Aljalal-Phys.102-27 March 2004-Ch21-page 1 Chapter 21 Entropy and the Second Law of hermodynamics
Aljalal-Phys.102-27 March 2004-Ch21-page 2 21-1 Some One-Way Processes Egg Ok Irreversible process Egg Wrong way Why not? Does not violate conservation of energy!
Aljalal-Phys.102-27 March 2004-Ch21-page 3 21-1 Some One-Way Processes Irreversible process A 30 0 C B 50 0 C ok A 40 0 C B 40 0 C A 40 0 C B 40 0 C Wrong way A 30 0 C B 50 0 C Why not? Does not violate conservation of energy
Aljalal-Phys.102-27 March 2004-Ch21-page 4 21-1 Some One-Way Processes he direction of the process is setup by a property called entropy S If an irreversible process occurs in a closed system, the entropy S of the system always increases, it never decreases. Some times, the change in entropy is called the arrow of time
Aljalal-Phys.102-27 March 2004-Ch21-page 5 21-1 Some One-Way Processes Entropy decreases Process does not occur Egg Wrong way
Aljalal-Phys.102-27 March 2004-Ch21-page 6 21-2 Change in Entropy Change in entropy Final state S = S -S = f i f i dq Initial state Heat transfer to the system For reversible processes emperature In kelvins Entropy is measured in J/K
Aljalal-Phys.102-27 March 2004-Ch21-page 7 21-2 Change in Entropy S = f i dq For reversible processes If the change in temperature (in kelvins) is small compared to the heat transferred to the system f dq 1 S = i S Q avg avg f i dq Average temperature
Aljalal-Phys.102-27 March 2004-Ch21-page 8 21-2 Change in Entropy Checkpoint 1
Aljalal-Phys.102-27 March 2004-Ch21-page 9 21-2 Change in Entropy Entropy is a state property. he change in entropy between state i and f depends only on these states and not on the way the system takes from one state to the other. Pressure i Same S A B f Volume State properties: Pressure Volume emperature Internal energy Entropy Work and heat are not state properties. hey depend on the path
Aljalal-Phys.102-27 March 2004-Ch21-page 10 21-2 Change in Entropy Example: he change in entropy of a free expansion process Free expansion Valve Valve Gas Vacuum P = 0 Gas Gas Initial Final Insulation Pressure i f We do not know the volume and pressure at points between the initial and final states Volume
Aljalal-Phys.102-27 March 2004-Ch21-page 11 21-2 Change in Entropy Example: he change in entropy of a free expansion process Since entropy is a state property, we will get the same change in entropy, if we replace the free expansion process (irreversible) with a reversible process that connects the initial and final states Isotherm Pressure i f In a free expansion process, f = i We will replace the free expansion process with an isothermal process Volume
Aljalal-Phys.102-27 March 2004-Ch21-page 12 21-2 Change in Entropy Example: he change in entropy of a free expansion process f Isotherm f dq 1 Q S = = dq = i i Vf For isothermal procss, Q = W = nr ln V i Vf S = nr ln Weight V (Lead shot) i Piston Pressure i f Gas hermal Reservoir at Insulating cylinder Volume
Aljalal-Phys.102-27 March 2004-Ch21-page 13 21-2 Change in Entropy o find the entropy change for an irreversible process occurring in a closed system, replace that process with any reversible process that connects the same initial and final states. Calculate the entropy change for this reversible process using S = f i dq
