Entropy, thermal engines

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Entropy, thermal engines

• Phys 2101
• Gabriela González

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Irreversible processes

• We can easily tell the direction of the arrow of
  time in irreversible processes which, if
  spontaneous, only happen one way
• spilling fluids
• gas expansion
• breaking solids
• Conservation of energy doesnt forbid these
  processes there is another quantity, entropy,
  which measures the degree of disorder
• S k log W

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Second Law of thermodynamics

• If an irreversible process occurs in a closed
  system, the entropy S of the system always
  increases it never decreases.
• If any process occurs in a closed system, the
  entropy of the system increases for irreversible
  processes and remains constant for reversible
  processes. It never decreases!
• ??S 0

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Entropy and heat
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Example

• An 8.0g ice cube at -10oC is put in a Thermos
  flask containing 100 cm3 of water at 20oC. The
  specific heat of ice is 2220 J/kgK, the specific
  heat of water is 4190 J/kgK, and the heat of
  fusion of ice-water is 330 kJ/kg.
• Whats the final equilibrium temperature?
• ?Q 0 cWmW(Tf-20oC) cice mice (0oC-(-10oC))
  mice Lf cW mice (Tf-0oC)
• 0 cw(mwmice)Tf-cwmw20oC cice mice10oC mice
  Lf
• 41900.108Tf - 41900.120
  22200.008100.008330000
• 452.5Tf-8380177.62640 ?Tf 12.24oC 285.39 K
  
• What is the change in entropy of the ice-water
  system?
• Use ?S ?dQ/T ?mcdT/T mc ln(Tf/Ti) for
  changes in temperature, and
• ?S ?dQ/T Q/T mLf/T for changes of phase
• ?Sw 0.1kg4190J/kgKln(285.39/293.15) ?11.24
  J/K
• ???Sice 8g(4.19J/gKln(285.39/273.15)
  2.22J/gKln(273.15/263.15) 330J/g/273.15K)
  11.88 J/K
• ???S 0.64 J/K

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Change in Entropy

• Isothermal process
• ?S ?dQ/T (1/T) ?dQ ?Q/TAdiabatic process
  ?Q0, so ?S0
• Any reversible process,
• dQ dEint dW
• nCV dT p dV
• dQ/T nCV dT/T pdV/T
• nCV dT/T pdV/(pV/nR)
• nCV dT/T nRdV/V
• ?S ?dQ/T
• nCV ln (Tf/Ti) nR ln (Vf/Vi)

• Eint (f/2)nRT n CV T
• W ? p dV
• pV nRT

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Ideal heat engines

• Ideal heat engines use a cycle of reversible
  thermodynamic processes. A heat engine
  transforms energy extracted as heat from thermal
  reservoirs, into mechanical work.
• Consider a Carnot engine a cycle with two
  isothermal processes at a high temperature TH
  (a?b) and and a low temperature TL (c ?d), and
  two adiabatic processes (b?c, d?a).
• ?Eint 0 Q W ? W
  Q QH-QL
• ?S 0
  QH/TH - QL/TL

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Thermal efficiency

• We use the heat QH to get work W done, so
  efficiency is defined as
• ? W/QH
• For a Carnot engine,
• W QH-QL
• QH/TH QL/TL
• so ?C (QH-QL)/ QH 1- QL/QH
  1- TL/TH lt 1

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Example

• Three ideal Carnot engines operate between (a)
  400K and 500K, (b) 500K and 600K, and (c) 400K
  and 600K.
• Rank them according to their efficiencies,
  greatest first.
• If they all extract the same amount of energy
  per cycle from the high temperature reservoir,
  rank them according to the work per cycle done by
  the engines, greatest first.