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課程名稱 : 物 理 University PhysicsPowerPoint Presentation

課程名稱 : 物 理 University Physics

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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

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

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

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

Heat Engines Second Law

Refrigerators and The Clausius Statement of the Second Law Second Law

冷凍機與第二定律之克勞秀士陳述

Refrigerators and The Clausius Statement of the Second Law Second Law

冷凍機與第二定律之克勞秀士陳述

Reversible and Irreversible Process Second Law

可逆與不可逆過程

Reversible Process Second Law

- 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.

The Carnot Cycle Second Law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

Efficiency of Carnot Cycle Second Law

- 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)

Efficiency of Carnot Cycle Second Law

- 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.

Entropy Second Law熵

Chemical Reaction

Entropy Second Law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.

Entropy Second Law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.

Entropy Second Law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

Entropy Second Law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.

Entropy Second Law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

Entropy Second Law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 :

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