Thermodynamics

Chapter 2 | 65Solution A well-insulated electric oven is being heated by its heating element.It is to be determined whether this is a heat or work interaction.Analysis For this problem, the interior surfaces of the oven form the systemboundary, as shown in Fig. 2–22. The energy content of the oven obviouslyincreases during this process, as evidenced by a rise in temperature. Thisenergy transfer to the oven is not caused by a temperature difference betweenthe oven and the surrounding air. Instead, it is caused by electrons crossing thesystem boundary and thus doing work. Therefore, this is a work interaction.System boundaryElectric ovenHeating elementEXAMPLE 2–6Heating of an Oven by Heat TransferAnswer the question in Example 2–5 if the system is taken as only the air inthe oven without the heating element.Solution The question in Example 2–5 is to be reconsidered by taking thesystem to be only the air in the oven.Analysis This time, the system boundary will include the outer surface ofthe heating element and will not cut through it, as shown in Fig. 2–23.Therefore, no electrons will be crossing the system boundary at any point.Instead, the energy generated in the interior of the heating element will betransferred to the air around it as a result of the temperature differencebetween the heating element and the air in the oven. Therefore, this is aheat transfer process.Discussion For both cases, the amount of energy transfer to the air is thesame. These two examples show that an energy transfer can be heat or work,depending on how the system is selected.Electrical WorkIt was pointed out in Example 2–5 that electrons crossing the system boundarydo electrical work on the system. In an electric field, electrons in a wire moveunder the effect of electromotive forces, doing work. When N coulombs of electricalcharge move through a potential difference V, the electrical work done isW e VNwhich can also be expressed in the rate form asW # e VI1W2(2–18)where W . e is the electrical power and I is the number of electrical charges flowingper unit time, that is, the current (Fig. 2–24). In general, both V and I varywith time, and the electrical work done during a time interval t is expressed as2W e VI dt1kJ21(2–19)When both V and I remain constant during the time interval t, it reduces toW e VI¢t1kJ2(2–20)FIGURE 2–22Schematic for Example 2–5.System boundaryElectric ovenHeating elementFIGURE 2–23Schematic for Example 2–6.W e = VI= I 2 R= V 2 /RFIGURE 2–24Electrical power in terms of resistanceR, current I, and potential difference V.RIV