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• ### Lect31

1. 1. Physics 101: Lecture 31, Pg 1 Physics 101:Physics 101: Lecture 31Lecture 31 Thermodynamics, part 2Thermodynamics, part 2 Review of 1st law of thermodynamics 2nd Law of Thermodynamics Engines and Refrigerators The Carnot Cycle
2. 2. Physics 101: Lecture 31, Pg 2 Quick ReviewQuick Review 1st Law of Thermodynamics: energy conservation Q = ∆U + W Heat flow into (or out of) system Increase (or decrease) in internal energy of system Work done by (or on) system V P U depends only on T (U = 3nRT/2 = 3PV/2) Point on P-V plot completely specifies state of system (PV = nRT) work done is area under curve for complete cycle ∆U=0 ⇒ Q=W
3. 3. Physics 101: Lecture 31, Pg 3 Second Law of ThermodynamicsSecond Law of Thermodynamics Not all processes that are allowed by energy conservation occur in nature. Why ? Example: Stone falls from height h: mgh -> ½ m v2 (just before impact) -> heat (contact with floor) This process is consistent with energy conservation. The reversed process: Stone lying on floor cools down and moves upward to height h, has never been observed in nature, although it is also allowed by energy conservation: Q->1/2 mv2 ->mgh Or: Ice melts but water does not spontaneously freeze, heat flows from hot to cold but never from cold to hot. ⇒ We need a new concept which makes these (reversed) processes highly unlikely.
4. 4. Physics 101: Lecture 31, Pg 4 New concept: Entropy (S)New concept: Entropy (S) A measure of “disorder” or probability of state of a system. A property of a system (=state function, just like P, V, T, U) related to number of different “states” of system Examples of increasing entropy: ice cube melts gases expand into vacuum Change in entropy: ∆S = Q/T (T in K !) SI unit: [J/K] » >0 if heat flows into system (Q>0) » <0 if heat flows out of system (Q<0)
5. 5. Physics 101: Lecture 31, Pg 5 Reversible vs. Irreversible changes in a thermodynamic system:Reversible vs. Irreversible changes in a thermodynamic system: Reversible changes are conceived to be those that would occur very slowly, giving all the molecules in the system time to 'adjust' to new conditions, and all state variables time to adjust while still remaining uniform throughout a system. Theoretically you could imagine stopping at any point and reversing the change slowly, recovering the previous thermodynamic state. Definition given by Fermi (1936), in Thermodynamics: "A transformation is said to be reversible when the successive states of the transformation differ by infinitesimals from equilibrium states.” ∆Srev = 0 Irreversible: Processes in which new entropy is “created”. A system spontaneously changes, or energy is transformed in a way that creates new entropy. This does not allow complete recovery of all aspects of previous thermodynamic states. ∆Sirrev > 0 Processes that happen spontaneously are irreversible.
6. 6. Physics 101: Lecture 31, Pg 6 11stst and 2and 2ndnd Law of Thermodynamics:Law of Thermodynamics: A Perpetuum Mobile (perpetual motion) of 1stA Perpetuum Mobile (perpetual motion) of 1st and 2and 2ndnd kind is impossible.kind is impossible. M.C. Escher “Waterfall” (1961) Or: The energy of the universe is constant, the entropy of the universe seeks to be maximal. R.Clausius (1822-1888) Perpetuum Mobile of 1st kind : A machine that is able to provide useful work without input of external energy (e.g. heat) and without change of the physical or chemical status of its parts does not exist (or a machine that creates energy continuously does not exist). Perpetuum Mobile of 2nd kind: A machine undergoing a cyclic process which does nothing more than convert heat into mechanical (or other) work does not exist.
7. 7. Physics 101: Lecture 31, Pg 7 Second Law of ThermodynamicsSecond Law of Thermodynamics The entropy change (Q/T) of the system+environment ≥ 0 never < 0 order to disorder The entropy of the universe increases whenever an irreversible process occurs. All real processes in nature are irreversible. Consequences: A “disordered” state cannot spontaneously transform into an more “ordered” state. No engine operating between two reservoirs can be more efficient than one that produces zero change in entropy. The latter is called a “Carnot engine” (no real engine can ever be perfectly reversible but Carnot is a useful idealization, since it represents the limiting case) . Heat cannot be transferred spontaneously from cold to hot.