Aljalal-Phys.102-27 March 2004-Ch21-page 14 21-2 Change in Entropy For an ideal gas S V = nr ln f + nc ln v Vi f i
Aljalal-Phys.102-27 March 2004-Ch21-page 15 21-2 Change in Entropy Derivation of S For small changes V f = nr ln + nc v ln Vi E = Q - W de int int = dq - dw f i dq = dw + de int dq = pdv + ncvd f f f dq pdv ncvd S = = + i i i f f f dq nrdv ncvd S = = + V i i i Vf f S = nr ln + nc v ln V i pv=nr i
Aljalal-Phys.102-27 March 2004-Ch21-page 16 21-2 Change in Entropy For a solid or liquid substance for which the temperature changes by (no phase transition) Specific heat S = m c ln f i S = C ln f Heat capacity i
Aljalal-Phys.102-27 March 2004-Ch21-page 17 21-2 Change in Entropy Derivation of S = m c ln f i f f f dq mcd d S = = = mc i i i S = m c ln f i
Aljalal-Phys.102-27 March 2004-Ch21-page 18 21-2 Change in Entropy For a solid or liquid substance that undergoes phase transition at temperature S = L Heat of transformation m example : liquid to gas S = L F m example : gas to liquid S = -- LF m
Aljalal-Phys.102-27 March 2004-Ch21-page 19 21-2 Change in Entropy Derivation of S = L m f i f dq 1 Q L m S = = dq = = i
Aljalal-Phys.102-27 March 2004-Ch21-page 20 21-2 Change in Entropy Sample Problem 21-1 Free expansion Valve Valve Gas Vacuum P = 0 Gas Gas Initial one mole of Nitrogen gas he change in entropy? Insulation Final V f = 2 V i For an ideal gas Vf f S = nr ln + nc v ln Vi i For free expansion f = i 2Vi S = nr ln = (1 mol)(8.31 J/mol K) ln 2 = 5.76 J/K V i
Aljalal-Phys.102-27 March 2004-Ch21-page 21 21-2 Change in Entropy Sample Problem 21-2 Initial Final equilibrium L 20 0 C R 60 0 C L 40 0 C B 40 0 C Copper m = 1.5 Kg Copper m = 1.5 Kg he change in entropy of the two-block system? S = mc ln = (1.5kg)386J/kg K) ln f L R Li = mc ln f S = (1.5kg)386J/kg K) ln L Ri S= S + S = 2.4 J/K R 40+273 +27 3 =-35. 86J/K 20 40+273 +27 =38. 23J/ K 60 3
Aljalal-Phys.102-27 March 2004-Ch21-page 22 21-2 Change in Entropy Checkpoint 2
Aljalal-Phys.102-27 March 2004-Ch21-page 23 21-3 he Second Law of hermodynamics Second Law of hermodynamics If a process occurs in a closed system, the entropy of the system increases for irreversible processes and remains constant for reversible processes. It never decreases closed system S > 0 irreversible processes S = 0 reversible processes
Aljalal-Phys.102-27 March 2004-Ch21-page 24 21-4 he Second Law of hermodynamics In real world, almost all processes are irreversible. Processes in which the system s entropy remains constant are always idealization.
Aljalal-Phys.102-27 March 2004-Ch21-page 25 21-4 Entropy in the Real World: Engines High-emperature reservoir H Q H Heat engine W Working substance Q L Steam engine Water-gas L Low-emperature reservoir Car engine Gasoline-air
Aljalal-Phys.102-27 March 2004-Ch21-page 26 21-4 Entropy in the Real World: Engines Ideal engine In an ideal engine all processes are reversible and no wasteful energy transfer occur due, say, friction and turbulence Carnot Engine. Pronounced Car-no Carnot engine is an ideal engine. Carnot engine is the best engine in using energy as heat to do useful work. You can not build more efficient heat engine than a Carnot engine.
