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PHYSICS 231 INTRODUCTORY PHYSICS I Lecture 19 Last Lecture • First Law of Thermodynamics Q U W by the gas • Work done by/on a gas W by the gas PV W on the gas Some Vocabulary P • Isobaric V • P = constant • Isovolumetric • V = constant • W = 0 • Isothermal • T = constant • U = 0 (ideal gas) P V P V • Adiabatic • Q = 0 P V P-V Diagrams P Path moves to right: Wby the gas = Area under curve P V Path moves to left: Wby the gas = - Area under curve V (Won the gas = - Wby the gas) Work from closed cycles Clockwise cycle: WA->B->A= Area Counterclockwise cycle: WA->B->A= -Area U 0 in closed cycles (work done by gas) Example 12.8a Consider an ideal gas undergoing the trajectory through the PV diagram. In going from A to B to C, the work done BY the gas is _______ 0. C P A B V a) > b) < c) = Example 12.8b In going from A to B to C, the change of the internal energy of the gas is _______ 0. C P A B V a) > b) < c) = Example 12.8c In going from A to B to C, the amount of heat added to the gas is _______ 0. C D P A B V a) > b) < c) = Example 12.8d In going from A to B to C to D to A, the work done BY the gas is _______ 0. C D P A B V a) > b) < c) = Example 12.8e In going from A to B to C to D to A, the change of the internal energy of the gas is _______ 0. C D P A B V a) > b) < c) = Example 12.8f In going from A to B to C to D to A, the heat added to the gas is _______ 0. C D P A B V a) > b) < c) = Consider a monotonic ideal gas. Example 12.7 a) What work was done by P (kPa) the gas from A to B? A 75 20,000 J b) What heat was added to the gas between A and B? 50 20,000 J c) What work was done by the gas from B to C? 25 -10,000 J C d) What heat was added to the gas between B and C? -25,000 J 0.2 e) What work was done by the gas from C to A? 0 f) What heat was added to the gas from C to A? 15,000 J B V (m3) 0.4 0.6 Example 12.7 (Continued) g) What was total work done by gas in cycle? P (kPa) |Qin| A WAB + WBC + WCA = 10,000 J h) What was total heat added to gas in cycle? C B QAB + QBC + QCA = 10,000 J This does NOT mean that the engine is 100% efficient! |Qin| = QAB + QCA= 35,000 J |Qout| = |QBC| = 25,000 J V (m3) |Qout| Exhaust!!! Weng = |Qin|-|Qout| Heat Engines • Described by a cycle with: Qhot= heat that flows into engine from source at Thot Qcold= heat exhausted from engine at lower temperature, Tcold W= work done by engine Qhot engine W Qcold • Efficiency is defined: engine: Qhot Qcold W Qcold e 1 Qhot Qhot Qhot using W Qhot Qcold 2nd Law of Thermodynamics (version 1) No heat engine can be 100% efficient The most efficient engine is the Carnot Engine (an idealized engine), for which: Qcold Tcold Qhot Thot eCarnot W Qcold Tcold 1 1 Qhot Qhot Thot (T in Kelvin) In practice, we always have e eCarnot Carnot Cycle Example 12.9 An ideal engine (Carnot) is rated at 50% efficiency when it is able to exhaust heat at a temperature of 20 ºC. If the exhaust temperature is lowered to -30 ºC, what is the new efficiency. e = 0.585 Refrigerators Just a heat engine run in reverse! • Pull Qcold from fridge • Exhaust Qhot to outside Coefficient of Performance: fridge Qcold COP(cooling) W W Qcold Most efficient is Carnot refrigerator: COP(cooling) COPCarnot Qhot Tcold Thot Tcold Note: Highest COP for small T differences Heat Pumps Same as refrigerator, except • Pull Qcold from environment • Exhaust Qhot to inside of house Coefficient of Performance: Qhot COP(heating) W Qhot heat pump Qcold Again, most efficient is Carnot: COP(heat) COPCarnot W Thot Thot Tcold Like Refrigerator: Best performance for small T Example 12.10 A modern gas furnace can work at practically 100% efficiency, i.e., 100% of the heat from burning the gas is converted into heat for the home. Assume that a heat pump works at 50% of the efficiency of an ideal heat pump. If electricity costs 3 times as much per kw-hr as gas, for what range of outside temperatures is it advantageous to use a heat pump? Assume Tinside = 295 ºK. 5 T 295 245.8K -27 C 6 Entropy • Measure of Disorder of the system (randomness, ignorance) • S = kBlog(N) N = # of possible arrangements for fixed E and Q 1000 900 800 700 600 500 400 300 200 100 0 (0,12) (1,11) (2,10) (3,9) (4,8) (5,7) (6,6) (7,5) (8,4) (9,3) (10,2) (11,1) (12,0) Relative probabilities for 12 molecules to arrange on two halves of container. 2nd Law of Thermodynamics (version 2) The Total Entropy of the Universe can never decrease. On a macroscopic level, one finds that adding heat raises entropy: S Q / T Defines temperature in Kelvin! Why does Q flow from hot to cold? • Consider two systems, one with TA and one with TB • Allow Q > 0 to flow from TA to TB • Entropy changes by: S = Q/TB - Q/TA • This can only occur if S > 0, requiring TA > TB. • System will achieve more randomness by exchanging heat until TB = TA Carnot Engine Carnot cycle is most efficient possible, because the total entropy change is zero. It is a “reversible process”. For real engines: Qcold Qhot S Senvironment 0 Tcold Thot W Qcold Tcold e 1 1 eCarnot Qhot Qhot Thot Example 12.11a An engine does an amount of work W, and exhausts heat at a temperature of 50 degrees C. The chemical energy contained in the fuel must be greater than, and not equal to, W. a) True b) False Example 12.11b A locomotive is powered by a large engine that exhausts heat into a large heat exchanger that stays close to the temperature of the atmosphere. The engine should be more efficient on a very cold day than on a warm day. a) True b) False Example 12.11c An air conditioner uses an amount of electrical energy U to cool a home. The amount of heat removed from the home must be less than or equal to U. a) True b) False Example 12.11d A heat pump uses an amount of electrical energy U to heat a home. The amount of heat added to a home must be less than or equal to U. a) True b) False