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Unformatted text preview: Chapter 5 The First Law of Thermodynamics Chapter 5 THE FIRST LAW OF THERMODYNAMICS Closed System Energy Balance: General Systems 51C No. This is the case for adiabatic systems only. 52C Warmer. Because energy is added to the room air in the form of electrical work. 53C Warmer. If we take the room that contains the refrigerator as our system, we will see that electrical work is supplied to this room to run the refrigerator, which is eventually dissipated to the room as waste heat. 54C Energy can be transferred to or from a control volume as heat, various forms of work, and by mass transport. 55 Water is heated in a pan on top of a range while being stirred. The energy of the water at the end of the process is to be determined. Assumptions The pan is stationary and thus the changes in kinetic and potential energies are negligible. Analysis We take the water in the pan as our system. This is a closed system since no mass enters or leaves. Applying the energy balance on this system gives kJ 35.5 = = + = ∆ = + ∆ = 2 2 1 2 out in pw, in energies etc. potential, kinetic, internal, in Change system mass and work, heat, by nsfer energy tra Net out in kJ 10 kJ 5 kJ 5 . kJ 30 U U U U U Q W Q E E E QQ QQ Q QQ QQ Q Therefore, the final internal energy of the system is 35.5 kJ. 5 1 Chapter 5 The First Law of Thermodynamics 56E Water is heated in a cylinder on top of a range. The change in the energy of the water during this process is to be determined. Assumptions The pan is stationary and thus the changes in kinetic and potential energies are negligible. Analysis We take the water in the cylinder as the system. This is a closed system since no mass enters or leaves. Applying the energy balance on this system gives Btu 52 = = ∆ ∆ = = ∆ = ∆ = 1 2 1 2 out out b, in energies etc. potential, kinetic, internal, in Change system mass and work, heat, by nsfer energy tra Net out in Btu 8 Btu 5 Btu 65 U U U U U U U Q W Q E E E QQ QQ Q QQ QQ Q Therefore, the energy content of the system increases by 52 Btu during this process. 57 A classroom is to be airconditioned using window airconditioning units. The cooling load is due to people, lights, and heat transfer through the walls and the windows. The number of 5kW window air conditioning units required is to be determined. Assumptions There are no heat dissipating equipment (such as computers, TVs, or ranges) in the room. Analysis The total cooling load of the room is determined from Q Q Q Q cooling lights people heat gain = + + where Q Q Q lights people heat gain 10 100 W 1 kW 40 360 kJ / h 4 kW 15,000 kJ / h 4.17 kW = × = = × = = = Substituting, . Q cooling 9.17 kW = + + = 1 4 417 Thus the number of airconditioning units required is 9.17 kW 5 kw / unit = → 1 83 . 2 units 5 2 Room 40 people 10 bulbs Q cool · 15,000 kJ/h Chapter 5 The First Law of Thermodynamics 58 An industrial facility is to replace its 40W standard fluorescent lamps by their 35W high efficiency...
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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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