1. Lecture 27Purdue University, Physics 2201 Lecture 27 Thermodynamics II Physics 220

2. Lecture 27Purdue University, Physics 2202 Overview –First Law of thermodynamics: Energy Conservation Q =  U + W –Heat Engines Efficiency = 1-Q C /Q H –Refrigerators Coefficient of Performance = Q C /(Q H – Q C ) Today –Second law of thermodynamics –Carnot Engine (Sadi Carnot 1796-1832) –Entropy –Disorder

3. Second law of thermodynamics Heat flow spontaneously from a warm object to a colder one. It is not possible for heat to flow spontaneously from a cold object to a warmer one. Lecture 27Purdue University, Physics 2203

4. The world’s best engine? Lecture 27Purdue University, Physics 2204

5. Lecture 26Purdue University, Physics 2205 THTH TCTC QHQH QCQC W HEAT ENGINE The objective: turn heat from hot reservoir into work The cost: “waste heat” 1st Law: Q H -Q C = W Efficiency e  W/Q H =W/Q H = (Q H -Q C )/Q H = 1-Q C /Q H Heat Engine: Efficiency

6. Lecture 26Purdue University, Physics 2206 THTH TCTC QHQH QCQC W HEAT ENGINE THTH TCTC QHQH QCQC W REFRIGERATOR System System taken in closed cycle   U system = 0 l Therefore, net heat absorbed = work done Q H - Q C = W (engine) Q C - Q H = -W (refrigerator) energy into blob = energy leaving blob Engines and Refrigerators

7. Lecture 26Purdue University, Physics 2207 THTH TCTC QHQH QCQC W REFRIGERATOR The objective: remove heat from cold reservoir The cost: work 1st Law: Q H = W + Q C Coefficient of performance K r  Q C /W = Q C /W = Q C /(Q H - Q C ) Refrigerator: Coefficient of Performance

8. Lecture 26Purdue University, Physics 2208 iClicker An ideal heat engine absorbs 36 kJ of heat and exhausts 18 kJ of heat every cycle. What is the efficiency of the engine? Q H -Q C = W Efficiency e  W/Q H A. 1 B. 0.5 C. 2 D. 0.25

9. Lecture 27Purdue University, Physics 2209 Carnot Cycle Idealized Heat Engine –No Friction –Reversible Process Isothermal Expansion Adiabatic Expansion Isothermal Compression Adiabatic Compression Q H all at T H Q C all at T C This is most efficient engine Lecture 279Purdue University, Physics 220

10. Ideal Efficiency For a Carnot engine Q H  T H and Q C  T C e = 1-Q C /Q H = 1 - T C /T H K r  Q C /W = Q C /W = Q C /(Q H - Q C ) = T C /(T H - T C) Carnot engine most efficient since it operates across greatest T difference for given T H and T C For refrigerator: Lecture 2710Purdue University, Physics 220

11. Lecture 27Purdue University, Physics 22011 Engine

12. Real Engines For ideal Carnot engine i.e. Greatest possible efficiency e = 1-Q C /Q H = 1 - T C /T H and Q H /T H = Q C /T C All operating engines are less efficient and Q H /T H < Q C /T C Define Entropy S = Q/T S C > S H Then S out of hot reservoir is less than S into cold reservoir. Lecture 2712Purdue University, Physics 220

13. Lecture 27Purdue University, Physics 22013 Q AT LOW T BRINGS ABOUT MORE DISORDER THAN IT CAUSED AT HIGH T

14. Entropy and Disorder T H Hot T C Cold In thermal conductivity the same heat flows from hot to cold Q H = Q C = Q S C = Q/T C > S H = Q/T H MORE S ADDED TO COLD THAN TAKEN FROM HOT TOTAL S INCREASE Q AT LOW T BRINGS ABOUT MORE DISORDER THAN IT CAUSED AT HIGH T Lecture 2714Purdue University, Physics 220

15. Lecture 27Purdue University, Physics 22015 New Concept: Entropy (S) A measure of “disorder” A property of a system (just like P, V, T, U) –related to number of different “states” of system Examples of increasing entropy: –ice cube melts –gases expand into vacuum Change in entropy:  S = Q/T >0 if heat flows into system (Q>0) <0 if heat flows out of system (Q<0)

16. Lecture 27Purdue University, Physics 22016 Entropy Question Some ice (-5 C) is used to cool a cup of water What happens to the entropy of the ice? A) IncreaseB) SameC) Decreases What happens to the entropy of the water? A) IncreaseB) SameC) Decreases What happens to the total entropy (water+ice)? A) Increase B) Same C) Decreases  S = Q/T  S = Q/T ice – Q/T water Heat enters ice: Q>0 Heat Leaves water: Q<0

17. Lecture 27Purdue University, Physics 22017 The entropy change (Q/T) of the system+ environment  0 –never < 0 –order to disorder Consequences –A “disordered” state cannot spontaneously transform into an “ordered” state –No engine operating between two reservoirs can be more efficient than one that produces 0 change in entropy. This is called a Carnot engine Second Law of Thermodynamics

18. Lecture 27Purdue University, Physics 22018 THTH TCTC QHQH QCQC W HEAT ENGINE The objective: turn heat from hot reservoir into work The cost: “waste heat” 1st Law: Q H -Q C = W Efficiency e  W/Q H =W/Q H = 1-Q C /Q H  S = Q C /T C - Q H /T H  0 Therefore, Q C /Q H  T C / T H  S = 0 for Carnot Q C /Q H = T C / T H for Carnot Therefore e = 1 - Q C /Q H  1 - T C / T H e = 1 - T C / T H for Carnot e = 1 is forbidden! e largest if T C << T H Engines and the 2nd Law

19. Lecture 27Purdue University, Physics 22019 Consider a hypothetical device that takes 1000 J of heat from a hot reservoir at 300K, ejects 200 J of heat to a cold reservoir at 100K, and produces 800 J of work. Does this device violate the second law of thermodynamics ? A) Yes B) No correct l W (800) = Q hot (1000) - Q cold (200) l Efficiency = W/Q hot = 800/1000 = 80% l Max eff = 1-T c /T h =1 - 100/300 = 67%  S H = Q H /T H = (-1000 J) / (300 K) = -3.33 J/K  S C = +Q C /T C = (+200 J) / (100 K) = +2 J/K  S TOTAL =  S H +  S C = -1.33 J/K  (violates 2 nd law) iClicker

20. Lecture 27Purdue University, Physics 22020 Summary of Concepts First Law of thermodynamics: Energy Conservation Q =  U + W Heat Engines Efficiency = 1-Q C /Q H Refrigerators Coefficient of Performance = Q C /(Q H - Q C ) Entropy  S = Q/T 2 nd Law: Entropy always increases! Carnot Cycle: Reversible, Maximum Efficiency e = 1 – T c /T h