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**Unformatted text preview: **Heat Engines and Second Law of Thermodynamics Chapter 22 Heat Engine • Energy is transferred from a source at a high temperature (Qh) • Work is done by the engine (Weng) • Energy is expelled to a source at a lower temperature (Qc) Cyclical process: ΔEint = 0 Its iniFal and ﬁnal internal energies are the same (P,V,T the same) 1st Law: since ΔE =0, Qnet = Weng The work done by the engine equals the net energy absorbed by the engine The work in a cyclic process is equal to the area enclosed by the curve of the PV diagram Eﬃciency of a heat engine In pracFce, when we build an engine, we want to know how much heat (e.g. chemical energy from gasoline) we have to provide to get a certain amount of useful work from the engine. Thermal eﬃciency of a heat engine is the ra'o of the network done by the engine during one cycle to the energy absorbed at the higher temperature during the cycle: ε=
W Qh − Qc Q = = 1− c QH Qh Qh That is, “what you get out compared to you put in” In principle, you might think that one could build a perfect engine, one that has eﬃciency 1 (or 100%). Such an engine would convert all heat received into useful work and no heat is wasted. All aVempts to build such an engine have failed and we now know that it is impossible, even in principle, to build such an engine. Second law of thermodynamics Second law of thermodynamics, Kelvin’s statement: “ It is impossible to built an engine whose sole purpose is to extract heat from a reservoir and convert it completely into work”. Let’s compare the ﬁrst and the second laws. The ﬁrst law
Energy is conserved: One cannot get more work from an engine than the amount of heat that is put in. (if you want Work done, you got to pay with another form of energy: Internal or Heat) “There is no free lunch” The second law: In fact, you can’t even “break even”. Some heat has to be expelled as waste heat. “You don’t even get what you paid for!” Carnot engine
the most eﬃcient engine If a perfect engine cannot be built, what would be an ideal engine, i.e. a most eﬃcient engine? Need to construct a cycle that only expels or receives heat at certain Fmes at ﬁxed temperatures. Use adiabaFc and isothermal transformaFons: a b: isothermal expansion, temperature Th b c: adiabaFc expansion c d: isothermal compression, temperature Tc d a: adiabaFc compression a b c d a This hypotheFcal ideal engine is called the Carnot engine. Th Th Tc The Carnot engine (contd…) A B: isothermal expansion, temperature Th ΔEint = 0 ⇒ Qh = W = nRTh ln VB > 0.....(1)
VA B C: adiabaFc expansion Q = 0, Th (VBγ −1 ) = Tc (VCγ −1 )....( 2) C D: isothermal compression, temperature Tc ΔEint = 0 ⇒ Qc = W = − nRTc ln
γ γ Q = 0, Tc (VD −1 ) = Th (V A −1 )....( 4) VC < 0....( 3) VD D A: adiabaFc compression From (2) & (4): Th ⎛ VC ⎞ = Tc ⎜ VB ⎟ ⎝⎠
γ −1 ⎛V ⎞ =⎜ D⎟ ⎝V ⎠
A γ −1 ⇒ VC VB = VD VA
VB < 0....(5 ) VA With this, (3) becomes: Divide (5) by (1): Therefore, Qc Tc = Qh Th ΔEint = 0 ⇒ Qc = W = − nRTc ln We know, eﬃciency: ε = 1 − Qc Qh ε = 1− Tc Th It can be shown that this is the maximum limit of eﬃciency any heat engine can achieve. In fact, any real engine has eﬃciency signiﬁcantly lower that the value given by this formula. Example 1 A Carnot engine, with eﬃciency = 0.5 is made to operate between a max temperature of 500K, and an unknown minimum temperature. Find the temperature of the lower reservoir. ε = 1−
Tc → Th (1 − ε ) = Tc → Tc = 500 K (1 − 0.5) = 250 K Th Example 22.1 (textbook) An engine transfers 2.00 × 103 J of energy from a hot reservoir during a cycle and transfers 1.50 × 103 J as exhaust to a cold reservoir. a) Find the eﬃciency of the engine. b) How much work does this engine do in one cycle? (Ans. 25%, 5.0 × 103 J) Reversible and irreversible processes • A reversible process is one carried out inﬁnitely slowly, so that the process can be considered as a series of equilibrium states, and the whole process could be done in reverse with no change in magnitude of the work done or heat exchanged – And one for which the system can be returned to its iniFal state along the same path – A Carnot engine is a reversible engine – Of course this cannot be done since it would take an inﬁnite amount of Fme • An irreversible process does not meet these requirements – Most natural processes are irreversible – Reversible process are idealiza@ons, but some real processes are good approximaFons Entropy We have seen that the zeroth law of thermodynamics gives the concept of temperature and the ﬁrst law gives the concepts of internal energy. Both temperature and internal energy are state properFes. The second law gives us another state property called the entropy (usually denote by S). Consider an inﬁnitesimal process in which a system changes from one equilibrium state to another. If dQr is the amount of heat transferred when the system follows a reversible path between the states, then change in the entropy dS is equal to this amount of heat