PHY 182 1st Edition Lecture 5Outline of Last Lecture I. What is Heat?II. The First Law of ThermodynamicsIII. Thermal Properties of MatterOutline of Current Lecture I. Specific Heat of GasesII. Introduction to Adiabatic ProcessesIII. How to Solve Calorimetry ProblemsCurrent LectureSpecific Heat of Gases- The heat required to change the temperature of a gas depends on the process by which the gas changes state.- The two specific heats for gases that we will be dealing with are the molar specific heat for an isobaric (constant pressure) process and an isochoric (constant volume) process.- In a gas process where neither pressure nor volume nor volume are held constant, heat and change in temperature are not directly related, you must use Q=ΔEth-W to solve for heat.- A gas's molar specific heats depend on if it is a monatomic or diatomic gas. In both cases, the difference between the two specific heats is always the same, 8.3, which is also equal to R (the universal gas constant).- To find the change in the thermal energy of a gas that goes through an ideal-gas process,you can use the equation ΔEth=nCvΔT (Cv: molar specific heat for constant volume process).- The heat added to or removed from a system during a process depends on the PV diagram path.Introduction to Adiabatic Processes- The work done during an adiabatic process is given by the equation W=nCvΔT.These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.- If a gas is compressed during an adiabatic process, its change in temperature will be positive. If a gas is expanded, its change in temperature will be negative.How to Solve Calorimetry (An example)- 0.3 kilograms of coffee at 70°C is poured into a 0.12 kilogram aluminum cup that is at 20°C. What is the equilibrium temperature?- Assume that the coffee has a specific heat of 4190 J/kgK (which is actually the specific heat of water). Also look up the specific heat for aluminum; it is approximately 910 J/kgK.- Write the Q expressions for the coffee and the cup.Qcup= mcΔT= (.12)(Tfinal-20)(910)Qcoffee= mcΔT= (.3)(Tfinal-70)(4190)- Next, set the two expressions equal to each other then solve for TfinalYou will find that Tfinal is equal to 66°C. This is the equilibrium
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