CHAPTER7_SECTION2 - Chapter 7 Entropy Entropy Change of...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 7 Entropy Entropy Change of Incompressible Substances 7-48C No, because entropy is not a conserved property. 7-49 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 well-insulated 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 A-3). 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 7-27 WATER Copper 50 kg 120 L Chapter 7 Entropy 7-50 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 well-insulated 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 A-3). 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, = ∆...
View Full Document

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.

Page1 / 25

CHAPTER7_SECTION2 - Chapter 7 Entropy Entropy Change of...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online