Capters 19 & 20 Heat and te First Law of Termodynamics Te Zerot Law of Termodynamics Te First Law of Termodynamics Termal Processes Te Second Law of Termodynamics Heat Engines and te Carnot Cycle Refrigerators, Air Conditioners, and Heat Pumps
Heat and te First Law of Termodynamics Te Zerot Law of Termodynamics We ave already discussed te zerot law, and include it ere for completeness: If object A is in termal equilibrium wit object C, and object B is separately in termal equilibrium wit object C, ten objects A and B will be in termal equilibrium if tey are placed in termal contact.
Heat and te First Law of Termodynamics Internal Energy: refers to te total energy of all te molecules in a object (or system). Internal energy of an ideal gas depends only on its temperature,t. Termal (Heat) Energy: is tat portion of te internal energy tat canges te temperature of te system. (It is te transfer of energy due to a difference in temperature) Work: is te transfer of energy from one object to anoter by mecanical means. Te First Law of Termodynamics is a statement of te conservation of energy. It states tat te cange in te Internal Energy (U) of a system is equal to te difference between eat () added to a system and te work (W) done by te system. If a system s volume is constant, and eat is added, its internal energy increases.
Heat and te First Law of Termodynamics If a system does work on te external world, and no eat is added, its internal energy decreases. Combining tese gives te First law of Termodynamics. Te cange in a system s internal energy is related to te eat and te work W as follows:
Heat and te First Law of Termodynamics Exercise 1(a): Jogging along a beac, you do 4.3 10 5 J of work and give off 3.8 10 5 J of eat. Wat is te cange in your internal energy? Exercise 1(b): Switcing over to working, you give off 1.2 10 5 J of eat and your internal energy decreases by 2.6 10 5 J. How muc work ave you done wile walking?
19-7 Termal Processes (i) We will assume tat all processes we discuss are quasi-static tey are slow enoug so tat te system is always in equilibrium. (ii) We also assume tey are reversible: For a process to be reversible, it must be possible to return bot te system and its surroundings to exactly te same states tey were in before te process began. An idealized reversible process. (a) Te gas is compressed as te piston is slowly moved down; in order for te temperature to remain constant, eat,, goes from te gas to its surroundings. (b) As te gas expands back to its initial position, it draws eat,, from its surroundings, returning te gas and te surrounding to teir initial states. Te piston is assumed frictionless.
19-7 Termal Processes Work done in olume canges at Constant Pressure (Isobaric Process) Consider a gas at pressure, P, contained in a cylinder fitted wit a movable piston of cross-sectional area, A sown below. Let s calculate te Work done by te gas as it expands against te piston. Te gas exerts a force on te piston, F = PA. Te work done by te gas to move te piston troug a displacement, dx, is: dw F dx PAdx Pd W dw f i Pd P F I d P Figure sows Pressure P versus olume on a P plot for constant pressure process. Work done = Area of saded region W = P( f i ) = P
19-7 Termal Processes Work done at constant olume (Isovolumetric Process) Consider te case of adding eat to a system of gas at constant volume as sown in te figure, causing its Pressure to increase. Since, tere is no displacement of any of te walls, implies no work is done (W = 0). Te P plot sows constant volume process in wic Pressure increases from P i to P f Tus for a constant volume process: W = 0 Ten From First Law of Termodynamics it implies tat: U U W
19-7 Termal Processes Exercise 2: Find te Total Work A gas undergoes te tree-part process as sown in te figure, connecting te states A and B. (i) Find te total work done by te gas during te tree-part process. (ii) Find te work done by te gas if te process connecting A and B were a constant-pressure expansion at 120 kpa.