Aljalal-Phys.102-27 March 2004-Ch21-page 27 21-4 Entropy in the Real World: Engines Carnot Engine Weight (Lead shot) Piston Pressure a Start at state a Gas hermal Reservoir at H Insulating cylinder H Volume L
Aljalal-Phys.102-27 March 2004-Ch21-page 28 21-4 Entropy in the Real World: Engines Remove Shot Carnot Engine Piston Stroke a b isothermal Gas Insulating cylinder Pressure a Q H hermal Reservoir at H b H Volume L
Aljalal-Phys.102-27 March 2004-Ch21-page 29 21-4 Entropy in the Real World: Engines Carnot Engine Remove Shot Piston Stroke b c adiabatic Gas Insulating cylinder Pressure a b c H Volume L
Aljalal-Phys.102-27 March 2004-Ch21-page 30 21-4 Entropy in the Real World: Engines Carnot Engine Put shot Piston Stroke c d isothermal Gas Insulating cylinder Pressure a hermal Reservoir at L d b c H Q L Volume L
Aljalal-Phys.102-27 March 2004-Ch21-page 31 21-4 Entropy in the Real World: Engines Carnot Engine Put shot Piston Stroke d a adiabatic Gas Insulating cylinder Pressure a d b c H Volume L
Aljalal-Phys.102-27 March 2004-Ch21-page 32 21-4 Entropy in the Real World: Engines Pressure Entropy change for a Cronot engine a P-V diagram Q H emperature H L a d Q H Q L -S diagram b c Entropy d b c H Q L Volume L
Aljalal-Phys.102-27 March 2004-Ch21-page 33 21-4 Entropy in the Real World: Engines Work done by a Cronot engine during a cycle First law of hermodynamics E int = Q - W For a cyclic process E int = 0 Pressure a Q H Area inside 0 = Q - W W = Q W = Q H + Q L W = Q H - Q L d b c H Q L Volume L
Aljalal-Phys.102-27 March 2004-Ch21-page 34 21-4 Entropy in the Real World: Engines Entropy change of a Cronot engine per cycle S = S H + S L QH S= - H Q L L Heat is removed emperature H L a d Q L Q H b c Entropy For a cyclic process S = 0 H 0 = - Q H H Q = H Q L L Q L L
Aljalal-Phys.102-27 March 2004-Ch21-page 35 21-4 Entropy in the Real World: Engines Entropy change of a Cronot engine per cycle emperature H L a d Q H Q L b c Q H H = Q L L Entropy Since H > L, more energy is extracted from the high temperature reservoir than delivered to the lowtemperature reservoir H > L Q H > Q L
Aljalal-Phys.102-27 March 2004-Ch21-page 36 21-4 Entropy in the Real World: Engines hermal efficiency of an engine energy we get W ε= = any engine energy we pay for Q H hermal efficiency of a Carnot engine Q - Q H L εc = Q H Q L = 1- L Q H = 1- H L εc= 1- Carnot engine H
Aljalal-Phys.102-27 March 2004-Ch21-page 37 21-4 Entropy in the Real World: Engines hermal efficiency of a Carnot engine L εc= 1- ε C = 1 when L = 0 K or H =, H which are impossible conditions No series of processes are possible whose sole result is the transfer of energy as heat from thermal reservoir and the complete conversion of this energy to work H here are no perfect engines Q H Perfect engine W
Aljalal-Phys.102-27 March 2004-Ch21-page 38 21-4 Entropy in the Real World: Engines hermal efficiency of a Carnot engine L εc= 1- Always less than unity (less than 100%) H No real engine can have efficiency greater than ε C Real engines have efficiencies less than ε C because the processes that form their cycles are irreversible If your car were powered by a Carnot engine, it would have an efficiency of bout 55%; its actual efficiency is probably about 25%.
