Page 1 :
SEARCH { CYBERWORLDFORU.BLOGSPOT.COM } ON GOOGE, , ALL STUDY MATERIALS IN FREE OF COST, HERE YOU WILL GET FOLLOWING MATERIALS :, , , , , , , , , , , , HANDWRITTEN NOTES, PRINTED NOTES { DETAILED , REVISION , SHORT }, IMPORTANT QUESTIONS, BOOKS PDF, BOOKS SOLUTIONS, TEST PAPERS, CHAPTERWISH PREVIOUS YEAR PAPERS, SAMPLE PAPERS, CONCEPT MAPS, HOT QUESTIONS, NEWS AND UPDATES, , MATERIALS ARE AVILABLE FOR FOLLOWING :, , , , , , , , , , CLASS 9TH, CLASS 10TH, CLASS 11TH, CLASS 12TH, IIT-JEE, NEET, NTSE, KVPY, OTHER COMPETITIVE EXAMS

Page 3 :
HEAT & THERMODYNAMICS, , HEAT & THERMODYNAMICS, tF, , 1. THERMAL PROPERTIES OF MATTER, 212, , •, , Thermal Expansion, , •, , Heat & Clariometry, , •, , Heat Transfer, , Temperature (°F), , In this topic, we discuss various phenomenas involving, thermal and how does a matter behave on experiencing the, flow of thermal energy. Primarily we study, , O, , 1.1 Temperature and Heat, Temperature : Temperature is a relative measure of hotness, or coldness of a body., , Ice point  273.15 K, Steam point  373.15 K, , Conversion : t(k) = t°C + 273.15, , Comparing it with the celcius scale, number of scale division, in both the scales is same., , Heat : Heat is a form of energy flow (i) between two bodies, or (ii) between a body and its surroundings by virtue of, temperature difference between them, , t c  0C t k  273.15, , 100, 100, , SI Unit : Joule (J), Conversion : 1cal = 4.186 J, Heat always flows from a higher temperature system, to a lower temperature & system., , 1.2 Measurement of Temperature, Principle : Observation of Thermometric property with the, change in temperature and comparing it with certain, reference situations., •, , Reference situation is generally ice point or steam point., 1.2.1 Celcius and Fahrenheit Temperature Scales, In Celsius Scale, , In Fahrenheit Scale, , Ice point  0°C, , Ice point  32°F, , Steam point  1000°C, , Steam point  212°F, , It implies that 100 division in celcius scales is equivalent to, 180 scale divisions in fahrenheit scale., Hence , , t, t f  32,  c, 180, 100, , tC, , 100, , It is kelvin scale, , Commonly Used Unit : °C or °F, , Commonly Used Unit : Calorie (Cal), , Temperature (°C), , 1.2.2 Absolute Temperature Scale, , SI Unit : Kelvin (k), , •, , tF=180, , •, , Kelvin scale is called as absolute scale, because it is, practically impossible to go beyond OK in the negative, side., Steam, Point, , 373.15 K, , 100°C, , 212.0°C, , Ice, Point, , 273.15 K, , 0°C, , 0°C, , OK, , –27315°C, , –27315°C, , Absolute, zero, , Kelvin Scale, 1K, , Celcius Scale, 1°C, , Fahrenheit Scale, 1.8°F, , Comparison of Temperature Scales, , 1.2.3 Thermometers, Instrument used to measure temperature of any system is, called as thermometer., Examples : Liquid in Glass thermometer, Platinum Resistance, Thermometer, Constant Volume Gas Thermometers.