19-7 Termal Processes Work done at constant Temperature (Isotermal Process) If te temperature is constant, te pressure varies inversely wit te volume. For an ideal gas: P = nrt = constant Te grap sows isoterms for 4 different temperatures Consider an ideal gas contained in a cylinder fitted wit a movable piston of cross sectional area A, allowed to expand from i to f at constant temperature Te grap sows te P plot Since T is constant, U = 0 Tus any eat,, added to te system does work, W, on te surrounding (e.g. moves a piston troug a displacement, x) Work done = Area of saded region
19-7 Termal Processes Te work done by te expending gas is te area under te curve (saded region): For an ideal gas, te work done (area under te curve) can be found by calculus using te ideal gas law, (P = nrt): Work done = Area of saded region dw F dx PAdx P d W dw f i P d F I nrt d nrt f I d nrt ln f i For an Isotermal process: Since T is constant, Cange in Internal Energy, U = 0 Ten from First Law of Termodynamics it implies: U = W = W Tus any eat,, added to te system does work, W, on te surrounding
19-7 Termal Processes Adiabatic Process: An adiabatic process is one in wic no eat flows into or out of te system: = 0. Te adiabatic P- curve is similar to te isotermal one, but is steeper. One way to ensure tat a process is adiabatic is to insulate te system. Adiabatic Compression Anoter way to ensure tat a process is effectively adiabatic is to ave te volume cange occur very quickly. In tis case, eat as no time to flow in or out of te system. Same principle applies to a Diesel Engine: Fuel-air mixture in a cylinder is rapidly compressed by te piston igniting te mixture to run te engine Adiabatic Expansion
18-3 Termal Processes For an Adiabatic Process: No eat enters or leaves te system. Since = 0, Ten from First Law of Termodynamics it implies: U = W U = W Summary of te different types of termal processes:
Exercise 3: Heat Flow 19-7 Termal Processes A cylinder olds 0.50 mol of an ideal gas at a temperature of 310 K. As te gas expands isotermally from an initial volume of 0.31 m 3 to 0.45 m 3, determine te amount of eat tat must be added to te system to maintain a constant temperature. Exercise 4: Adiabatic Compression Wen a certain gas is compressed adiabatically, te amount of work done on it is 640 J. Find te cange in te internal energy of te gas.
20-1 Te Second Law of Termodynamics Te Second Law of Termodynamics: Wen objects of different temperatures are brougt into termal contact, te spontaneous flow of eat is always from te ig temperature object to te low temperature object. Spontaneous eat flow never proceeds in te reverse direction (i.e. from a cold object to a ot object) However, if any of tese processes occurred in reverse, it would not violate te conservation of energy (Te First Law of Termodynamics). A eat engine: is a device tat converts eat energy into mecanical work. A classic example is te steam engine. Fuel eats te water; te vapor expands and does work against te piston; te vapor condenses back into water again and te cycle repeats.
20-2 Heat Engines and te Carnot Cycle All eat engines ave: a ig-temperature reservoir, T, wic supplies eat to te engine. a low-temperature reservoir, T c, were eat is released (Exaust). does mecanical work, W. te engine operates in a cyclic process, i.e. cange in internal energy is zero since it returns to te starting state after every cycle. Te Figure sows a scematic diagram of a Heat Engine
20-2 Heat Engines and te Carnot Cycle Operation of a Heat Engine: A Heat Engine absorbs an amount of eat,, from te ot reservoir, T, during eac cycle. Of tis eat, some appears as useful work, W, and te rest of te eat is given off as waste eat, c, to te cold reservoir, T c. Tus from Conservation of Energy: OR: W W c c Te efficiency, e, of any eat engine is defined as te ratio of te work done, W, to te eat input at te ig temperature, H : e W Te efficiency can also be written: e W c 1 c In order for te engine to run, tere must be a temperature difference; oterwise eat will not be transferred.
20-3 Heat Engines and te Carnot Efficiency Tis is an idealization; no real engine can be perfectly reversible. However, if a system passes slowly from te initial to te final state troug a series of equilibrium states, ten te process is nearly reversible. e W c 1 c From te efficiency formula, For an engine to be 100% efficient implies tat C must be zero. Tat is no eat sould be expelled to te cold reservoir or all eat input must be converted to useful work. Tis is impossible according to te second law of termodynamics Te maximum-efficiency eat engine is described in Carnot s teorem: If an engine operating between two constant-temperature reservoirs is to ave maximum efficiency, it must be an engine in wic all processes are reversible. In addition, all reversible engines operating between te same two temperatures, T H and T C, ave te same efficiency. No real Engine operating between te same two temperatures, T H and T C, can be more efficient tan a Carnot Engine operating between te same two temperatures.
