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Physics 101: Lecture 28, Pg 1 Physics 101: Lecture 28 Thermodynamics II l Today’s lecture will cover Textbook Chapter 15.6-15.9 Final.

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Presentation on theme: "Physics 101: Lecture 28, Pg 1 Physics 101: Lecture 28 Thermodynamics II l Today’s lecture will cover Textbook Chapter 15.6-15.9 Final."— Presentation transcript:

1 Physics 101: Lecture 28, Pg 1 Physics 101: Lecture 28 Thermodynamics II l Today’s lecture will cover Textbook Chapter 15.6-15.9 Final

2 Physics 101: Lecture 28, Pg 2Recap: è 1st Law of Thermodynamics è energy conservation Q =  U + W Heat flow into system Increase in internal energy of system Work done by system V P l U depends only on T (U = 3nRT/2 = 3pV/2) l point on p-V plot completely specifies state of system (pV = nRT) l work done is area under curve l for complete cycle  U=0  Q=W

3 Physics 101: Lecture 28, Pg 3 THTH TCTC QHQH QCQC W HEAT ENGINE THTH TCTC QHQH QCQC W REFRIGERATOR system l 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 green blob = energy leaving green blob Engines and Refrigerators

4 Physics 101: Lecture 28, Pg 4 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 = Heat Engine: Efficiency

5 Physics 101: Lecture 28, Pg 5 Heat Engine ACT l Can you get “work” out of a heat engine, if the hottest thing you have is at room temperature? A) YesB) No T H 300K T C = 77K QHQH QCQC W HEAT ENGINE

6 Physics 101: Lecture 28, Pg 6 THTH TCTC QHQH QCQC W REFRIGERATOR The objective: remove heat from cold reservoir The cost: work 1st Law: Q H = W + Q C coeff of performance K r  Q C /W = Q C /W = Refrigerator: Coefficient of Performance

7 Physics 101: Lecture 28, Pg 7 New concept: Entropy (S) l A measure of “disorder” l A property of a system (just like p, V, T, U) è related to number of number of different “states” of system l Examples of increasing entropy: è ice cube melts è gases expand into vacuum l Change in entropy: è  S = Q/T »>0 if heat flows into system (Q>0) »<0 if heat flows out of system (Q<0)

8 Physics 101: Lecture 28, Pg 8ACT A hot (98 C) slab of metal is placed in a cool (5C) bucket of water. What happens to the entropy of the metal? A) IncreaseB) SameC) Decreases What happens to the entropy of the water? A) IncreaseB) SameC) Decreases What happens to the total entropy (water+metal)? A) Increase B) Same C) Decreases  S = Q/T

9 Physics 101: Lecture 28, Pg 9 Second Law of Thermodynamics l The entropy change (Q/T) of the system+environment  0 è never < 0 è order to disorder l Consequences è A “disordered” state cannot spontaneously transform into an “ordered” state è No engine operating between two reservoirs can be more efficient than one that produces 0 change in entropy. This is called a “Carnot engine”

10 Physics 101: Lecture 28, Pg 10 Carnot Cycle l Idealized Heat Engine è No Friction   S = Q/T = 0 è Reversible Process »Isothermal Expansion »Adiabatic Expansion »Isothermal Compression »Adiabatic Compression

11 Physics 101: Lecture 28, Pg 11 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 Engines and the 2nd Law

12 Physics 101: Lecture 28, Pg 12 Example Consider a hypothetical refrigerator that takes 1000 J of heat from a cold reservoir at 100K and ejects 1200 J of heat to a hot reservoir at 300K. 1. How much work does the refrigerator do? 2. What happens to the entropy of the universe? 3. Does this violate the 2nd law of thermodynamics? THTH TCTC QHQH QCQC W

13 Physics 101: Lecture 28, Pg 13 Preflight 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 ? 1. Yes 2. No

14 Physics 101: Lecture 28, Pg 14 Act Which of the following is forbidden by the second law of thermodynamics? 1. Heat flows into a gas and the temperature falls 2. The temperature of a gas rises without any heat flowing into it 3. Heat flows spontaneously from a cold to a hot reservoir 4. All of the above

15 Physics 101: Lecture 28, Pg 15 Summary l First Law of thermodynamics: Energy Conservation  Q =  U + W l Heat Engines è Efficiency = = 1-Q C /Q H l Refrigerators è Coefficient of Performance = Q C /(Q H - Q C ) Entropy  S = Q/T l 2 nd Law: Entropy always increases! l Carnot Cycle: Reversible, Maximum Efficiency e = 1 – T c /T h l Ch 15 Problems è 1, 5, 9, 11, 15, 21, 31,41


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