Page 4 :
HEAT & THERMODYNAMICS, •, , •, , Liquid in Glass thermometer and Platinum Resistance, thermometer give uniform readings for ice point & steam point, but go non uniform for different liquids and different materials., Constant volume gas thermometer gives same readings, respective of which gas. It is based on the fact that at low, pressures and constant volume, P × T for a gas., Pressure, , Gas B, , 0°C, , All gases converge to absolute zero at zero pressure., 1.3 Thermal Expansion, It is widely observed, that most materials expand on heating, and contract on colling., This expansion is in all dimensions., Experimentaly it has been observed that fractional change, in any dimension is proportional to the change in, temperature., , Gas A, , –273.15°C, , •, , Temperature, (°C), , Linear Expansion, , x,  KT, x, , constant (k), , L,   T, L, , Coefficient of Linear expansion () :, Increase in length per unit length per degree rise in temp., , L, , Area Expension, , A,   T, A, , L, , Volume Expansion, , Coefficient of Area Expansion () :, Increase in area per unit area per degree rise in temp., , V,   T, V, , Coefficient of volume expansion () :, Increase in area per unit volume per degree rise in temp., , V, , Units of ,, = /°C or /K, •, , In general with change in volume the density will also, change., , •, ,  for metals generally higher than  for non-metals, , •, , is nearly constant at high temperatures but all low temp, it depends on temp., , 6, –5, , –1, , r(10 K ), , 250, , T(K), , 500

Page 5 :
HEAT & THERMODYNAMICS, Coefficient of volume expansion of Cu as a function of, temperature., •, , we know, V,  T  compressive strain, V, , For ideal gases  is inversely propertional to temp. at, constant pressure, nRT, V T, , , P, V, T, , v, , , T, , , , •, , As an exception, water contracts on heating from 0°C to, 4°C andhence its density increases from 0°C to 4°C. Thus, is called as anamolous expansion, , •, , Practical applications in railway tracks metal tyres of cart, wheels, bridges and so many other applications., 1.4 Heat & Calorimetry, When two systems at different temperatures are connected, together then heat flows from higher temperature to lower, temperature till the time their temperatures do not become, same., , 1 gm/cc, Density, , •, , 4°C, , (a), , Principle of calorimetry states that, neglecting heat loss to, surroundings, heat lost by a body at higher temperature is, equal heat gained by a body at lower temperature., , (b), , heat gained = heat lost, , In general, , Whenever heat is given to any body, either its temperature, changes or its state changes., , 3,   3  , 2, , 1.4.1 Change in Temperature, , Proof : Imagine a cube of length, l that expands equally in, all directions, when its temperature increases by small T;, , When the temp changes on heating,, , We have, , Heat supplied change in temp (T), , Then, , l = lT, , amount of substance (m/n), , Also, 3, , 3, , 3, , 2, , 2, , 2, , V = (l l) – l = l + 3l l + 3ll + l – l, , nature of substance (s/C), , 3, , 2, , = 3l l, , , ...(1), , 2, , So, 3V, V , l = 3VT, l, , , , V,  3T, V, , , ,  = 3, , V,  l2 ], [Using, l, , s = specific heat capacity per kg, T = Change in temp, or, , In case, thermal expansion is prevented inside the rod by, fixing its ends rigidly, then the rod acquires a compressive, strain due to external fones at the ends corresponding stress, set up in the rod is called thermal stress., , H = nCT, n = Number of moles, , ...(2), , Similarly we can prove for area expansion coefficient, , H = msT, m = Mass of body, , 3, , In Equation (1) we ignore 3ll & l as l is very small as, compared to l., , •, , ...(3), , T  T, , , , 4°C, , ol,    Thermal stress, V, , Also, , C = Specific/Molar heat Capacity per mole, T = Change in temp, •, , Specific Heat Capacity : Amount of heat required to raise, the temperature of unit mass of the substance through one, degree., Units, SI  J/KgK, , SH Oe = 1 cal/g°C, , Common  Cal/gC°, , SH O ice = 0.5 cal/g°C, , 2, , 2

Page 6 :