20-3 Carnot Engine and te Carnot Efficiency Te Carnot Engine: Te Carnot Engine makes use of a reversible cycle (Series of reversible process wic takes te system from te initial to te final state troug a series of equilibrium states and returns to te initial state. Te Carnot Engine utilizes Carnot cycle illustrated in te figure c 1. Process AB: Te gas is first expanded isotermally at constant temperature T, wit te addition of eat, from te ot reservoir 2. Process BC: Te gas is expanded Adiabatically, no eat is excanged but te temperature drops to T c 3. Process CD: Te gas is isotermally compressed at constant temperature T c, and eat c flows out to te cold reservoir 4. Process DA: Finally te gas is compressed Adiabatically back to its original state, no eat is excanged but te temperature rises to T A B D C
20-3 Carnot Engine and te Carnot Efficiency For a Carnot Engine: efficiency depends only on te temperatures of te ot and cold reservoirs, T and T c. [Proof: Use Ideal gas law] Terefore te ratio of te temperatures must be te same as te ratio of te transferred eats, and c. Terefore, te Maximum efficiency of a eat engine (Carnot Efficiency) can be written: e max T T T c 1 T T c Te Maximum work a eat engine can do is ten: W max e max 1 T T c If te two reservoirs are at te same temperature, te efficiency is zero; te smaller te ratio of te cold temperature to te ot temperature, te closer te efficiency will be to 1.
18-7 Refrigerators, Air Conditioners, and Heat Pumps Wile eat will flow spontaneously only from a iger temperature to a lower one, it can be made to flow te oter way if work is done on te system. Refrigerators, air conditioners, and eat pumps all use work to transfer eat from a cold object to a ot object. If we compare te eat engine and te refrigerator, we see tat te refrigerator is basically a eat engine running backwards it uses work to extract eat from te cold reservoir (te inside of te refrigerator) and exausts to te kitcen. Note tat: More eat is exausted to te kitcen tan is removed from te refrigerator.
18-7 Refrigerators, Air Conditioners, and Heat Pumps An ideal refrigerator would remove te most eat from te interior of te refrigerator wile requiring te smallest amount of work. Coefficient of Performance, COP: is defined as te ratio eat, C removed from a low-temperature area (inside a refrigerator) to te work done to remove te eat. COP is a measure of te efficiency of a refrigerator Typical refrigerators ave COP values between 2 and 6. Bigger is better!
18-7 Refrigerators, Air Conditioners, and Heat Pumps An Air Conditioner is essentially identical to a refrigerator; te cold reservoir is te interior of te ouse or oter space being cooled, and te ot reservoir is outdoors. Exausting an air conditioner witin te ouse will result in te ouse becoming warmer, just as keeping te refrigerator door open will result in te kitcen becoming warmer. Finally, a Heat Pump is te same as an air conditioner, except wit te reservoirs reversed. Heat is removed from te cold reservoir outside, and exausted into te ouse, keeping it warm. Note tat te work te pump does actually contributes to te desired result (a warmer ouse) in tis case.
Exercise 5: Heat into Work A eat engine wit an efficiency of 24% performs 1250 J of work. Find: (i) te eat absorbed from te ot reservoir, (ii) te eat given off to te cold reservoir, Exercise 6: Car Efficiency A car engine as an efficiency of 20% and discarges 92 kj of eat per second as waste eat from te engine. (i) How muc eat input (J/s) is required from te ot reservoir? (ii) How muc mecanical work (J/s) is performed during operation?
Exercise 7: Carnot Efficiency A eat engine operating at maximum efficiency of 24% as its cold reservoir at a temperature of 295 K. (i) Wat is te temperature of te ot reservoir? (ii) Wat is te maximum work performed if 570 J of eat is rejected? Exercise 6: A Pony Claim An engine manufacturer claims tat an Engine s eat input is 9.0 kj/s at a temperature of 435 K and its eat output is 4.0 kj/s at a temperature of 285 K. Do you believe tese claims?