8. 8. Physics 101: Lecture 31, Pg 8 TH TC QH QC W HEAT ENGINE TH TC QH QC W REFRIGERATOR system Engines and Refrigerators System taken in closed cycle ⇒ ∆Usystem = 0 Therefore, net heat absorbed = work done QH - QC = W (engine) QC - QH = -W (refrigerator) energy going into “green blob” = energy leaving “green blob”
9. 9. Physics 101: Lecture 31, Pg 9 TH TC QH QC W HEAT ENGINE The objective: turn heat from hot reservoir (QH) into work The cost: “heat is wasted” 1st Law: QH -QC = W efficiency e ≡ W/QH =W/QH = (QH-QC)/QH = 1-QC/QH
10. 10. Physics 101: Lecture 31, Pg 10 TH TC QH QC W REFRIGERATOR The objective: remove heat from cold reservoir (QC) The cost: work needs to be done 1st Law: QH = W + QC coefficient of performance CPr ≡ QC/W = QC/W = QC/(QH - QC)
11. 11. Physics 101: Lecture 31, Pg 11 TH TC QH QC W HEAT ENGINE The objective: turn heat from hot reservoir into work. The cost: “heat is wasted” 1st Law: QH -QC = W efficiency e ≡ W/QH =W/QH = 1-QC/QH ∆S = QC/TC - QH/TH ≥ 0 ∆S = 0 for Carnot Therefore, QC/QH ≥ TC/ TH QC/QH = TC/ TH for Carnot Therefore e = 1 - QC/QH ≤ 1 - TC/ TH e = 1 - TC/ TH for Carnot => efficiency of a realistic engine can never be larger than eCarnot ! Engines and the 2nd Law
12. 12. Physics 101: Lecture 31, Pg 12 Concept QuestionConcept Question Consider a hypothetical device that takes 1000 J of heat from a hot reservoir at 300K, ejects 200 J of heat to a cold reservoir at 100K, and produces 800 J of work. Does this device violate the first law of thermodynamics ? 1. Yes 2. No This device doesn't violate the first law of thermodynamics because no energy is being created nor destroyed. All the energy is conserved. correct W (800) = Qhot (1000) - Qcold (200) Efficiency = W/Qhot = 800/1000 = 80%
13. 13. Physics 101: Lecture 31, Pg 13 Concept QuestionConcept Question Consider a hypothetical device that takes 1000 J of heat from a hot reservoir at 300K, ejects 200 J of heat to a cold reservoir at 100K, and produces 800 J of work. Does this device violate the second law of thermodynamics ? 1. Yes 2. No . correct W (800) = Qhot (1000) - Qcold (200) Efficiency = W/Qhot = 800/1000 = 80% Max eff = 1 - 100/300 = 67% = eCarnot e > eCarnot is forbidden by second law : ∆ S = ∆SH+∆SC=200/100 J/K– 1000/300 J/K < 0
14. 14. Physics 101: Lecture 31, Pg 14 Concept QuestionConcept Question Consider a hypothetical refrigerator that takes 1000 J of heat from a cold reservoir at 100K and ejects 1200 J of heat to a hot reservoir at 300K. 1. How much work does the refrigerator do? 2. What happens to the entropy of the universe? 3. Does this violate the 2nd law of thermodynamics? Answers: 200 J Decreases yes TH TC QH QC W QC = 1000 J QH = 1200 J Since QC + W = QH, W = 200 J ∆SH = QH/TH = (1200 J) / (300 K) = 4 J/K ∆SC = -QC/TC = (-1000 J) / (100 K) = -10 J/K ∆STOTAL = ∆SH + ∆SC = -6 J/K decreases (violates 2nd law)
15. 15. Physics 101: Lecture 31, Pg 15 Heat Capacities of an Ideal GasHeat Capacities of an Ideal Gas As discussed in Chapter 12, the heat needed to raise the temperature of a solid or liquid is given by: Q=cm ∆T where c is the heat capacity of the material. Gases: Volume and/or pressure change when temperature changes (this effect can be safely neglected in case of solids and liquids). ⇒ Heat capacity of a gas depends on if T changes at constant V, cV, or constant P, cP,: V=const.: ∆U=Q=cV m DT=3/2 n R ∆T => CV=cVm/n = 3/2 R P=const.: ∆U=Q-P∆V=cP m ∆T-n R ∆T=3/2 n R ∆T => CP=cP m/n = 5/2 R CV and CP are the molar specific heat capacities of an ideal monatomic gas.