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Unformatted text preview: Chapter 7 Entropy Entropy Change of Incompressible Substances 748C No, because entropy is not a conserved property. 749 A hot copper block is dropped into water in an insulated tank. The final equilibrium temperature of the tank and the total entropy change are to be determined. Assumptions 1 Both the water and the copper block are incompressible substances with constant specific heats at room temperature. 2 The system is stationary and thus the kinetic and potential energies are negligible. 3 The tank is wellinsulated and thus there is no heat transfer. Properties The density and specific heat of water at 25 ° C are ρ = 997 kg/m 3 and C p = 4.18 kJ/kg. ° C. The specific heat of copper at 27 ° C is C p = 0.386 kJ/kg. ° C (Table A3). Analysis We take the entire contents of the tank, water + copper block, as the system . This is a closed system since no mass crosses the system boundary during the process. The energy balance for this system can be expressed as U E E E out in ∆ = ∆ = energies etc. potential, kinetic, internal, in Change system mass and work, heat, by nsfer energy tra Net NN NN N NN NN N or, = ∆ + ∆ water Cu U U )] ( [ )] ( [ 1 2 1 2 = + water Cu T T mC T T mC where kg 120 ) m 0.120 )( kg/m 997 ( 3 3 = = = V m water ρ Using specific heat values for copper and liquid water at room temperature and substituting, C 25) ( C) kJ/kg kg)(4.18 (120 C 80) ( C) kJ/kg kg)(0.386 (50 2 2 = ⋅ + ⋅ T T T 2 = 27.0 ° C The entropy generated during this process is determined from ( 29 ( 29 ( 29 ( 29 kJ/K 3.355 K 298 K 300.0 ln K kJ/kg 4.18 kg 120 ln kJ/K 3.140 K 353 K 300.0 ln K kJ/kg 0.386 kg 50 ln 1 2 water 1 2 copper = ⋅ = = ∆ = ⋅ = = ∆ T T mC S T T mC S ave ave Thus, kJ/K 0.215 = + = ∆ + ∆ = ∆ 355 . 3 140 . 3 water copper total S S S 727 WATER Copper 50 kg 120 L Chapter 7 Entropy 750 A hot iron block is dropped into water in an insulated tank. The total entropy change during this process is to be determined. Assumptions 1 Both the water and the iron block are incompressible substances with constant specific heats at room temperature. 2 The system is stationary and thus the kinetic and potential energies are negligible. 3 The tank is wellinsulated and thus there is no heat transfer. 4 The water that evaporates, condenses back. Properties The specific heat of water at 25 ° C is C p = 4.18 kJ/kg. ° C. The specific heat of iron at room temperature is C p = 0.45 kJ/kg. ° C (Table A3). Analysis We take the entire contents of the tank, water + iron block, as the system . This is a closed system since no mass crosses the system boundary during the process. The energy balance for this system can be expressed as U E E E out in ∆ = ∆ = energies etc. potential, kinetic, internal, in Change system mass and work, heat, by nsfer energy tra Net NN NN N NN NN N or, = ∆...
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This homework help was uploaded on 04/18/2008 for the course EML 3007 taught by Professor Chung during the Spring '08 term at University of Florida.
 Spring '08
 Chung

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