Lecture 15 - UMD Physics · Lecture 15 • Clarifications about lecture 14 • heat engines and...
Transcript of Lecture 15 - UMD Physics · Lecture 15 • Clarifications about lecture 14 • heat engines and...
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Lecture 15
• Clarifications about lecture 14
• heat engines and refrigerators using ideal gas as working substance
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Clarification about lecture 14 (I)• For both gases being
monoatomic or both being diatomic, use
• ...but not for one monoatomic and the other diatomic (as in example)
E1f
N1= E2f
N2= Etot
N1+N2!
E1f = N1N1+N2
Etot; E2f = N2N1+N2
Etot
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Clarification about lecture 14 (II)• collision is reversible:
transfer energy from system 2 (faster atom) to system 1 (slower atom)
• ...but less likely to find faster atoms in system 2 than in system 1
• on the average, collisions transfer energy from system 1 to system 2
T1 > T2
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Clarification about lecture 14 (III)• entropy is measure of disorder...and never decreases...
• Initial state more ordered (less entropy): less energetic atoms on one side, more energetic on the other....than equilibrium: same energy for atoms on both sides
• ...but initial state less ordered than if heat from cold to hot (more energetic side becomes even more energetic)
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Refrigerators• closed cycle uses external work to
remove heat from cold reservoir and exhaust heat to hot reservoir (2nd law does not allow spontaneous): e.g. air-conditioner or kitchen...make air that is cooler than environment even colder
• exhaust more heat than removed from inside (cool room by leaving refrigerator door open?)
• coefficient of performance:
• perfect refrigerator ( ) forbidden by 2nd law (informal statement # 3): real refrigerator uses work ( )
!Eth = 0 (cyclical) : QH = QC + Win
Win = 0; K =!K <!
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Example
• 1.0 L of 20 degree Celsius water is placed in a refrigerator. The refrigerator’s motor must supply an extra 8.0 W power to chill the water to 5 degree Celsius in 1.0 hr. What is the refrigerator’s coefficient of performance?
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No perfect Heat Engine ( )• connect perfect engine to refrigerator: no net work for 2
combined, but heat transferred from cold to hot (not allowed by 2nd law)
• informal statement # 4: no perfect heat engine, must waste heat...
(! = WoutQH
>! 1 from energy conservation)
! = WoutQH
< 1
! = 1
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Example of Proof by Contradiction
• want to prove statement “A” (e.g., there exists a perfect engine) is not true
• assume A is true, find a violation of basic law (e.g. ,2nd law of thermodynamics) assumption is incorrect, A is not true
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• Using only energy conservation and heat not transferred from cold to hot, deduce heat engines and refrigerators exist; must use closed-cycle processes; no perfect...
• upper limit on ?!, K
Unanswered questions
Summary
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Ideal gas Heat Engines
Ideal gas summary I
• closed cycle trajectory: clockwise for Wout > 0Wout = Wexpand ! |Wcompress| = area inside closed curve
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Ideal gas summary II• depends only on T
• identify each process, draw pV diagram
• use ideal gas law to know n, p, V, T at one point
• use ideal gas law and equations for specific processes for p, V, T at beginning/end of each process
• calculate for each process
• by adding `s: confirm by area within curve
• add positive values of Q to find
• check:
Strategy for heat engine problems
(!Eth)net = 0, ! < 1, signs of Ws and Q...
QH
WsWout
Q, Ws and !Eth
Eth