Subjects to be discussed: Defn T: Av KE KE directly proportional to absolute T, KE---kT
2
e discussed: KE proportional to absolute T, KE---kT C & B↔C Then A↔B y transfer due delta T ternal E: sum of all KE+PE tion ΔQ/Δt = kA (ΔT/Δx)--- k:thermal conductivity, (..) thermal gradient heat flow in a fluid medium no medium needed for propagation, elctromagnetic radiation, P~ AT 4 at Capacity: heat needed to raise T of 1 kg of a substance by 1 Cº. indep of T. Is it? (J kg -1 K -1 ) ilibrium: reached at T1=T2 tent Heat of Fusion: heat required to melt a unit mass of A at its (normal) m es : no V (point newtonian particles), no interaction (zero intermolec forces), lisions, vacuum between particles kT v 2 = (v1 2 + v2 2 + v3 2 +...)N or RMS speed ~ average speed (but actually e but close). Average velocity = 0 nal E of Ideal gas (U) depends ony on T or KE since no PE kT (total int E for N ideal gas atoms) of molec x collisions with walls) molecular defn roscopic defn) tte Law: P ~ 1/V at n,T const. PV=const, V= const/P V ~ T at const P,n V = const x T Law: V ~ n at const T, P. V= const X n = const x T = const x n → V ~ nT/P.... V= nRT/P... PV=nRT Equation of state
description
Subjects to be discussed: Defn T: Av KE KE directly proportional to absolute T, KE---kT 0th Law: A↔C & B↔C Then A↔B Heat: energy transfer due delta T U: total internal E: sum of all KE+PE Heat conduction Δ Q/ Δ t = kA ( Δ T/ Δ x)--- k:thermal conductivity, (..) thermal gradient - PowerPoint PPT Presentation
Transcript of Subjects to be discussed: Defn T: Av KE KE directly proportional to absolute T, KE---kT
Subjects to be discussed:
• Defn T: Av KE• KE directly proportional to absolute T, KE---kT• 0th Law: A↔C & B↔C Then A↔B• Heat: energy transfer due delta T• U: total internal E: sum of all KE+PE• Heat conduction ΔQ/Δt = kA (ΔT/Δx)--- k:thermal conductivity, (..) thermal gradient• convection: heat flow in a fluid medium• radiation: no medium needed for propagation, elctromagnetic radiation, P~ AT4
• Specific Heat Capacity: heat needed to raise T of 1 kg of a substance by 1 Cº. assume C is indep of T. Is it? (J kg-1 K-1)• Thermal Equilibrium: reached at T1=T2• Specific Latent Heat of Fusion: heat required to melt a unit mass of A at its (normal) melting point Q= m L• Phase changes• Idael Gases: no V (point newtonian particles), no interaction (zero intermolec forces), elastic collisions, vacuum between particles• ½ mv2 = 3/2 kT v2= (v12 + v22 + v32 +...)N or RMS speed ~ average speed (but actually not the same but close). Average velocity = 0 • Total Internal E of Ideal gas (U) depends ony on T or KE since no PE U = 3/2 N kT (total int E for N ideal gas atoms)• P ~ (speed of molec x collisions with walls) molecular defn• P= F/A (macroscopic defn)• Boyle Mariotte Law: P ~ 1/V at n,T const. PV=const, V= const/P• Charles Law V ~ T at const P,n V = const x T• Avogadro’s Law: V ~ n at const T, P. V= const X n • V= const/P = const x T = const x n → V ~ nT/P.... V= nRT/P... PV=nRT Equation of state
• U =3/2 NkT = 3/2 nRT= 3/2 PV for point-monoatomic particles. if polyatomic U is bigger• Mechanical work done by gas: w= PΔV (area under PV curve at const n, T)• work done by the sys (negative sign, internal E decreases), work done on the sys positive• Isobaric: const P, isochoric: const V, isothermal: const T, adiabatic: ΔQ=0• 1st LAW: ΔU= q + w U: state function-capital letters, q and w are not—lower case• SOLVE EXAMPLE FROM PETRUCCI, isothermal +adiabatic expansion: waht happens to T, w, q and U ?• 2nd LAW: in a heat engine, absorbed heat can not be completely converted into work. (no %100 ordered flow of energy) or: in a spontaneous process: ΔSunverse>0• ΔS= ΔQ/T• Heat does not flow from cold to hot objects• Effiicieny of a Heat Engine Eff= W/Qh= (Qh – Qc)/ Qh < 1• The most efficient engine: Carnot Heat engine Eff= (Th-Tc)/Th
