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CHAPTER5_SECTION2 - Chapter 5 The First Law of...

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Chapter 5 The First Law of Thermodynamics Closed System Energy Analysis: Solids and Liquids 5-45 A number of brass balls are to be quenched in a water bath at a specified rate. The rate at which heat needs to be removed from the water in order to keep its temperature constant is to be determined. Assumptions 1 The thermal properties of the balls are constant. 2 The balls are at a uniform temperature before and after quenching. 3 The changes in kinetic and potential energies are negligible. Properties The density and specific heat of the brass balls are given to be = 8522 kg/m 3 and C p = 0.385 kJ/kg. C. Analysis We take a single ball as the system. The energy balance for this closed system can be expressed as E E E Q U m u u Q mC T T in out out ball out Net energy transfer by heat, work, and mass system Change in internal, kinetic, potential, etc. energies ( ) ( ) 2 1 1 2 The total amount of heat transfer from a ball is m V D Q mC T T out 3 1 2 6 8522 0 05 0 558 0 558 0 385 120 74 9 88 ( ) ( . . ( ) ( . )( . ( ) . kg /m m) 6 kg kg kJ /kg. C) C kJ /ball 3 3 Then the rate of heat transfer from the balls to the water becomes ( ( . ) Q n Q total ball ball balls /min) kJ /ball 100 9 88 988 kJ / min Therefore, heat must be removed from the water at a rate of 988 kJ/min in order to keep its temperature constant at 50 C since energy input must be equal to energy output for a system whose energy level remains constant. That is, E E E in out when system 0 . 5-42 Brass balls, 120 C Water bath, 5 C
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Chapter 5 The First Law of Thermodynamics 5-46 A number of aluminum balls are to be quenched in a water bath at a specified rate. The rate at which heat needs to be removed from the water in order to keep its temperature constant is to be determined. Assumptions 1 The thermal properties of the balls are constant. 2 The balls are at a uniform temperature before and after quenching. 3 The changes in kinetic and potential energies are negligible. Properties The density and specific heat of aluminum at the average temperature of (120+74)/2 = 97 C = 370 K are = 2700 kg/m 3 and C p = 0.937 kJ/kg. C (Table A-3). Analysis We take a single ball as the system. The energy balance for this closed system can be expressed as E E E Q U m u u Q mC T T in out out ball out Net energy transfer by heat, work, and mass system Change in internal, kinetic, potential, etc. energies ( ) ( ) 2 1 1 2 The total amount of heat transfer from a ball is kJ/ball 62 . 7 C ) 74 120 ( C) kJ/kg. 937 . 0 )( kg 1767 . 0 ( ) ( kg 1767 . 0 6 m) 05 . 0 ( ) kg/m 2700 ( 6 2 1 3 3 3 T T mC Q D V m out Then the rate of heat transfer from the balls to the water becomes kJ/min 762 ) kJ/ball 62 . 7 ( balls/min) 100 ( ball ball Q n Q total Therefore, heat must be removed from the water at a rate of 762 kJ/min in order to keep its temperature constant at 50
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