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## PHYSICS 231 INTRODUCTORY PHYSICS I

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**PHYSICS 231INTRODUCTORY PHYSICS I**Lecture 19**Last Lecture**• First Law of Thermodynamics • Work done by/on a gas**V**V V V Some Vocabulary P • Isobaric • P = constant • Isovolumetric • V = constant • W = 0 • Isothermal • T = constant • U = 0 (ideal gas) • Adiabatic • Q = 0 P P P**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 (work done by gas) in closed cycles**Example 12.8a**C P 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. B A V > < =**Example 12.8b**C P In going from A to B to C, the change of the internal energy of the gas is _______ 0. B A V > < =**Example 12.8c**C D P In going from A to B to C, the amount of heat added to the gas is _______ 0. B A V > < =**Example 12.8d**C D P In going from A to B to C to D to A, the work done BY the gas is _______ 0. B A V > < =**Example 12.8e**C D P In going from A to B to C to D to A, the change of the internal energy of the gas is _______ 0. B A V > < =**Example 12.8f**C D P In going from A to B to C to D to A, the heat added to the gas is _______ 0. B A V > < =**Example 12.7**Consider a monotonic ideal gas. a) What work was done by the gas from A to B? b) What heat was added to the gas between A and B? c) What work was done by the gas from B to C? d) What heat was added to the gas between B and C? e) What work was done by the gas from C to A? f) What heat was added to the gas from C to A? P (kPa) A 75 20,000 J 50 20,000 J B 25 -10,000 J C V (m3) -25,000 J 0.2 0.4 0.6 0 15,000 J**Example 12.7 (Continued)**|Qin| P (kPa) g) What was total work done by gas in cycle? h) What was total heat added to gas in cycle? A WAB + WBC + WCA = 10,000 J B C QAB + QBC + QCA = 10,000 J V (m3) This does NOT mean that the engine is 100% efficient! |Qout| Exhaust!!! |Qin|= QAB + QCA= 35,000 J |Qout| = |QBC| = 25,000 J Weng = |Qin|-|Qout|**Qhot**engine W Qcold Heat Engines • Described by a cycle with: Qhot= heat that flows into engine from source at ThotQcold= heat exhausted from engine at lower temperature, Tcold W= work done by engine • Efficiency is defined: using**2nd Law of Thermodynamics(version 1)**The most efficient engine is the Carnot Engine (an idealized engine), for which: No heat engine can be 100% efficient (T in Kelvin) In practice, we always have**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**Qhot**fridge W Qcold Refrigerators • Just a heat engine run in reverse! • Pull Qcold from fridge • Exhaust Qhot to outside Coefficient of Performance: Most efficient is Carnot refrigerator: Note: Highest COP for small T differences**Qhot**heatpump W Qcold Heat Pumps • Same as refrigerator, except • Pull Qcold from environment • Exhaust Qhot to inside of house Coefficient of Performance: Again, most efficient is Carnot: Like Refrigerator: Best performance for small DT**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.**Entropy**• Measure of Disorder of the system(randomness, ignorance) • S = kBlog(N) N = # of possible arrangements for fixed E and Q Relative probabilities for 12 molecules to arrange on two halves of container.**2nd Law of Thermodynamics(version 2)**On a macroscopic level, one finds that adding heat raises entropy: The Total Entropy of the Universe can never decrease. 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:DS = Q/TB - Q/TA • This can only occur if DS > 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:**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