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# 11th Physics Notes - Thermal Expansion for JEE Main

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### 11th Physics Notes - Thermal Expansion for JEE Main

1. 1. 9011041155 / 9011031155 • Live Webinars (online lectures) with recordings. • Online Query Solving • Online MCQ tests with detailed solutions • Online Notes and Solved Exercises • Career Counseling 1
2. 2. 9011041155 / 9011031155 Thermal Expansion Increase in the Dimensions of a substance on heating is called its Thermal Expansion. Expansion in Solids Almost all the solids (except bismuth, ice and cast iron) expand on heating. In solids all the atoms and molecules are closely bound to each other. So, they exert strong electromagnetic forces on each other. Due to this tight bonding between them, though they vibrate about their mean positions, they cannot leave it forever. When a solid is heated, the atoms and molecules absorb thermal energy and start vibrating with greater amplitude. 2
3. 3. 9011041155 / 9011031155 So, they move away from each other or intermolecular distance between them increases. This causes changes in different dimensions of the solid. 1. Linear Expansion Increase in the length of the solid on heating is called linear expansion. Let, o L0 = length of the solid at 0 C. o Lt = length of the solid at t C. t = rise in temperature of the solid. ∴ Lt – L0 = increase in the length. It is found that Lt – L0 α L0t ∴ Lt – L0 = α L0t .....(1) where α is coefficient of linear expansion of the solid. 3
4. 4. 9011041155 / 9011031155 L t L0 L0 t If L0 = 1m or cm and t = 1℃, then α = Lt – L0 Thus, temperature coefficient of linear expansion of a solid is increase in its length per unit original length at 0℃, per unit rise in its temperature. Its unit is per ℃ (/℃) and values for solids are ranging -6 -5 from 10 to 10 / ℃. From equation (1) Lt = L0 + L 0 α t ∴ Lt = L0 (1 + α t) ..... (2) 4
5. 5. 9011041155 / 9011031155 Let, L1 = length of the solid at t1℃ L2 = length of the solid at t2℃ From equation (2), L1 = L0 (1 + α t1) and L2 = L0 (1 + α t2) L2 L1 1 t2 1 t1 ∴ L2 / L1 = (1 + α t2) (1 + α t1) -1 -1 Binomial expansion of (1 + α t1) will be 1 – α t1 + terms containing higher powers of α. As α is very small, its higher powers can be neglected. -1 ∴ (1 + α t1) = 1 – α t1 5
6. 6. 9011041155 / 9011031155 ∴ L2/L1 = (1 + α t2) (1 – α t1) 2 ∴ L2/L1 = 1 – α t1 + α t2 – α t1t2 ∴ L2/L1 = 1 + α (t2 – t1) ∴ L2 = L1 + L1α (t2 – t1) ... (3) ∴ L2 – L1 = L1 α (t2 – t1) L2 L1 L1 t2 t1 Thus, by knowing L1, L2 and rise in temperature (t2 – t1), a can be calculated approximately. 6
7. 7. 9011041155 / 9011031155 2. Areal or Superficial Expansion Let, A0 = area of the solid at 0℃. At = area of the solid at t℃. t = rise in temperature of the solid. ∴ At – A0 = increase in the area. It is found that At – A0 α A0t ∴ At – A0 = β A0t ..... (1) where β is coefficient of areal expansion of the solid. A t A0 A0 t If A0 = 1m2 or cm2 and t = 1℃, then β = At – A0 Thus, temperature coefficient of areal expansion of a solid is increase in its area per unit original area 7
8. 8. 9011041155 / 9011031155 at 0℃, per unit rise in its temperature. Its unit is per ℃(/ ℃). From equation (1) At = A0 + A0 βt ∴ At = A0 (1 + βt) ..... (2) Every time it is not essential to measure area of the solid at 0℃, to calculate β. It can be calculated at any temperature with the formula : A2 A1 A1 t2 t1 8
9. 9. 9011041155 / 9011031155 3. Volumetric Expansion Let, V0 = volume of the solid at 0℃. Vt = volume of the solid at t℃. t = rise in temperature of the solid. ∴ Vt – V0 = increase in the volume. It is found that Vt – V0 α V0t ∴ Vt – V0 = γ V0t ..... (1) where b is coefficient of areal expansion of the solid. V1 V0 V0 t 3 3 If V0 = 1m or cm and t = 1℃, then γ = Vt – V0 Thus, temperature coefficient of volume expansion of a solid is increase in its volume per unit original 9
10. 10. 9011041155 / 9011031155 volume at 0℃, per unit rise in its temperature. Its unit is per ℃ (/℃). From equation (1) Vt = V0 + V0 β t ∴ Vt = V0 (1 + γ t)..... (2) Every time it is not essential to measure volume of the solid at 0℃, to calculate γ. It can be calculated at any temperature with the formula : V2 V1 V1 t2 t1 10
