How does a cycle produce zero As? I thought that the whole thing about cycles was an entropy that the designers try to minimize ( MP 1B. 13 d) Entropy changes in the"hot brick problem' examine in a more quantitative manner the changes that occurred when we put the two bricks together, as depicted on the left-hand side of the figure below. The process by which the two bricks come to the same temperature is not a reversible one, so we need to devise a reversible path. To do this imagine a large number of heat reservoirs at varying temperatures spanning the range TH T +dT, as in the right hand side of the figures. The bricks are put in contact with them sequentially to raise the temperature of one and lower the temperature of the T M Temperature equalization of two bricks Reservoirs used in reversible state transformation other in a reversible manner. The heat exchange at any of these steps is do= CdT. For the high temperature brick, the entropy change is Tm Cdt hot brick C In where C is the heat capacity of the brick ( J/kg). This quantity is less than zero. For the cold brick TM CaT ld brick =CIn -M The entropy change of the two bricks is bricks C In =CIn 0 The process is not reversible The essential difference between the free expansion in an insulated enclosure and the s<g e) Difference between the free expansion and the reversible isothermal expansion of an ideal reversible isothermal expansion of an ideal gas can also be captured clearly in terms of entropy changes. For a state change from initial volume and temperature V, T, to final volume and(the same)temperature V,, T the entropy change is do 2 Pdv As= dS 1B-9How does a cycle produce zero ∆S? I thought that the whole thing about cycles was an entropy that the designers try to minimize. (MP 1B.13) d) Entropy changes in the “hot brick problem” We can examine in a more quantitative manner the changes that occurred when we put the two bricks together, as depicted on the left-hand side of the figure below. The process by which the two bricks come to the same temperature is not a reversible one, so we need to devise a reversible path. To do this imagine a large number of heat reservoirs at varying temperatures spanning the range TH − dT,............,TL + dT , as in the right hand side of the figures. The bricks are put in contact with them sequentially to raise the temperature of one and lower the temperature of the TH TL TM TM TH ........ TL TH - dT TL + dT Temperature equalization of two bricks Reservoirs used in reversible state transformation other in a reversible manner. The heat exchange at any of these steps is dQ = CdT . For the high temperature brick, the entropy change is: TM CdT = C ln  TM  ∆Shot brick = ∫  TH T  TH  where C is the heat capacity of the brick (J/kg). This quantity is less than zero. For the cold brick, TM CdT = C ln  TM   ∆S . cold brick = ∫TL T  TL  The entropy change of the two bricks is 2 ∆Sbricks = C  ln   TM   + ln   TM     = C ln TM > 0.   TH   TL  T THL The process is not reversible. e) Difference between the free expansion and the reversible isothermal expansion of an ideal gas The essential difference between the free expansion in an insulated enclosure and the reversible isothermal expansion of an ideal gas can also be captured clearly in terms of entropy changes. For a state change from initial volume and temperature V T1 to final volume and (the 1, same) temperature V T1 the entropy change is 2 , 2 2 dU 2 PdV ∆S = ∫1 dS = ∫1 T + ∫1 , T 1B-9