Physics 207, Lecture 21, Nov. 12Fluids: A tricky problemExample problem: Air bubble risingPV diagrams: Important processes1st Law of Thermodynamics1st Law: Work & HeatSlide 71st Law: Work (Area under the curve)Combinations of Isothermal & Adiabatic ProcessesIsothermal processesFirst Law of ThermodynamicsWork, Heat & ThemodynamicsQ : Latent heat and specific heatMechanical equivalent of heatExerciseSlide 16Latent HeatLatent Heats of Fusion and VaporizationExercise Latent HeatEnergy transfer mechanismsSlide 22Thermal ConductivitiesExercise 2 Thermal ConductionExercise Thermal ConductionSlide 26Minimizing Energy TransferAnti-global warming or the nuclear winter scenarioPhysics 207, Lecture 20, Nov. 10Physics 207: Lecture 22, Pg 1Physics 207, Lecture 21, Nov. 12Goals:Goals:•Chapter 16Chapter 16 Use the ideal-gas law. Use pV diagrams for ideal-gas processes.•Chapter 17Chapter 17 Employ energy conservation in terms of 1st law of TD Begin understanding the concept of heat. Demonstrate how heat is related to temperature change Apply heat and energy transfer processes in real situations Recognize adiabatic processes.•AssignmentAssignment HW9, Due Wednesday, Nov. 19th HW10, Due Sunday, Wednesday, Read through 18.3Physics 207: Lecture 22, Pg 2Fluids: A tricky problemA beaker contains a layer of oil (green) with density ρ2 floating on H2O (blue), which has density ρ3. A cube wood of density ρ1 and side length L is lowered, so as not to disturb the layers of liquid, until it floats peacefully between the layers, as shown in the figure.What is the distance d between the top of the wood cube (after it has come to rest) and the interface between oil and water?Hint: The magnitude of the buoyant force (directed upward) must exactly equal the magnitude of the gravitational force (directed downward). The buoyant force depends on d. The total buoyant force has two contributions, one from each of the two different fluids. Split this force into its two pieces and add the two buoyant forces to find the total forcePhysics 207: Lecture 22, Pg 3Example problem: Air bubble risingA diver produces an air bubble underwater, where the absolute pressure is p1 = 3.5 atm. The bubble rises to the surface, where the pressure is p2 = 1 atm. The water temperatures at the bottom and the surface are, respectively, T1 = 4°C, T2 = 23°CWhat is the ratio of the volume of the bubble as it reaches the surface,V2, to its volume at the bottom, V1? (Ans.V2/V1 = 3.74)Is it safe for the diver to ascend while holding his breath? No! Air in the lungs would expand, and the lung could rupture. This is addition to “the bends”, or decompression sickness, which is due to the pressure dependent solubility of gas. At depth and at higher pressure N2 is more soluble in blood. As divers ascend, N2 dissolved in their blood stream becomes gaseous again and forms N2 bubbles in blood vessels, which in turn can obstruct blood flow, and therefore provoke pain and in some cases even strokes or deaths. Fortunately, this only happens when diving deeper than 30 m (100 feet). The diver in this question only went down 25 meters. How do we know that?Physics 207: Lecture 22, Pg 4PV diagrams: Important processesIsochoric process: V = const (aka isovolumetric)Isobaric process: p = constIsothermal process: T = constconstant TpVVolumePressure2211 TpTp12IsochoricVolumePressure2211 TVTV12IsobaricVolumePressure2211VpVp 12IsothermalPhysics 207: Lecture 22, Pg 51st Law of ThermodynamicsThermal energy Eth : Microscopic energy of moving molecules and stretched molecular bonds. ΔEth depends on the initial and final states but is independent of the process.Work W : Energy transferred to the system by forces in a mechanical interaction.Heat Q : Energy transferred to the system via atomic-level collisions when there is a temperature difference. Work W and heat Q depend on the process by which the system is changed.The change of energy in the system, ΔEth depends only on the total energy exchanged W+Q, not on the process.ΔEth =W + QW & Q with respect to the systemPhysics 207: Lecture 22, Pg 61st Law: Work & HeatWork done on system (an ideal gas) Won system < 0 Moving left to right [where (Vf > Vi)]If ideal gas, PV = nRT, and given Pi & Vi fixes TiWby system > 0 Moving left to rightfinalinitial)curveunder area( dVpWPhysics 207: Lecture 22, Pg 71st 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 22, Pg 81st Law: Work (Area under the curve)Work depends on the path taken in the PV-diagram :(a) Wa = W1 to 2 + W2 to 3 (here either P or V constant) Wa (on) = - Pi (Vf - Vi) + 0 > 0(b) Wb = W1 to 2 + W2 to 3 (here either P or V constant) Wb (on) = 0 - Pf (Vf - Vi) > Wa > 0(c) Need explicit form of P versus V but Wc (on) > 0 213123Physics 207: Lecture 22, Pg 9Combinations of Isothermal & Adiabatic Processes 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 casePV = const.All real processes are not.Example: Opening a valve between two chambers, one with a gas and one with a vacuum.Isothermal PV= const.=nRTPhysics 207: Lecture 22, Pg 10Isothermal processesWork done when PV = nRT = constant P = nRT / Vfinalinitial)curveunder area( dVpWfifiVVVV/ nRT/ nRT VdVVdVW)/VV( nRTifnW Physics 207: Lecture 22, Pg 11First Law of ThermodynamicsAll engines employ a thermodynamic cycleW = ± (area under each pV curve)Wcycle = area shaded in turquoiseWatch sign of the work!Physics 207: Lecture 22, Pg 12Work, Heat & ThemodynamicsSomething in common: a thermodynamic cycle with work and heatPhysics 207: Lecture 22, Pg 13Q : Latent heat and specific 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 / °C kg )Q = M c ΔTMolar specific heat C of a substance is the energy required to raise the temperature of 1 mol by 1 K.Q = n C ΔTIf a phase transition involved then the heat
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