Lecture 241st Law of Thermodynamics1st Law: Work & HeatSlide 4Paths on the pV diagram (with an idea gas)Isothermal processes (in an ideal gas)Adiabatic Processes (in an ideal gas)Work and Ideal Gas Processes (on system)Slide 10Slide 11Both Q and W can change TWhat about Q? What are the relationships between DETh and T.Heat and Latent HeatQ : Latent heat and specific heatLatent HeatMechanical equivalent of heatExerciseSlide 20Heat and Ideal Gas Processes (on system)Energy transfer mechanismsSlide 26Thermal ConductivitiesExercise Thermal ConductionSlide 31Slide 32Minimizing Energy TransferAnti-global warming or the nuclear winter scenarioSlide 35Physics 207: Lecture 24, Pg 1Lecture 24Goals:Goals:•Chapter 17Chapter 17 Employ heat (Q) and energy transfer in materials Recognize adiabatic processes (i.e., Q=0)•Chapter 18Chapter 18 Follow the connection between temperature, thermal energy, and the average translational kinetic energy molecules Understand the molecular basis for pressure and the ideal-gas law. To predict the molar specific heats of gases and solids.•AssignmentAssignment HW11, Due Wednesday 9:00 AM For Thursday, Read through all of Chapter 18Physics 207: Lecture 24, Pg 21st Law of ThermodynamicsWork W and heat Q depend on process by which the system is changed (path dependent).The change of energy in the system, ΔEth depends only on the total energy exchanged W+Q, not on the process.ΔEth =W + Q(if K & U =0 )W & Q with respect to the systemPhysics 207: Lecture 24, Pg 31st Law: Work & HeatWork done on system (an ideal gas….notice minus sign, V is in reference to the system) Won system < 0 Moving left to right [where (Vf > Vi)]If ideal gas, pV = nRTWby system > 0 Moving left to rightfinalinitialon)curveunder area( dVpWPhysics 207: Lecture 24, Pg 41st Law: Work & HeatWork: Depends on the path taken in the pV-diagram(It is not just the destination but the path…) Won system > 0 Moving right to leftPhysics 207: Lecture 24, Pg 6(1) Isobaric(2) Isothermal (3) Isochoric(4) Adiabatic Paths on the pV diagram (with an idea gas)pV3421T1T2T3T4W = - p V ????W = 0???? IdealgasPhysics 207: Lecture 24, Pg 7Isothermal processes (in an ideal gas)Work done when PV = nRT = constant P = nRT / Vfinalinitial)curveunder area( dVpWfifiVVVV/ nRT/ nRT VdVVdVW)/VV( nRTifnW pV3T1T2T3T4For this we need access to thermal energyPhysics 207: Lecture 24, Pg 8Adiabatic Processes (in an ideal gas)An adiabatic process is process in which there is no thermal energy transfer to or from a system (Q = 0)A reversible adiabatic process involves a “worked” expansion in which we can return all of the energy transferred.In this case PV = const.All real processes are not.We need to know Cp & CVpV2134T1T2T3T4VCCpPhysics 207: Lecture 24, Pg 9Work and Ideal Gas Processes (on system)Isothermal)/VV( nRTifnW Isobaric)V-V( pifWIsochoric0W)(12constconst2121VVPdVWVVVVVd VVFYI: Adiabatic (and reversible) PV = const.Physics 207: Lecture 24, Pg 10Two process are shown that take an ideal gas from state 1 to state 3. (“by” means “by the system on the world”)Compare the work done by process A to the work done by process B.A. WA > WBB. WA < WB C. WA = WB = 0D. WA = WB but neither is zeroON BYA 1 3 W12 = 0 (isochoric)B 1 2 W12 = -½ (p1+p2)(V2-V1) < 0 -W12 > 0B 2 3 W23 = -½ (p2+p3)(V1-V2) > 0 -W23 < 0B 1 3 = ½ (p3 - p1)(V2-V1) > 0 < 0p1p2p3Physics 207: Lecture 24, Pg 11Two process are shown that take an ideal gas from state 1 to state 3. Compare the work done by process A to the work done by process B.A. WA > WBB. WA < WB C. WA = WB = 0D. WA = WB but neither is zeroON BYA 1 3 W12 = 0 (isochoric)B 1 2 W12 = -½ (p1+p2)(V2-V1) < 0 -W12 > 0B 2 3 W23 = -½ (p2+p3)(V1-V2) > 0 -W23 < 0B 1 3 = ½ (p3 - p1)(V2-V1) > 0 < 0Physics 207: Lecture 24, Pg 12Both Q and W can change TWe know how work changes the mechanical energy of a solid “system”Here our system is an ideal gas….only the temperature can changeFor real materials we can change the temperature or the stateWe must quantify the response of a system to thermal energy transfer (Q)Physics 207: Lecture 24, Pg 13What about Q? What are the relationships between ETh and T.Latent HeatLatent HeatSpecific HeatPhysics 207: Lecture 24, Pg 14Heat and Latent HeatLatent heat of transformation L is the energy required for 1 kg of substance to undergo a phase change. (J / kg)Q = ±MLSpecific heat c of a substance is the energy required to raise the temperature of 1 kg by 1 K. (Units: J / K kg )Q = M c ΔTMolar specific heat C of a gas at constant volume is the energy required to raise the temperature of 1 mol by 1 K.Q = n CV ΔTIf a phase transition involved then the heat transferred is Q = ±ML+M c ΔTPhysics 207: Lecture 24, Pg 15Q : Latent heat and specific heatThe molar specific heat of gasses depends on the process pathCV= molar specific heat at constant volumeCp= molar specific heat at constant pressureCp= CV+R (R is the universal gas constant)Physics 207: Lecture 24, Pg 17Latent Heat Most people were at least once burned by hot water or steam. An equal amount (by mass) of boiling water and steam contact your skin. Which is more dangerous, the water or the steam?Physics 207: Lecture 24, Pg 18Mechanical equivalent of heatHeating liquid water: Q = amount of heat that must be supplied to raise the temperature by an amount T . [Q] = Joules or calories. calorie: energy to raise 1 g of water from 14.5 to 15.5 °C(James Prescott Joule found the mechanical equivalent of heat.) 1 cal = 4.186 J1 kcal = 1 cal = 4186 JSign convention:+Q : heat gained- Q : heat lostPhysics 207: Lecture 24, Pg 19ExerciseThe specific heat (Q = M c ΔT) of aluminum is about twice that of iron. Consider two blocks of equal mass, one made of aluminum and the other one made of iron, initially in thermal equilibrium.Heat is added to each block at the same constant rate until it reaches a temperature of 500 K. Which of the following statements is true? (a) The iron takes less time than the
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