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Absolute Zero

Absolute Zero. Physics 313 Professor Lee Carkner Lecture 15. Exercise #14 Carnot Cycle. Isothermal heat = work W = (1.5)(8.31)(700)ln(2X10 -3 /4X10 -4 ) Net work depends of efficiency h = W/Q H = 0.6 Can get output heat from first law Q L = Q H - W = 14043-8426 = 5617 J.

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Absolute Zero

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  1. Absolute Zero Physics 313 Professor Lee Carkner Lecture 15

  2. Exercise #14 Carnot Cycle • Isothermal heat = work • W = (1.5)(8.31)(700)ln(2X10-3/4X10-4) • Net work depends of efficiency • h = W/QH = 0.6 • Can get output heat from first law • QL = QH - W = 14043-8426 = 5617 J

  3. Carnot and Temperature • How are the heat exchanges related to the temperature? • For ideal gas: QH/QL = TH/TL [(ln V2/V3)/ln V4/V1)] • The volume term equals 1 (can relate V’s from the adiabatic processes)

  4. Temperature Scale • Temperature can be related to the heat transfers of a Carnot engine • Using the triple point of water • Called the thermodynamic temperature • Can make a “Carnot Thermometer” by running a Carnot engine at unknown T and T for triple point of water

  5. Absolute Zero • If you lower TL, you lower QL • Defines absolute zero • Absolute zero defined this is way is:

  6. Efficiency • Can write the efficiency of a Carnot engine as: • Increase the efficiency by increasing TH and decreasing TL • ro • For a Carnot refrigerator the coefficient of performance is:

  7. Entropy • The limits on efficiency for engines and refrigerators are expressions of entropy • Entropy represents a preferred direction for processes

  8. Heat and Temperature • We saw that: • If we include the signs of the heat: • This is true for any Carnot cycle • Any curve can be represented as the sum of many Carnot cycles

  9. Entropy Defined • For any reversible cycle: • The integral along any reversible (non-closed) path represents the change in entropy: dS = dQ/T

  10. Creating Entropy • How might we change the entropy of a system • Consider work done on a substance in contact with a heat reservoir at temperature T • The ratio of work to the temperature of the reservoir is the entropy change • Note:

  11. Ideal Gas Entropy • To calculate entropy need expression for dQ dQ = CVdT +PdV DS =  CV (dT/T) +  nR (dV/V) • Similarly for: DS = n  cP (dT/T) - nR ln (Pf/Pi)

  12. T and S • Heat can be expressed as: • Heat is the area under the curve on a TS diagram

  13. The TS Diagram • How are standard processes plotted on a TS diagram? • Isotherm • Adiabatic • No entropy change, so vertical line

  14. Other Processes • Isobar • Curved line with slope: • Isochor • Curved line with slope:

  15. TS Diagram Isentrope T Isochor Isobar Isotherm S

  16. Entropy and Isotherms • We write change in entropy as: • If T is constant • The change in entropy for an isothermal process depends only on the temperature and the total heat exchange

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