11. 11. 9011041155 / 9011031155 Relation between α, β and γ Consider a thin square plate of a solid having side L0 at 0℃. ∴ Surface area at 0℃ = A0 = 2 L0 Let the plate be heated to t℃. Let, Lt = length of the plate at t℃ ∴ At = area of the plate at t℃ = Lt 2 For linear expansion, Lt = L0 (1 + α t) 2 ∴ Lt = 2 L0 (1 + α t) 2 i.e. At = A0 (1 + α t) 2 2 ∴ At = A0 (1 + 2αt + α t ) 2 2 As α is very small α t can be neglected. ∴ At = A0 (1 + 2αt) But, At = A0 (1 + β t) ∴ 2α = β 11 2
12. 12. 9011041155 / 9011031155 Consider a uniform of a solid having side L0 at 0℃. ∴ Volume of the cube at 0℃ = V0 = 3 L0 Let the plate be heated to t℃. Let, Lt = length of the plate at t℃ ∴ Vt = volume of the plate at t℃ = Lt 3 For linear expansion, Lt = L0 (1 + α t) 3 ∴ Lt = 3 L0 (1 + α t) 3 i.e. Vt = V0 (1 + α t) 2 2 3 3 3 ∴ Vt = V0 (1 + 3αt + 3α t + α t ) 2 2 3 3 As α is very small 3α t , a t can be neglected. ∴ Vt = V0 (1 + 3αt) But, Vt = V0 (1 + γ t) ∴ 3α = γ 12
13. 13. 9011041155 / 9011031155 ∴ 6α = 3β = 2γ This is the relation between the three expansion coefficients. 13
14. 14. 9011041155 / 9011031155 Boiling point / Steam point The temperature at which pure water freezes at standard atmospheric pressure is called as ice point or freezing point. The temperature at which pure water boils at standard atmospheric pressure is called as boiling point or steam point. 1. Celsius scale On this scale the ice point (melting point of pure ice) is marked as 0ºC and the steam point (boiling point of water) is marked as 100ºC, both taken at normal atmospheric pressure. The interval between these points is divided into 100 equal parts. Each of these division is called as one degree Celsius and is written as 1ºC. 14
15. 15. 9011041155 / 9011031155 2. Fahrenheit scale On this scale the ice point (melting point of pure ice) is marked as 32ºF (boiling point of water) steam point is marked as 212ºF. The interval between these two reference points is divided into 180 equal parts. Each division is called as degree Fahrenheit and is written as ºF. The graph of Fahrenheit temperature (tf) versus Celsius temperature (tc). It is a straight line whose equation is, tf 32 180 tc 0 100 t f 32 180 tc 0 100 .............(1 (a)) Tk 273.15 100 15 .........(1(b))
16. 16. 9011041155 / 9011031155 Ideal gas equation Boyle’s law states that at constant temperature, the volume of a given mass of a gas is inversely proportional to its pressure. Mathematically this law may be expressed as, V 1 , at constant temperature ......(2) P Charle’s law states that at constant pressure, volume of a given mass of a gas is directly proportional to its absolute temperature. Mathematically this law may be expressed as, V ∝ T, at constant pressure .... (3) Combining (2) and (3) we have, PV = constant .....(4) T For one mole of a gas the constant of proportionality is R 16
17. 17. 9011041155 / 9011031155 PV T R .............(5) PV = RT---This relation is called as ideal gas equation. R is universal gas constant.= 8.31 JK-1 mol-1 PV = nRT ......... (6) From equation (6) PV ∝ T. This relationship allows a gas to be used to measure temperature in a constant volume gas thermometer keeping the volume of a gas constant, it gives P ∝ T. 17
18. 18. 9011041155 / 9011031155 Specific heat Capacity Specific heat of a substance is defined as the quantity of heat required to rise the temperature of unit mass of a substance through 1ºC (or 1 K). Q ∝ M ∆T Q ∝ cM ∆T ............. (7) where c is called as specific heat (or specific heat capacity) of a substance. From equation (8) we get c Q .............. (9) m T If m = 1 kg and ∆T = 10C then c = Q. SI unit of specific heat is Jkg-10C-1 or Jkg-1 K-1, CGS Unit is cal g-1 K-1 or Jkg-1 K-1. The specific heat of water is 4.2 Jkg-10C-1; it means that 4.2 Jkg-10C-1 of energy must be added to 1g of water to rise its temperature by 18