Aljalal-Phys.102-27 March 2004-Ch21-page 39 21-4 Entropy in the Real World: Engines Stirling engine It is an ideal engine Pressure Q a Q H It has a smaller efficiency than that of a Carnot engine operating between the same temperatures ε < ε S C d b Q L c Q H L V a V b Volume
Aljalal-Phys.102-27 March 2004-Ch21-page 40 21-4 Entropy in the Real World: Engines Checkpoint 3 400 ε C =1- =0.2 500 600 ε C =1- =0.25 800 400 ε C =1- =0.33 600
Aljalal-Phys.102-27 March 2004-Ch21-page 41 21-4 Entropy in the Real World: Engines Sample Problem 21-3 Carnot engine H = 850 K L = 300 K W = 1200 J per cycle Each cycle takes 0.25 s What is the efficiency of this engine? 300 ε C =1- = 0.65 = 65% 850 What is the average power of this engine? W 1200 J P= = =4.8KW t 0.25 s
Aljalal-Phys.102-27 March 2004-Ch21-page 42 21-4 Entropy in the Real World: Engines Sample Problem 21-3 How much energy Q H is extracted as heat form the high-temperature reservoir every cycle? W ε= Q H W 1200 J Q H = = =1855 J ε 0.647 How much energy Q L is delivered as heat to the low-temperature reservoir every cycle? W = Q H - Q L QL = Q H - W = 1855 J 1200 J = 655 J
Aljalal-Phys.102-27 March 2004-Ch21-page 43 21-4 Entropy in the Real World: Engines Sample Problem 21-3 What is the entropy change of the working substance for the energy transfer to it from the high-temperature reservoir? QH 1855 J S= H = = 2.18 J/K 850 K H What is the entropy change of the working substance for the energy transfer to it from the low-temperature reservoir? QL S H= = L -655 J 300 K = -2.18 J/K
Aljalal-Phys.102-27 March 2004-Ch21-page 44 21-4 Entropy in the Real World: Engines Sample Problem 21-4 Is it possible for an engine operated between the boiling and freezing points of water to have an efficiency of 75%? L (0+237)K ε=1- =1- = 0.27=27% (100+273) K H It can not be
Aljalal-Phys.102-27 March 2004-Ch21-page 45 21-4 Entropy in the Real World: Refrigerator Refrigerator High-emperature reservoir H High-emperature reservoir H Q H Q H Refrigerator W Heat engine W Q L Q L L L Low-emperature reservoir Low-emperature reservoir
Aljalal-Phys.102-27 March 2004-Ch21-page 46 21-4 Entropy in the Real World: Refrigerator Ideal refrigerator In an ideal refrigerator, all processes are reversible and no wasteful energy transfer occur due, say, friction and turbulence Coefficient of performance We want to extract as much energy Q L as possible from the low-temperature reservoir (what we want) for the least amount of work W (what we pay) Coefficient of performance What we want Q L K= = What we pay W
Aljalal-Phys.102-27 March 2004-Ch21-page 47 21-4 Entropy in the Real World: Refrigerator Coefficient of performance What we want Q L K= = What we pay W For typical room air conditioner K 2.5 For typical household refrigerator K 5 A Carnot refrigerator operates in the reverse of the Carnot engine QH QL W = Q H - Q L = Q L Q L L K C= = = W Q - Q - H H L H L L L K C= - H L Coefficient of performance of a Carnot refrigerator
Aljalal-Phys.102-27 March 2004-Ch21-page 48 21-4 Entropy in the Real World: Refrigerator L K C= - H L Coefficient of performance of a Carnot refrigerator he value of K C is higher the closer the temperatures of the two reservoirs to each other. For H > L, K C > 1.
Aljalal-Phys.102-27 March 2004-Ch21-page 49 21-4 Entropy in the Real World: Refrigerator Why perfect refrigerators are impossible? H A perfect refrigerator transfer heat Q form a cold reservoir to a warm reservoir without the need for work Q Perfect Refrigerator Q L
Aljalal-Phys.102-27 March 2004-Ch21-page 50 21-4 Entropy in the Real World: Refrigerator Why perfect refrigerators are impossible? Let us assume it is possible to have a perfect refrigerator Q S= - H Q L Since H > L S < 0 his violates the second law of thermodynamics. he change in entropy for the closed system (refrigerator + reservoirs) can not be negative Perfect refrigerators do not exist Perfect refrigerators violate the second law of thermodynamics
Aljalal-Phys.102-27 March 2004-Ch21-page 51 21-4 Entropy in the Real World: Refrigerator the second law of thermodynamics closed system S > 0 irreversible processes S = 0 reversible processes Another formulation of the second law of thermodynamics No series of processes are possible whose sole result is the transfer of energy as heat from a reservoir at a given temperature to a reservoir at a higher temperature
Aljalal-Phys.102-27 March 2004-Ch21-page 52 21-4 Entropy in the Real World: Refrigerator Checkpoint 4 Let the temperature change = d +d L a- K C= H -( L +d) 1 - d L b- K C= H -( L - d) L c- K C= + H d - L L d- K C= 2 - H d - L 4 3