Second Law of Thermodynamics
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1 Dr. Alain Brizard College Physics II (PY 211) Second Law of Thermodynamics Textbook Reference: Chapter 20 sections 1-4. Second Law of Thermodynamics (Clausius) Heat flows naturally from a hot object to a cold object; heat will not flow spontaneously from a cold object to a hot object. Heat Engine: any device that changes heat into mechanical work. Heat Pump: any device that allows heat to flow from a cold object to a hot object. Reversible Cycles A cycle Γ is defined as a sequence of processes that trace a closed curve on the PV plane (see Figure below), i.e., the system begins and ends in the same thermodynamic state. A cycle evolves either in the clockwise direction or in the counter-clockwise direction; a cycle is reversible if it traces the same closed curve independently of the cycle s evolution. 1
2 Since a cycle begins and ends in the same thermodynamic state (P 0,V 0,n 0,T 0 ), the change of internal energy for a (reversible) cycle is zero and, thus, the First Law of Thermodynamics implies that U Γ 0 W Γ Q Γ, i.e., the work done by a system as it evolves along the cycle Γ is equal to the heat absorbed by the system during the cycle Γ. Note that W Γ positive if cycle evolves in clockwise direction negative if cycle evolves in counter-clockwise direction Second Law of Thermodynamics (Kelvin-Planck) No device is possible whose sole purpose is to transform a given amount of heat completely into work. The Second Law of Thermodynamics implies that the heat Q Γ absorbed during a cycle must be expressed as Q Γ Q out, where is the heat input and Q out is the heat output during the cycle Γ. If work is done by the system during the cycle (i.e., W Γ > 0), we find that >Q out and represents heat absorbed by the system at a high temperature and Q out represents heat released by the system at a low temperature. The ideal efficiency ε Γ of the cycle Γ is defined as ε Γ W Γ 1 Q out. The Second Law of Thermodynamics, therefore, states that the ideal efficiency of any cycle describing a heat engine can never be 100 %. A heat pump, in contrast, involves a system absorbing heat from a cold object (at a low temperature ), while work W Γ < 0 is being done on the system, and releasing heat Q out + W Γ to a hot object (at a high temperature ). The ideal efficiency of a heat pump (i.e., a refrigerator) is defined in terms of the coefficient of performance (CP) CP Γ W Γ Q out. 2
3 The Figure below shows the difference between a heat engine and a heat pump. Carnot Cycle Sadi Carnot ( ) discovered an idealized reversible cycle with the highest ideal efficiency. The diagram below shows the four-step Carnot cycle (C : a b c d a). The first step a b represents an isothermal expension from (,P a )to(,p b )at constant temperature, where P a nr P b. The second step b c represents an adiabatic expansion from (,P b ) on the 3
4 isothermal to (V c,p c ) on the isothermal, where P b P c ( ) 5 The third step c d represents an isothermal compression from (V c,p c )to(,p d ) at constant temperature, where P c V c nr P d. The fourth step d a represents an adiabatic compression from (,P d ) on the isothermal to (,P a ) on the isothermal, where P a P d ( ) 5 The two adiabatic processes yield the following identity V c Step a b: Isothermal expansion U ab 0 and W ab nr ln ( ) Vb Q ab > 0 Step b c: Adiabatic expansion Q bc 0 and U bc 3 2 nr ( ) W bc < 0 Step c d: Isothermal compression U cd 0 and W cd nr ln ( ) Vc Q cd < 0 Step d a: Adiabatic compression Q da 0 and E da 3 2 nr ( ) W da > 0 For the Carnot cycle C : a b c d a, wefind [ ( ) Vb U C 0 and W C nr ln ln ( )] Vc Q C 4
5 We can now calculate the ideal efficiency of the Carnot cycle ε C W C 1 Q out 1 ln(v c / ) ln( / ) If we now use the identity shown above, we find ln(v c / )ln( / ) and thus Carnot s Theorem 0 < ε C 1 < 1 (since 0 < < ). All reversible engines operating between the same two constant temperatures and < have the same efficiency. Any irreversible engine operating between the same two fixed temperatures will have an efficiency less than this. Otto Cycle The Otto cycle involves a four-step sequence involving an adiabatic expansion (a b), an isovolumetric process (b c), an adiabatic compression (c d), and an isovolumetric process (d a); here, V c and while T a ( / ) 2/3 T b and T d ( / ) 2/3 T c. The efficiency of the Otto cycle is given as ε Otto 1 Q out 1 Va 3 1 T b T a 1 T c T d < ε C 1 T c T a 5
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