課程名稱 : 物 理 University Physics

1. 課程名稱 : 物 理University Physics 臺北科技大學光電系 何 文 章 99. 02. 23

2. Contents (I) • Ch 20 Kinetic Theory氣體動力論 • Ch 21 Entropy and the second law of Thermodynamics 熵與熱力學第二定律 • Ch 22 Electrostatics靜電 • Ch 23 The Electric Filed電場 • Ch 24 Gauss’s Law高斯定律 • Ch 25 Electric Potential電位 • Ch 26 Capacitors and Dielectric電容器與介電質 • Ch 27 Current and Resistance電流與電阻 • Ch 28 Direct Current Circuits直流電路 • Ch 29 The Magnetic Filed磁場 • Ch 30 Sources of The Magnetic Filed 磁場來源

3. Contents (II) • Ch 31 Electromagnetic Induction 電磁感應 • Ch 32 Inductance and Magnetic Materials • 電感及磁性材料 • Ch 33 Alternating Current Circuits 交流電路 • Ch 34 Maxwell’s Equations : Electromagnetic Waves 馬克士威方程式；電磁波 • Ch 35 Light : Reflection and Refraction 光：反射與折射 • Ch 36 Lenses and Optical Instruments 透鏡與光學儀器 • Ch 37 Wave Optics (I) 波動光學(一) • Ch 38 Wave Optics (II) 波動光學(二) • Ch 39 Special Relativity 特殊相對論 • Ch 40 Early Quantum Theory 早期量子理論 • Ch 41 Wave Mechanical 波動力學

4. Ch 21 Entropy and the second law of Thermodynamics熵與熱力學第二定律 • Heat Engines and The Kelevin-Planck Statement of the Second Law 熱機與第二定律之凱爾文蒲朗克陳述 • Refrigerators and The Clausius Statement of The Second Law 冷凍機與第二定律之克勞秀士陳述 • Reversible and Irreversible Process 可逆與不可逆過程 • The Carnot Cycle 卡諾循環 • Gasoline Engine – Otto Cycle 汽油機 - 鄂圖循環 • Entropy熵 • Entropy and the Second Law 熵與第二定律

5. Heat Engines and The Kelevin-Planck Statement of the Second Law 熱機與第二定律之凱爾文蒲朗克陳述 蒸汽機 汽油機 柴油機

6. Heat Engines and The Kelevin-Planck Statement of the Second Law 熱機與第二定律之凱爾文蒲朗克陳述 + QH, - QC

7. Heat Engines and The Kelevin-Planck Statement of the Second Law 熱機與第二定律之凱爾文蒲朗克陳述

8. Heat Engines

9. Refrigerators and The Clausius Statement of the Second Law 冷凍機與第二定律之克勞秀士陳述

10. Refrigerators and The Clausius Statement of the Second Law 冷凍機與第二定律之克勞秀士陳述

11. Reversible and Irreversible Process 可逆與不可逆過程

12. Reversible Process • A reversible process proceeds quasistatically from the initial equilibrium state to the final equilibrium state. There is no friction and no transfer of heat associated with a finite temperature difference.

13. The Carnot Cycle 卡諾循環 * 等溫膨脹 絕熱壓縮 等溫 絕熱 絕熱膨脹 Net Work 淨 功 等溫壓縮

14. The Carnot Cycle 卡諾循環 ΔU = Q - W a → b : ΔU =nCV△T= 0, QH = Wab, Wab > 0 Wab = nRTH ln(Vb/Va) ΔU = nCVΔT W = ∫PdV PV = nRT b → c : Q =0, Wbc > 0 ΔU = -Wbc ∵ΔU ～ ΔT↘ ∴TH →TL Wbc = -ΔU = - nCVΔT = - nCV(TL-TH) d → a : ? c → d : ΔU = nCV△T= 0, Wcd = nRTC ln(Vd/Vc) < 0 QL = Wcd

15. Efficiency of Carnot Cycle • Isothermal Processes • The process a to b, the heat absorbed from the hot reservoir is • The process c to d, the heat discharged into the cold reservoir is • Adiabatic Process • PV = constant, → TVγ-1 = constant. ……(I) ……(II)

16. Efficiency of Carnot Cycle • The arguments of the logarithms in Eqs. (I), (II) are the same. The ratio of these equations is therefore • The efficiency of a heat engine is ε= W/QH = QH - QC / QH = 1 – (QC / QH ) • For the Carnot engine, we see that the Carnot efficiency εC is • The Carnot efficiency depends only on the Kelvin temperatures of the reservoirs. The ratio of the heat transfers is equal to the ratio of the Kelevin temperature of the reservoirs.

18. Gasoline Engine – Otto Cycle 汽油機 - 鄂圖循環 四衝程 進氣衝程 壓縮衝程 動力衝程 排氣衝程

19. 壓縮衝程 進氣衝程 點火 動力衝程 排氣衝程 排氣 p0 Adiabatic expansion 動力衝程 點火 排氣 壓縮衝程 進氣衝程 排氣衝程

20. ΔU =Q-W, W = ∫pdV, B → C D → A ΔU = 0 γ = Cp/Cv

21. Entropy 熵 Chemical Reaction

22. Entropy 熵 Each cycle : ΔQH/TH+ ΔQC/TC=0 A finite number of cycles : ΣΔQ/T = 0

23. Entropy 熵 • Consider two equilibrium states a and b. The paths I and • II indicate just two of many possible paths the system can • follow from a to b. (II) (I) (II) (I) The integral of dQR/T is independent of the path between the Equilibrium states.

24. Entropy 熵 • We define an infinitesimal change in entropy as • For a finite change in entropy in a reversible process →The change in entropy depends only on the initial and final equilibrium states, not on the thermodynamic path.

25. Entropy 熵 from an initial to a finial equilibrium state. ΔU = Q-W For an ideal gas, , and PV = nRT Integrating from an initial to a finial equilibrium state we find ← Change in entropy Of an ideal gas

26. Entropy 熵 ΔT = 0 ΔQ = 0 ΔU = 0

27. Entropy 熵 Example : 21.3 An insulated metal rod whose ends are in thermal contact with two reservoirs. When a steady state has been reached, during a certain time interval a quantity of heat Q = 200 J is transferred from the hot reservoir at TH = 60℃ to the cold reservoir at TC = 12℃. What are the change in entropy of each reservoir? Sol: The change in entropy of hot reservoir is △SH = -Q/TH = - (200 J) / (273 +60)K = - 0.6 J/K The change in entropy of cold reservoir is △SC = Q/TC = (200 J) / (273 +12)K = 0.7 J/K The net change in entropy of two reservoir are △S = △SH + △SC = -Q/TH + Q/TC = 0.1 J/K.

28. Entropy 熵 Example : 21.5 A copper ball of mass m=0.5 kg and specific heat c = 390 (J/kg．K) is at a temperature T1 = 90℃. The ball is thrown into a large lake a T2 = 10℃, which stays constant. Find the change in entropy of the ball. Sol : For an infinitesimal change in temperature, the heat transferred from the ball is dQ = mcdT. The rateed change entropy is dS = dQ/T = mcdT/T. The temperature of the ball change from T1 to T2, So

29. Entropy 熵 Exercise 6 What is the change in entropy of 300 g of water as its temperature increases from 10℃ to 25℃? The specific heat of water is 4.19 (kJ/kg‧K). Sol :