HEAT & THERMODYNAMICS, •, , Molar Heat Capacity : Amount of heat required to raise the, temperature of unit mole of the substance through one degree, , On heating., , Units, , Step - 2 : Ice melts to H2O(l) keeping the temp constant, , Step - 1 : Temp changes to 0°C first, , SI  J/mol K, , Step - 3 : Temp. inverses to 100°C, , Common  Cal/gc°, •, , Heat Capacity : Amount of heat required to raise the, temperature of a system through one degree, , , , H = ST, , Step - 4 : H2O(l) boils to steam keeping the temp constant, Step - 5 : Further temp increases, Temp, , where S = Heat Capacity, Units, SI  J/K, Common  Cal/C°, •, , For H2O specific heat capacity does change but fairly very, less., , •, , Materials with higher specific heat capacity require a lot of, heat for some a given in temperature, , Heat, , •, , The slope is inversely proportional to heat capacity., , •, , Length of horizontal line depends upon mL for the process., 1.4.3 Pressure dependence on melting point and boiling point, , 1.4.2 Change in state, When the phase changes on heating, Then, Heat supplied  amount of substance which changes the, state (M),  nature of substance (L), , , •, , For some substance melting point decreases with increases, pressure and for other melting point increases, , •, , Melting poing increases with increase in temperature. We, can observe the above results through phaser diagrams., P, (atm), , H = mL, Where L = Latent Heat of process, , •, , Latent Heat : Amount of heat required per mass to change, the state of any substance., Units, , •, , B, , Solid, , O Vapour, A, , T(°C), , A, , Line AO , , Sublimation curve, , The change in state always occurs at a constant, temperature., , Line OB , , Fusion curve, , Line OC , , Vapourization curve, , For example, , Point O , , Triple Point, , Point C , , Critical temperature, , Solid, , , , Liq, , Lf, , Liq, , , , Gas, , Lv, , Lv = Latent heat of vaporization, , Liq, , C, , O Vapour, , Common  Cal/g, , Lf = Latent Heat of fusion, •, , Solid, , C, , Liq, , For H2O, , SI  J/Kg, , P, (atm), , B, , For CO2, , T(°C), , Triple Point : The combination pressure and temperature, at which all three states of matter (i.e. solids, liquids gases, co-exist., For H2O it is at 273.16K and 0.006 Atm., , In case any material is not at its B.P or M.P, then on heating, the temperature will change till the time a particular state, change temperature reaches., , Critical Point : The combination of pressure & temp, beyond which a vapour cannot be liquified is called as, critical point., , For Example : If water is initially at –50°C at 1 Atm pressure, in its solid state., , Corresponding temperature, pressure are called as critical, temperature & critical pressure.

Page 7 :
HEAT & THERMODYNAMICS, •, , •, , From the phasor diagram, we can see that melting point, decreases with increases in pressure for H2O., , •, , Larger the thermal conductivity, the greater will be rate of, heat energy flow for a given temperature difference., , Based on this is the concept of reglation., , •, , Kmetals > Knon metals, , Reglation : The phenomena of refreezing of water melted, below the normal melting point due to addition of pressure., , •, , Thermal conductivity of insulators is very low. Therefore,, air does not let the heat energy to be conducted very easily., , It is due to this pressure effect on melting point that cooking, is tough on mountains and lasier in pressure cooker., , •, , For combinations of rods between two ends kept at different, temperatures, we can use the concept of equivalent thermal, conductivity of the composite rod., , 1.5 Heat Transfer, , For example :, , There are three modes of heat transfer., •, , Conduction, , •, , Convection, , •, , Radiation, , T1, , TC > TD, , TC, , L, , •, ,  TD , , The term, , T, , C, ,  TD , L, , in the above equation is called as, , The term Q, (i.e.) rate of flow of heat energy can also be, named as heat current, , •, , The term (L/KA) is called as thermal resistance of any, conducting rod., Thermal Resistance : Obstruction offered to the flow of, heat current by the medium, , ...