by the other. (2) Mixing of two substances when temperature changes only : It means no phase change. Suppose two substances having masses m 1 and If bodies are of same material [where m = Mass of the bullet, v = Velocity of the bullet, heat of the bullet]  Rise in temperature t  L ID W = JQ mix H  mix  (5) The kinetic energy of a bullet fired from a gun gets converted into heat on striking the target. By this heat the temperature of bullet increases by . From (1) Temperature of mixture ( ) is always ≥ lower temperature ( ) and ≤ higher temperature ( ), i.e.,  L  mix   H. 60 Joule erg  4.2  107 cal cal energy. c1  c 2 1   2 2 (3) Mixing of two substances when temperature and phase both changes or only phase changes: A very common example for this category is ice-water mixing. Suppose water at temperature  °C is mixed with ice at 0 °C, first ice will melt and then it's temperature rises to attain thermal equilibrium. Hence; Heat given = Heat taken W i  mW CW (W  mix )  mi Li  miCW (mix  0)   mix  m i Li CW mW  m i m W W  (i) If mW  mi then  mix  W  Li CW 2 (ii) By using this formulae if  mix   i then take mix  0C Heating Curve If to a given mass (m) of a solid, heat is supplied at constant rate P and a graph is plotted between temperature and time, the graph is as shown in figure and is called heating curve. From this curve it is clear that E C T2 Temp. (3) Value of J  4.2 Heat lost = Heat gained i.e. principle of calorimetry represents the law of conservation of heat E3 This is Joule’s law and J is called mechanical equivalent of heat. (1) From W = JQ if Q = 1 then J = W. Hence the amount of work done necessary to produce unit amount of heat is defined as the mechanical equivalent of heat. (2) J is neither a constant, nor a physical quantity rather it is a conversion factor which used to convert Joule or erg into calorie or kilo calories vice-versa. B T1 A O D Boiling Melting point Melting t1 t2 Boiling point t3 Fig. 12.21 t4 Time 564 Thermometry, Thermal Expansion and calorimetry the atmosphere to melt down. So, in the mountains, when snow falls, one does not feel too cold, but when ice melts, he feels too cold. (1) In the region OA temperature of solid is changing with time so, [as Q = Pt] Q  mc S T  P t  mc S T But as (T/t) is the slope of temperature-time curve amount of heat from teeth. (2) In the region AB temperature is constant, so it represents change of state, i.e., melting of solid with melting point T. At A melting starts and at B all solid is converted into liquid. So between A and B substance is partly solid and partly liquid. If L is the latent heat of fusion. Q  mL F or 1 F [as Q  P(t 2  t1 ) ] F 1 ] tan 0 D YG (3) In the region BC temperature of liquid increases so specific heat (or thermal capacity) of liquid will be inversely proportional to the slope of line BC 1 Slope of line BC (4) In the region CD temperature is constant, so it represents the change of state, i.e., boiling with boiling point T. At C all substance is in liquid state while at D in vapour state and between C and D partly liquid and partly gas. The length of line CD is proportional to latent heat of vaporisation 2 i.e., L  Length of line CD [In this region specific heat  U V  It is more painful to get burnt by steam rather than by boiling water at same temperature. This is so because when steam at 100°C gets converted to water at 100°C, then it gives out 536 calories of heat. So, it is clear that steam at 100°C has more heat than water at 100°C (i.e., boiling of water). the same temperature then expansion of both will be equal because thermal expansion of isotropic solids is similar to true photographic enlargement. It means the expansion of cavity is same as if it has been a solid body of the same material. But if same heat is given to the two spheres, due to lesser mass, rise in temperature of hollow sphere will be U i.e. Latent heat of fusion is proportional to the length of line of zero i.e., c L  temperatures close to 0K is known as cryogenics while that dealing with the measurement of very high temperature is called as pyrometry.  A solid and hollow sphere of same radius and material, heated to L  length of line AB slope. [In this region specific heat   Branch of physics dealing with production and measurement of ID or 60 melts down, it absorbs large 1 Slope of line OA P(t 2  t1 ) m (obtained from ice). This is because, when ice-cream i.e. specific heat (or thermal capacity) is inversely proportional to the slope of temperature-time curve. LF  that of water E3 cs   There is more shivering effect of ice-cream on teeth as compared to 1 ] tan 0 ST (5) The line DE represents gaseous state of substance with its temperature increasing linearly with time. The reciprocal of slope of line will be proportional to specific heat or thermal capacity of substance in vapour state.   Q  more As    . mc    Hence its expansion will be more.  Specific heat of a substance can also be negative. Negative specific heat means that in order to raise the temperature, a certain quantity of heat is to be withdrawn from the body. e.g. Specific heat of saturated vapours.  Specific heat for hydrogen is maximum (3.5 cal / gm o C) and it is minimum for radon and actinium ~ 0.022 cal / gm  C .  The minimum possible temperature is 0 K.  Amount of steam at 100°C required to just melt m gm of ice at 0°C is m/8 gm.  If we put the beaker containing water in melting ice, the water in the beaker will cool to 0°C but will never freeze. Water Melting Ice  A pressure in excess of 25 atm is required to make helium solidfy.  After snow falls, the temperature of the atmosphere becomes very At 1 atm pressure, helium remains a liquid down to absolute zero. low. This is because  Boiling temperature of water, if pressure is different from normal the snow absorbs the heat from pressure is t = [100°C – (760 – P in mm)  0.037]°C Boiling  Confusing S.I. and C.G.S. units It is advised to do questions on calorimetry in C.G.S. as calculations becomes simple. If the final answer is in joules, then convert cal into joules.  Invar and quartz have very small values of co-efficient of linear expansion.  In S.I. nomenclature " degree" is not used with the kelvin scale; e.g. 273°K is wrong while 273 K is correct to write. 60  Magnetic thermometer is recommended for measuring very low temperature (2K).  The most sensitive thermometer is gas thermometer.  In winters, generally fog disappear before noon. Because, the atmosphere warms up and tends to be unsaturated. The condensed vapours reevaporates and the fog disappears.  Standardisation of thermometer is obtained with gas thermometer. Because coefficient of expansion of gas is very large. ST U D YG U evaporation and hence, cooling. They do not sweat. ID  Dogs hang their tongues in order to expose a surface to the air for E3  Dew formation is more probable on a cloudiness calm night.