for the reversible process divided by the absolute temperature of the system: dSr = Qr T In a ﬁnite process in which temperature may change, the change in entropy is given by: ΔS = ∫ f i dS = ∫ f i Qr T Entropy (contd…) For Carnot process, ΔS == ∫A dQ + ∫C dQ = Qh − Qc B D Th Tc Th Tc But we have found that Qc Tc = , which Qh Th can be wriVen as Qc Qh =c Tc Th Therefore for the Carnot process, ΔS = 0. Change in entropy for reversible process: 1. Isothermal process: 2. AdiabaFc process 3. Cyclic process ΔS = ΔQr/T ΔS = 0 ΔS = 0 (since ΔQr = 0) (again ΔQr = 0) entropy increases entropy decreases Note that : energy absorbed ΔQ > 0 energy expelled ΔQ < 0 Example Ice has a latent heat of fusion Lf=3.33E5 J/K melts at a temperature T=0oC . (a) Calculate the change in entropy (ΔS) of this substance when a mass m=5kg of the substance melts. If the ice is held at constant volume, and its temperature is raised to 50oC, what is the change in entropy of the system? ⎛ Tf ⎞ ⎛ 273 + 50 ⎞ 5g ΔS = nCv ln ⎜ ⎟ = ( ) ⋅ ( 27 J / mol ⋅ K ) ⋅ ln ⎜ 18 g / mol ⎝ 273 ⎟ ⎠ ⎝ Ti ⎠ ⎛ 323 ⎞ ΔS = 0.278mol ⋅ 27( J / mol ⋅ K ) ⋅ ln ⎜ = 1.26 J / K ⎝ 273 ⎟ ⎠ Entropy – a measure of disorder When a boVle of perfume is opened, a person standing on the other side of the room can smell the aroma soon ajer the boVle is opened. (The molecules from the boVle eventually and randomly make their way across the room to the person) How come the molecules from the perfume do not all join together and make their way back into the boVle? Why do they seem to spread in a disorderly fashion across the room? • • • • • The resoluFon to this quesFon is that the molecules are capable of doing going back into the boVle, (It violates no fundamental laws) but the probability of it happening is miniscule. The explanaFon is Entropy. Entropy always tends to go towards a more disorderly state of the system Entropy: a quanFtaFve measure of the disorder of a system Entropy: a measure of the energy in a system or process that is unavailable (“used
up” energy) to do work. In a reversible thermodynamic process, entropy is expressed as the heat absorbed or emiVed divided by the absolute temperature Microstates and macrostates Microstate: a parFcular conﬁguraFon of the individual consFtuents in the system Macrostate: descripFon of the system’s condiFons from a macroscopic point of view. Take four coins and toss them, calling the number of heads and tales that show on each toss. Macrostates 4 heads 3 heads, 1 tale 2 heads, 2 tales 1 head, 3 tales 4 tales Possible Microstates HHHH HHHT, HHTH, HTHH, THHH HHTT, HTHT, THHT, HTTH, THTH, TTHH TTTH, TTHT, THTT, HTTT TTTT Number of Microstates 1 4 6 4 1 16 possible microstates 5 possible macrostates • • • • Each microstate is equally probable. Thus, the number of microstates that give the same macrostate corresponds to the relaFve probability of that state occurring. The macrostate of two heads and two tales is the most probable state in this example. Out of a total of 16 possible microstates, six correspond to tossing two heads and two tales, so the probability of throwing two heads and two tales is 6 out of 16, or 38%. Microstates and Macrostates (contd…) There are a total of 1030 microstates and 100 macrostates possible. For 99 heads and 1 tail, there are 100 microstates possible since each coin could come up tails once. But any macrostate of 99 heads and 1 tail is equivalent. The probability of all heads is ~1 in 1030. NOT a good bet! The probability of 50 heads and 50 tails is 1.0 x 1029/ 1030 = 0.10 or 10%. Therefore as the number of coins increases, we see the probability of obtaining an orderly arrangement of say, 100 heads becomes extremely unlikely. Microstates and Macrostates (contd…) • That was for 100 coins. Can you imagine the numbers when you have 1030 air molecules in a room. • The most probable arrangement for the air molecules is to take up the whole space and move randomly. This is described by the Maxwellian distribuFon. • On the other hand a very orderly arrangement of all the air molecules concentrated in one corner of the room and all moving with the same velocity is extremely unlikely. • This is why perfume molecules do not go back into the boVle. They are allowed to by the ﬁrst law of thermodynamics, but the second law tells us that the probability that they will do so is essenFally zero. The the total number of microstates for a given macrostate is referred to as the thermodynamics probability and is ojen denoted by W. Boltzmann gave a formula for entropy of a system (macrostate) in terms of its microstates: S = kBlnW This equaFon indicates mathemaFcally that entropy is a measure of microscopic disorder. Second law of thermodynamics revisited If we have a closed system, i.e. a system that does not exchange heat or parFcles with the environment (example: engine + heat reservoir), we have in general: ΔS ≥ 0
This is an alternaFve statement of the Second Law of Thermodynamics ...

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