19. 19. 9011041155 / 9011031155 1ºC. In case of gas slight change in temperature is accompanied with considerable changes in both i.e. in volume and pressure. If gas is heated at constant pressure, volume changes and hence some work is done on surrounding in expansion. Hence more heat is required. Therefore specific heat at constant pressure is greater than specific heat at constant volume. 1. Specific heat of a gas at constant volume (cv) is defined as the quantity of heat required to rise the temperature of unit mass of a gas through 1K (or 10C) when its volume is kept constant. 2. Specific heat of a gas at constant pressure (cp) is defined as the quantity of heat required to rise the 19
20. 20. 9011041155 / 9011031155 temperature of unit mass of a gas through 1K (or 10C) when its pressure is kept constant. For unit mass of a gas, the specific heats are called principal specific heats. i. Molar specific heat of a gas at constant volume (Cv) is defined as the quantity of heat required to rise the temperature of one mole of the gas through 1 K (or 10C), when is volume is kept constant. ii. Molar specific heat of gas at constant pressure (CP) is defined as the quantity of heat required to rise the temperature of one mole of the gas through 1 K (or 10C) when its pressure is kept constant. The number of molecules in one mole of a gas is given by Avogadro’s number N = 6.025 × 1023 molecules per mole 20
21. 21. 9011041155 / 9011031155 = 6.025 × 1026 molecules per kilomole. SI unit of molar specific heat is J/K and mole K. The molar specific heat = molecular weight × principle specific heat. Cp = M × cp and Cv = M cv .............. (10) At standard pressure the temperature at which a substance changes its state from solid to liquid is called as its melting point. Melting point of water is 0ºC. At standard pressure the temperature at which a substance changes it state from liquid to gas is called a boiling point. The boiling point of water is 100ºC. The triple point of water is that point where water in a solid, liquid and gas states coexists in equilibrium and this occurs only at a unique temperature and a pressure. 21
22. 22. 9011041155 / 9011031155 Latent heat of a substance is the quantity of heat required to change the state of unit mass of the substance without changing its temperature. Conduction It is mode of transfer of heat through a medium, without actual migration of the particles of the medium. Temperature gradient 22
23. 23. 9011041155 / 9011031155 It is the rate of change of temperature with distance for a metal rod in steady state. The figure shows, a section of a metal rod of length ∆x. In steady state, if the temperature of the left end is θ + ∆θ and that of the right and is θ, the temperature gradient is ∆θ / ∆x. In steady state, the rod doesn’t absorb any heat. So, the amount of heat entering the rod per second (Q) is same as the amount of heat leaving the rod per second. 23
24. 24. 9011041155 / 9011031155 Experimentally it is found that, the heat flowing through a conductor or the amount of heat conducted by a conductor is directly proportional to area (A) of the conductor, temperature gradient (∆θ / ∆x) and time t for which the heat is flowing. i.e. Q ∝ A (∆θ / ∆x) t ∴ Q = kA (∆θ / ∆x)t where k is coefficient of thermal conductivity. Q k A If A 1, x x t 1 and t 1, then k=Q 24
25. 25. 9011041155 / 9011031155 So, coefficient of thermal conductivity is the amount of heat flowing through a conductor, of unit area, in unit time, when unit temperature gradient is maintained along its length. Its units are cal / cm ℃ s in C.G. S., kcal / m ℃ s in M.K.S. and J / m ºK s in S.I. 1 1 -3 -1 Its dimensions are [M L T K ] 25
26. 26. 9011041155 / 9011031155 Convection Transfer of heat energy because of actual migration of particles from 1 point to another point is known as Convection 26
27. 27. 9011041155 / 9011031155 Radiation Transfer of heat energy through vacuum is known as radiation Newton’s law of cooling The rate of fall of temperature of a body is directly proportional to excess temperature of a body over the surrounding (provided that the excess is small) 27
28. 28. 9011041155 / 9011031155 • • Ask Your Doubts For inquiry and registration, call 9011041155 / 9011031155. – Team Ednexa 28