ΔU = Q W. Tue Dec 1. Assign 13/14 Friday Final: Fri Dec 11 2:30PM WALTER 145. Thermodynamics 1st Law. 2 nd Law. Heat Engines and Refrigerators

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1 Tue Dec 1 Thermodynamics 1st Law ΔU = Q W 2 nd Law SYS Heat Engines and Refrigerators Isobaric: W = PΔV Isochoric: W = 0 Isothermal: ΔU = 0 Adiabatic: Q = 0 Assign 13/14 Friday Final: Fri Dec 11 2:30PM WALTER Sheets handwritten about 30% Thermal Physics 16 MC, 4 problems Exam Conflict? Notify me ASAP Study: Homework, PRS, Pre-class Examples in text Review problems Thursday finish up details on Chapter 15, then review Q/A Sessions (tentative)

2 Textbook Question: This semester we started using the OpenStax Textbook (1) I purchased a hardcopy of the textbook (2) I downloaded a copy of the textbook and referenced it during the course (3) I downloaded a copy of the textbook but did not use it during the course (4) I did not use any textbook

3 1st Law of Thermodynamics! Sect 15.3 ΔU = Q W SYS SIGNS! Q: added to system If added, Q is + If removed, Q is - W: Work done BY system (W SYS ) If system does (+) work on surroundings, W sys is (+) If surroundings do (+) work on system, W sys is (-)

4 Transitions between States PV diagram (ideal gas) State is a point same point, same state same state, same P,V,n,T,U Cycle loop back to initial state end at same U! For cycle: ΔU = 0

5 Work Ideal Gas! Sect 15.2 W = Force*displacement W = (PA) * s W = P (ΔV) If P constant, isobaric transition ΔU = Q ( PΔV ) Area under Curve = Work Holds in general!

6 Given the PV diagram to the right and the information that ΔU AB is positive, what can you conclude about Q AB and W AB? (1) Q>0, W>0 (2) Q>0, W = 0 (3) Q>0, W<0 (4) Q=0, W>0 (5) Q=0, W=0 (6) Q=0, W<0 (7) Q<0, W>0 (8) Q<0, W=0 (9) Q<0, W<0 P A C B V ΔU = Q W From the graph, W is +. The only way ΔU can be + is if Q is + and greater than W.

7 Cycles P A B Cycle loop back to initial state end at same U! For cycle: ΔU = 0 D C ΔU NET = ΔU AB + ΔU BC + ΔU CD + ΔU DA Total W add up transitions W NET = W AB + W BC + W CD + W DA Total Q add up transitions Q NET = Q AB + Q BC + Q CD + Q DA V

8 The NET work done by the system during the cycle shown is: (1) positive (2) zero (3) negative Positive work done from left to right. Negative work from right to left. Larger area under curve left to right greater magnitude of work. What if reversed direction of arrows?

9 Given the cycle shown, the heat absorbed by the system is: (1) positive (2) zero (3) negative ΔU = 0; ΔU = Q-W Q = ΔU + W = 0 + (some positive number) Q must be positive

10 In the cycle below, Q AB is positive and ΔU BC is positive. What is the sign of ΔU CA? (Can you identify the work. heat and change in internal energy for each process? Don t forget about the First Law.) (1) positive 2) zero 3) negative ΔU AB positive (W=0, Q+) ΔU BC is positive ΔU Cycle = ΔU AB + ΔU BC + ΔU CA =0

11 Some Examples of Cycles in Heat Engines Diesel Cycle physics/thermo/engines/dieselg.php Stirling Cycle Anim.adp Otto Cycle thermo/engines/ottog.php Carnot Cycle ~xmwang/mygui/ CarnotG.html

12 Heat Engines Work Hot Reservoir, Cold Reservoir Conserve Energy Cycle: ΔU = 0 Q cycle = W cycle Q H -Q C = W Q H = W + Q C Eff = W/Q H Always < 1 unless in %

13 Example A heat engine has an efficiency of 64% and produces 5500 J of work. Determine: a) the input heat Q H =W/e =5500J/0.64 = 8590 J W = Q Q e = H W Q H C b) The rejected heat or heat exhausted Q C = Q H -W = 8590J -5500J = 3090 J Many "faces" of efficiency e = W Q H e = Q Q Q H H C e = 1 H Q / Q C 1

14 What is the efficiency of a heat engine that pulls 8000J from the hot reservoir and expels 6000J in the form of exhaust? (1) 10% (2) 25% (3) 33% (4) 50% (5) 60% (6) 75% (7) 133% W = 8000J 6000J eff = W/Q H = 2000J/8000J = 25%

15 A certain heat engine operates by absorbing a fixed amount of heat from the hot reservoir each cycle and expelling a certain amount of heat to the cold reservoir. The engine is redesigned such the energy absorbed from the hot reservoir each cycle is the same, but the energy expelled to the cold reservoir each cycle decreases. What happens to the efficiency of the engine? 1) it increases 2) it stays the same 3) it decreases e = 1 H Q / Q C 1 Q C gets smaller, Q C /Q H gets smaller, 1-Q C /Q H gets larger

16 Carnot Engine Absolute best engine if all processes reversible Reversible both sys and surr returned to original state All real processes irreversible For theoretically best engine (Carnot Engine): Q C T = C T eff = C MAX, Carnot 1 QH TH T H T in KELVIN!

