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Lecture 10   temperature. thermometers. thermal expansion.
 

Lecture 10 temperature. thermometers. thermal expansion.

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Lecture 10 temperature. thermometers. thermal expansion.

Lecture 10 temperature. thermometers. thermal expansion.

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    Lecture 10   temperature. thermometers. thermal expansion. Lecture 10 temperature. thermometers. thermal expansion. Presentation Transcript

    • Lecture 10 Temperature. Thermometers. Thermal expansion.
    • Matter is very complex A simple cup of coffee contains ~ 1023 atoms Numerically impossible to follow trajectory of each atom, use Newton’s laws for force, acceleration etc.. Two options 1. Statistical: probability distribution of molecule velocities 2. Macroscopic: a few variables (temperature, pressure, volume…) characterize the bulk properties of matter These variables are called state variables.
    • Temperature and thermal equilbrium Temperature ~ average kinetic energy of atoms Two pieces of matter in thermal contact exchange energy until their temperatures (T ) are the same (0th law of thermodynamics) Example: Molecules in coffee transfer energy to the air molecules. Eventually all coffee molecules have the same average kinetic energy as the molecules of air in this room Reaching equilibrium (same T ) requires transfer of energy This energy in transit is called heat Q.
    • ACT: Temperature When a series of blocks are connected thermally, heat starts to flow between the blocks as shown. What does this tell us about relative temperatures of the blocks? 1 Q A. T1 = T2 = T3 = T4 B. T1 > T2 > T3 > T4 C. T1 < T2 < T3 < T4 2 Q’ 3 Q’’ 4 This is how the sequence is established empirically. Also: You just applied the 2nd law of thermodynamics....
    • Temperature scales: Celsius, Fahrenheit Celsius Based on the boiling and freezing points of water. 0°C = freezing point of water 100°C = boiling point of water Fahrenheit Based on the boiling and freezing points of alcohol. Connection to Celsius: 0°C = 32°F 100°C = 212°F 9 TF = TC + 32 5
    • Constant volume gas thermometers For gases, p is proportional toT for constant volume (Charles’ law). p p = constant T T gas (He) Liquid (Hg)
    • Temperature scales: Kelvin p(T) for different gases: p Extrapolation of ALL lines points to T = −273.15°C T (°C) Kelvin scale 0 K = −273.15°C (lowest energy state, quantum motion only) 2nd fixed point: 273.16 K (=0.01°C) is the triple-point for H 2O (ice, water, steam coexist) TK =TC + 273.15 With this choice, 1°C = 1K (equal increments)
    • Thermal expansion of a bar A bar of length L0 expands ΔL when temperature is increased by ΔT. Experimentally, ∆L = α L0 ∆T α = coefficient of linear expansion (depends on material) Basis for many thermometers e.g. liquid mercury Critically important in many engineering projects (expansion joints)
    • In-class example: Bimetallic strips Which of the following bimetallic strips will bend the furthest to the right when heated from room temperature to 100°C? α (µJ/m K) Invar Al Al Invar Brass C Al Brass D Ag Au E Al/Invar have the largest difference in α. Aluminum expands more than invar, so A will bend to the right. 1.3 Brass B A 24 Invar Al Al 21 Au 14 Ag 19 DEMO: Bimetallic strips
    • A couple of applications Bimetallic strips can be used as mechanical thermometers. About invar: Fe-Ni alloy with very small α Train tracks have expansion joints (gaps) to prevent buckling in hot weather. (origin of the “clickety-clac, clickety-clac”) High speed trains cannot afford the vibrations produced by these gaps. Alloys with small α to build tracks are a key development.
    • Area Thermal Expansion b b +Δb ΔT a a +Δa Afinal = (a + ∆a )(b + ∆b ) = ab + a ∆b + b ∆a + very small terms ≈ ab + a (bα∆T ) + b (a α∆T ) = ab + 2ab α∆T ∆ A ∆A = 2α A0 ∆T
    • ACT: Washer DEMO: Balls and rings A circular piece of metal with a round hole is heated so that its temperature increases. Which diagram best represents the final shape of the initial shape metal? A. Both inner, outer radii larger B. Inner radius smaller, outer radius larger C. Same size All lengths increase: a 2-D object grows in length and breadth
    • Opening tight jar lids αglass = 0.4-0.9 × 10-5 K−1 αbrass = 2.0 × 10-5 K−1 If you place the jar top under the hot water faucet, the brass expands more than the glass.
    • Volume Thermal Expansion c a c +Δc ΔT b a +Δa b +Δb Vfinal = (a + ∆a )(b + ∆b )(c + ∆c ) = abc + ab ∆c + ac ∆b + bc ∆a + smaller terms ≈ abc + 3abc (α∆T ) ≈ V0 + 3 0α∆T V ∆V ∆V = βV0 ∆T β = 3α Coefficient of volume expansion
    • The special case of water Most materials expand when temperature increases. Water between 0 and 4°C is the exception. V Maximum density Ice is less dense than cold (< 4°C ) water. Ice floats α has an important dependence on T 0 4°C T This prevents lakes from freezing from the bottom up, which would kill all forms of life.
    • Microscopic Model of Solids Consider atoms on a lattice, interacting with each other as if connected by a spring movie
    • Microscopic Model of Expansion (attempt 1) Model potential energy of atoms U = U 1 k (x − l0 )2 + const 2 vibration at high T vibration at low T l0 spacing l0 separation x Vibrating atoms spend equal time at x > l0 as at x < l0 → average separation of atoms = l0 , at both low and high T → material same size at low and high T → This spring model of matter does not describe expansion
    • Microscopic Model of Expansion (2) Change model to non-ideal spring U U ≠ 1 k (x − l 0 )2 2 asymmetric parabola vibration at high T vibration at low T spacing lT lhighT llowT x At high T, the average separation of atoms increases → solid expands (Now we need another parameter in model, asymmetry of parabola)
    • Why two models? Is this more useful compared to empirical ∆L = λ L0 ∆T ? The empirical formula can be used for simple cases. Example: for designing structures to take into account expansion. The microscopic model of expansion (non-ideal spring) can be used in wider number of cases. Example: thermal expansion while also under high-pressure in a chemical engineering plant
    • Thermal stress A rod of length L0 and cross-sectional area A fits perfectly between two walls. We want its length to remain constant when we increase the temperature. Constant length = zero net strain  ∆L   ∆L   ∆L  = + =0  ÷ ÷ ÷  L0  all  L0  thermal  L0  applied stress α∆T + F =0 YA F = −αY ∆T A Thermal stress (stress walls need to provide to keep length constant)
    • Application of thermal stress For very tight fitting of pieces (like wheels): Wheel is heated and then axle inserted. Wheel cools down and shrinks around axle, very tight.