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Physics 2053C – Fall 2001

Physics 2053C – Fall 2001. Chapter 15 Thermodynamics. Thermodynamics. Closed Systems Heat, Work and Internal Energy. Isothermal process (Temperature and Internal Energy are constant). Isobaric process (Pressure is constant, Work = P V). Adiabatic process (No heat added or lost).

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Physics 2053C – Fall 2001

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  1. Physics 2053C – Fall 2001 Chapter 15 Thermodynamics Dr. Larry Dennis, FSU Department of Physics

  2. Thermodynamics • Closed Systems • Heat, Work and Internal Energy. • Isothermal process (Temperature and Internal Energy are constant). • Isobaric process (Pressure is constant, Work = PV). • Adiabatic process (No heat added or lost). • Using thermodynamic cycles to describe engines that do work. • Efficiency of thermodynamic cycles. • ( = Work Done/Heat In).

  3. First Law of Thermodynamics Q = U + W • Heat added = change in Internal Energy + Work done. • Assumes a closed system. • Energy conservation applied to macroscopic systems.

  4. Thermodynamic Cycles A PA B Analysis starts by finding the conditions at each of end points and knowing the conditions for each transition. Pressure PC C D VA VB Volume PA = 2 PC = 4 atm VB = 2 VA= 16 L 2 moles of Krypton

  5. A PA B Pressure PC C D VA VB Volume Thermodynamic Cycles PAVA = nRTA TA= PAVA /(nR) TA= 4 atm * 8 L/(2*0.0821 L-atm/(mol-K)) TA= 195 K

  6. A PA B Pressure PC C D VA VB Volume Thermodynamic Cycles Q = U + W For each step in the cycle And the cycle as a whole.

  7. A PA B Pressure PC C D VA VB Volume Thermodynamic Cycles W = PV WAB = PAVAB WAB = PA(VB-VA) = 4 atm * (16 L – 8 L) = 32 L-atm WCD = PC(VD-VC) = 2 atm * (8 L – 16 L) = -16 L-atm

  8. A PA B Pressure PC C D VA VB Volume Thermodynamic Cycles U = 3/2 nR T UAB = 3/2 (nRTAB) = 3/2 (nRTB – nRTA ) = 3/2 (PBVB –PAVA ) UAB = 3/2 (PBVB –PAVA ) = 3/2 ( 4 atm * 16 L – 4 atm * 8 ) = 48 L-atm UCD = 3/2 (PDVD –PCVC ) = 3/2 ( 2 atm * 8 L – 2 atm * 16 ) = -24 L-atm

  9. A PA B Pressure PC C D VA VB Volume Thermodynamic Cycles U = 3/2 nRT UBC = 3/2 (PCVC –PBVB )= 3/2 ( 2 atm * 16 L – 4 atm * 16 ) = -48 L-atm UDA = 3/2 (PAVA –PDVD ) = 3/2 ( 4 atm * 8 L – 2 atm * 8 L) = 24 L-atm

  10. A PA B Pressure PC C D VA VB Volume Thermodynamic Cycles Efficiency = Work/(Heat In) = 16 L-atm/(80 L-atm + 24 L-atm) = 16 / 104 = 0.154

  11. QH W QL Engine Carnot Cycle •  = W/Qin = (QH-QL)/(QH) = (TH-TL)/(TH)  = 1 – (TL/TH ) TH TL

  12. CAPA #6 A heat engine does 7090 J of work in each cycle while absorbing 13.1 kcal of heat from a high-temperature reservoir. What is the efficiency (in percent) of this engine? Qin = 13.1 kcal = 13,100 cal = 54840 J E = Work/Qin = 7090 J / 54840 J = 0.1293 Or: 12.9%

  13. CAPA 7,8 & 9 #7: A Carnot engine works between two heat reservoirs at temperatures T hot = 323.0 K and T cold = 207.0 K. What is its efficiency?  = 1 – (TL/TH )  = 1 – (207/323)  = 1 – 0.641  = 0.359

  14. CAPA 7,8 & 9 #8: If it absorbs 119.0 J from the hot reservoir during each cycle, how much work does it do per cycle?  = (Work/QH ) Work =  * QH Work = 0.359 * 119.0 J = 42.7 J

  15. CAPA 7,8 & 9 #9: How much heat does it reject during each cycle? Work = QH – QL  QL = QH -Work QL = 119 – 42.7 = 76.3 J

  16. Second Law of Thermodynamics The total entropy of any system plus that of its environment increases as the result of any natural process. Natural processes tend to move toward a state of greater disorder. S = Q/T

  17. CAPA #10 A 220.0 kg block of ice at 0 oC is placed in a large lake. The temperature of the lake is just slightly higher than 0 oC, and the ice melts. What is the entropy change of the ice? S = Q/T = mLf/T = (220 kg* 3.33 x 105 J/kg) /273 K = 2.68 x 105 J/K

  18. CAPA #11 What is the entropy change of the lake? S = Q/T = -Qice/T = -(220 kg* 3.33 x 105 J/kg) /273 K = -2.68 x 105 J/K

  19. Next Time • Chapter 15 & Chapter 11 • Please see me with any questions or comments. See you on Wednesday.

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