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UntitledClassical Physics- at end of the 19thcentury:•Mechanics - Newton’s Laws•Electromagnetism - Maxwell’s Eqns•Optics - Geometric (particles) vs.Physical (waves)•Thermodynamics - Four Laws (0-3)•Gas Laws - Kinetic Theory- overlaps often led to important discoveries:•Maxwell’s Eqns EM radiation (optics)•Newton’s laws / kinetic theory microscopic/atomic description of macroscopic gas laws- around 1900, Theoretical problems:1) What is EM medium? Relativity2) Blackbody Radiation Quantum Physics- Experimental discoveries: Quantum Physics1895 X-rays Relativity1896 Radioactivity Atomic Physics1897 The electron Nuclear Physics Particle PhysicsBut first...Heat and Thermodynamics- study of Thermal Energy of systemsTemperature: a measure of thermal energy, units of Kelvins Room Temp ~ 290 KTemperature of an object is measuredby the change in some physicalproperty.Measuring device is called athermometer.Zeroth Law ofThermodynamicsIf bodies A and B are each in thermalequilibrium with a third body T, thenthey are in thermal equilibrium witheach other.Thermal equilibrium: all measureableproperties unchanging.Objects in thermal equilibrium are atthe same temperature.A BTTTemperature Scales•Daniel Fahrenheit (1686-1736)0°F = mixture of ice, water, salt100ºF = Human body temp (~98.6ºF)•Anders Celsius (1701-1744)0°C = Freezing point of H2O100°C = Boiling point of H2O•Lord Kelvin (1824-1907)H2O boil : 100°C = 212°F = 373 KH2O freeze : 0°C = 32°F = 273 KAbsolute zero : -273°C = -460°F = 0 KTC = TK - 273.15TF = (9/5)TC + 32Constant-Volume Gas Thermometerfi measure pressure of gas at fixed volumePressure = Force/Area (N/m2=Pa) (Pascals) 1 atm = 1.01x105 Pa = 14.7 lb/in2 = 760 mm of Hg = 760 torrT µ P at fixed V 0 Temperature (K)PressureVolume of GaspistonmIdeal-Gas TemperatureTk = (constant) x P at fixed V-Need one point:Triple point of H2O(ice/water/steam coexist)T3 = 273.16 K-Problem: different gases give different TBut as mass of gas reduced (mÆ0)and P3Æ0, they agree(approach “ideal” gas) 0 massTN2H2HeTemperature and HeatIf system S and environment E areAt different temperatures:Energy will transfer until theirtemperatures become equal.The transferred energy is calledHeat (symbol Q).TE > TS , Q > 0 Heat absorbed by S TS > TE , Q < 0 Heat lost by SSESEDefn: Require DQ = 1 calorieto raise 1 gm of H2O by DT = 1°C.1 calorie = 4.186 joules(heat is a form of energy) Specific HeatAmount of heat needed to raise thetemperature of m grams of asubstance by DT isDQ = c m DTwhere c is the specific heat (cals/g·°C)Specific heat of water = 1 cal/g·°C = 4186 J/kg·KMolar Specific HeatCan specify amount of substance inmoles: 1 mole = 6.02x1023 units = NA units (Avogadro’s number)( 1 mole of Al = 6.02x1023 atoms 1 mole of CO2 = 6.02x1023 molecules)The mass of a substance (in grams) ism = n Awhere n = (# moles) andA is the atomic (molecular) weight ofthe substance.1 mole of Carbon-12 has m = 12 grams.Some Examples:24.427900Aluminum24.563.5386Copper25.5108236Silver24.8184134Tungsten26.5207128LeadMol. Sp. Ht.(J/Mole˙K) ASpec. Heat(J/kg˙K)ElementNote the relativeconsistencyDQ = n C DTHeats of TransformationHeat may also change the phase (orstate) of a substance (at constant T).Matter exists in 3 common states:•Solid•Liquid•Gas (or vapor)Solid Liquid Gasmelt boilfreeze condense“fusion” “vaporization”Requires energy Releases energy Amount of energy/unit mass is Heat of transformation, L.e.g. for water:Heat of fusion LF = 79.5 cal/g = 333 kJ/kg = 6.01 kJ/moleHeat of vaporizationLV = 539 cal/g = 2256 kJ/kg = 40.7 kJ/moleHeat and WorkConsider this system:Pressure = Force/Area (P=F/A)If piston moves ds, then work doneby the gas:dW = F ds = P A ds = P dVTotal work done by the gas in movingfrom Vi to Vf :DW = ∫ P dVGASPiston(area A)insulationHeating padViVfFdsP-V DiagramsStudy effects of heat added/workdone by plotting P vs V of gas:The area under thecurve is the workdone.The work done depends on the specificpath from i to f.PVi Vf VifPVi Vf VifPVi Vf VifABThermodynamic CyclesIf volume decreases, the work done(by the gas) is negative.If we go from iÆf and back to i,the net work done by the gas is thearea inside the curve.PVfiPVifPVif+ work- workFirst Law ofThermodynamicsHeat DQ added to the system can havetwo effects:•Increase the internal energy ofthe system•Cause the gas to do workConservation of Energy says: DQ = DU + DW whereU is the internal energy of the system.fi 1st Law of Thermodynamics.State FunctionsA property of the state of the systemis often called a “State Function”.P, V, and T are state functions.So is U (the internal energy).Heat and Work are not.They are path-dependent, i.e. theydepend on how we go from i to f.However the combinationDQ - DW = DU does not depend on thepath.Various System ChangesConstant Volume Constant Pressure(isochoric) (isobaric) DW=0 DW=PDV DQ=DUConstant Temp Constant Heat(isothermal) (adiabatic) DU=0 DQ=0 DQ=DW DU=-DWCyclical Process(returns to original state) DU=0 DQ=DWPVifPVifP Vi fPVifPVifAll previous cases are “quasi-static”:Change occurs slow enough that thermalequilibrium can be considered true at alltimes.A non-quasi-static process:Adiabatic, free expansion:DQ = 0 (adiabatic)DW = 0 (nothing to work against)fi DU = 0Heat Transfer MechanismsHow does heat exchange occur?•Conduction•Convection•RadiationConduction•Occurs in systems where atoms stay in a fixed region.•Heat energy causes them to move,rotate, and/or vibrate.•Energy is transferred to adjacent atoms by interactions/collisions. Energy moves, not the atomsHeat conduction rate isPcond = DQ/Dt = k A DT/Dx• DT/Dx = (TH-TC)/Dxis the Temperature gradient.• Pcond is the Energy transferred pertime (SI units: Watts), sometimes called thermal current, I.• k is the Thermal Conductivity (SI units: Watts/m•K).TCTHDxArea ASome Thermal ConductivitiesSilver 428 W/m•KCopper 402Aluminum 235Lead 35Stainless Steel 14Hydrogen 0.18Helium 0.15Dry Air 0.026Window Glass 1.0White Pine 0.11Fiberglass 0.048Polyurethane Foam 0.024Using


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MSU PHY 215 - Classical Physics

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