17 An engine manufacturer claims to have an engine that draws heat from a reservoir at 375K. The engine does 5.0kJ of work each second and expels 4.0kJ of heat each second to a cold reservoir at 225K. Is this theoretically possible? (1) yes (2) no Claimed eff = W/Q H = W/(W+Q C ) = 5.0kJ / 9.0 kj = 0.56 Carnot eff = 1- (T C /T H ) = 1- (225K/375K) = 0.4 Claimed eff better than theoretical best

18 Clausius' Version of 2 nd Law of Thermodynamics If two objects are placed in thermal contact: hot object cools down cool one heats up Heat won t flow spontaneously from colder object to hotter object Why? Doesn t violate First Law of Thermodynamics (heat out of cold object = heat into hot object would certainly conserve energy). Must be a separate law!

19 Versions of the 2 nd Law of Thermodynamics Clausius: Heat won t flow spontaneously from colder object to hotter object. Kelvin-Planck: Heat can t be entirely converted into work. Entropy: Entropy is always increasing. Kinetic Theory: Disorder tends to increase. Heat Engine: It is impossible for a heat engine to be more efficient than a Carnot-cycle engine. Custodian: You can t get something clean without getting something else dirtier.

20 Third Law of Thermodynamics For Carnot heat engine: e T = C max, Carnot 1 TH You can have perfect efficiency, e = 1, but only if T C = 0 K. The Third Law of Thermodynamics says: There s no way to cool something to 0 K in a finite number of steps.

21 Summary of the Laws of Thermodynamics (with an attitude!) 0: There is a game. 1: You can t win, the best you can do is break even. 2: You can only break even at absolute zero. 3: You can t reach absolute zero.

22 Refrigerator Compress refrigerant Heat it Expand Cool it Q + W = Q C H Inside Q C Colder than in Refrig Expand Outside Q H Warmer than outside Motor Compress W

23 Refrigerators QH = QC + W Coefficient of Performance (COP) COP Q / C W Refrig = COP Refrig = QC / QH 1 Q / Q C H For Carnot Refrigerator Q C /Q H = T C /T H : COP Refrig = TC / TH 1 T / T C H

24 In cooling a warm leftovers, a refrigerator removes 60000J of heat from the food. The COP is 3.0. We eventually want to know the amount of heat exhausted into the room. First, though, which quantity does the 60000J represent? 1) Q C 2) Q H 3) W 60000J represents the amount of energy which must be removed from the cold reservoir.

25 In cooling warm leftovers, a refrigerator removes 60000J of heat from the food. If the COP is 3.0, what is the amount of heat exhausted to the room (moved into the hot reservoir)? 1) 20,000J 2) 60,000J 3) 80,000J 4) 180,000J 5) 240,000J COP = Q C /W 3.0 = J/W W = 20,000 J Q H = Q C +W = 60,000 J + 20,000 J

26 In cooling warm leftovers, a refrigerator removes 60000J of heat from the food. The COP is 3.0. If the power of the motor is 200W, what is the minimum time to cool the leftovers? 1) 100s 2) 200s 3) 300 s COP = Q C /W W = 20,000J P = Energy/Time Time = 100 s 200W = 20,000J/time

27 Heating Your Kitchen How long would a 3.40 kw space heater have to run to put into a kitchen the same amount of heat as a refrigerator (coefficient of performance = 3.29) does when it freezes 1.30 kg of water at 21.4 C into ice at 0 C? One way Another way Refrig Warms Kitchen Q Warms Kitchen Q C Q H Same Electrical Heater W Elect Energy = P*t Q H (Refrig) = Power (of elect heater) * time (elect heater run)

28 Each drawing shows a hypothetical heat engine or a hypothetical heat pump and shows the corresponding heats and works. Only one is allowed in nature. Which is it? 1) 2) 3) 4) 5)

29 Heat Pump Coefficient of Performance (COP) a bit different COP Heat pump = Q H W

30 Heat Engines, Refrigerators, Heat Pumps Heat Engine Natural flow of heat (hot to cold) can do work on surroundings Refrig or Heat Pump Unnatural flow of heat (cold to hot) requires work from surroundings Both cases: Q H = Q C + W What is related to what???? Think about Reservoirs, not the temp of something in the reservoir Which reservoir is hotter? Colder? Q C inside refrigerator; Q H Kitchen

31 Heat Engine, Heat Pump, and Refrigerators Summary Q H = W + Q C Device What we want What we buy Efficiency, e, or Coefficient of Performance ( COP ) Heat engine Carnot Heat pump Carnot Refrigerator Carnot W net Q H 0 < e = W net / Q H = 1 - (Q C /Q H ) < 1 e = 1 - (T C /T H ) = (ΔΤ/Τ Η ) Q H W net COP = Q H /W net = 1 + [Q C /(Q H - Q C )] > 1 COP = 1/[1 - (T C /T H )] Q C W net COP = 1/[(T H /T C ) - 1] COP = Q C /W net (which may be more or less than 1)

Thurs Dec 3. Assign 13/14 Friday Final: Fri Dec 11 2:30PM Walter 145. Thermodynamics 1st Law 2 nd Law Heat Engines and Refrigerators

Thurs Dec 3. Assign 13/14 Friday Final: Fri Dec 11 2:30PM Walter 145. Thermodynamics 1st Law 2 nd Law Heat Engines and Refrigerators Thurs Dec 3 Thermodynamics 1st Law 2 nd Law Heat Engines and Refrigerators Isobaric: W = PΔV Isochoric: W = 0 Isothermal: ΔU = 0 Adiabatic: Q = 0 Assign 13/14 Friday Final: Fri Dec 11 2:30PM Walter 145

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