(i), , Units  K/W, 1.5.2 Convection, , where Q = Rate of heat energy flow (J/s or W), 2, , A = Area of cross-section (m ), , The process in which heat is transferred from one point to, another by the actual movement of the heated material, particles from a place at higher temperature to another place, of lower temperature is called as thermal convection., , TCTD = Temperature of hot end and cold end respectively, (°C or K), L = Length of the rod (m), K = coefficient of thermal conductivity, Coefficient of Thermal Conductivity : It is defined as, amount of heat conducted during steady state in unit time, through unit area of any cross-section of the substance, under unit temperature gradient, the heat flow being normal, to the area., SI  J/mSk or W/mK., , T2, , •, , for uniform cross-section rods, , Units, , Leq, 2L, A, , SI  K/m, , The rate of heat energy flowing through the rod becomes, constant., L, , T1, , Units, , At steady state,, , C, , , , Temperature Gradient : The fall in temperature per unit, length in the direction of flow of heat energy is called as, Temperature Gradient, , TD, , Direction of, heat flow, , T, , T2, , Temperature Gradient., , A, , This is rate Q  kA, , L2, K 2, A, , where Keq for equivalent thermal conductivity of the, compositive., , 1.5.1 Conduction, Thermal conduction is the process in which thermal energy, is transferred from the hotter part of a body to the colder, one or from hot body to a cold body in contact with it, without any transference of material particles., , L1, K 1, A, , •, , If the medium is forced to move with the help of a fan or a, pump, it is called as forced convection., If the material moves because of the differences in density, of the medium, the process is called natural or free, convection., , •, , Examples of forced convection, Circulatory system, cooling system of an automobile heat, connector

Page 8 :
HEAT & THERMODYNAMICS, Examples of natural convection, , On integrating, , Trade winds, Sea Breeze/Land Breeze, Monsoons Burning, of Tea., , log (T2 – T2) = –Kt + C, or, , 1.5.3 Radiation, , This radiation of heat energy occurs in the form of EM, waves., , •, , These radiators are emitted by virtue of its temperature, like, the radiation by a red hot iron or light from a filament lamp., , •, , Every body radiates energy as well as absorbs energy from, surroundings., , –Kt, , where C1 = e, , c, , ...(6), , equation (6) enables you to calculate the time of cooling of, a body through a particular range of temperature., , It is a process of transmission of heat in which heat travels, directly from one place to another without the agency of, any intervening medium., •, , T2 = T1 + C1e, , T, (°C), , log, (T2–T1), , Time (minute), , Time, , •, , For small temp diff, the rate of cooling, due to conduction,, convection & radiation combined is proportional to, difference in temperature., , •, , Newton’s Law of cooling states that, the rate of loss of heat, , Approximation : If a body cools from Ta to TB in t times in, medium where surrounding temp is T0, then, , d, of the body is directly proportional to the differenct of, dt, , Ta  Tb,  T  Tb, ,  K a,  T0 , t,  2, , , •, , The proportion of energy absorbed depends upon the colour, of the body., , (a), , Newtons Law of cooling, , temp difference, Now, ,  ds,  k  T2  T1 , dt, , •, , Newton’s Law of cooling can be verified experimentally., , ...(4), , where k is a positive constant depending upon area and, nature of the surface of the body. Suppose a body of mass, m, specific heat capacity s is at temperature T 2 & T1 be the, temp of surroundings if dT2 the fall of temperature in time, dt., , T2, , T1, , loge (T2-T1), , •, , C, V, , Amount of heat lost is, dcs = msdT2, , , (b), , t, , Rate of loss of heat is given by, dT, dcs,  ms 2, dt, dt, , From Equation 4 and 5,  ms, , , , (a), , dT2,  k  T2  T1 , dt, , dT2, k, , dt   Kdt, T2  T1 ms, where K , , k, ms, , ...(5), , Set Up : A double walled vessel (v) containng water in, between two walls., A copper calorimeter (c) containing hot water placed inside, the double walled vessel. Two thermometers through the, carbs are used to not the temperature T 2 of H 2O in, calorimeter T 1 of water in between the double walls, respectively., Experiment : The temperature of hot water in the calorimeter, after equal intervals of time., Result : A graph is plotted between log (T2 – T1) and time, (t). The nature of the graph is observed to be a straight line, as it should be from Newton’s law of cooling.