Thermodynamics

186 | ThermodynamicsEXAMPLE 4–9Heating of a Gas by a Resistance HeaterA piston–cylinder device initially contains 0.5 m 3 of nitrogen gas at 400 kPaand 27°C. An electric heater within the device is turned on and is allowed topass a current of 2 A for 5 min from a 120-V source. Nitrogen expands atconstant pressure, and a heat loss of 2800 J occurs during the process.Determine the final temperature of nitrogen.Solution Nitrogen gas in a piston–cylinder device is heated by an electricresistance heater. Nitrogen expands at constant pressure while some heat islost. The final temperature of nitrogen is to be determined.Assumptions 1 Nitrogen is an ideal gas since it is at a high temperature andlow pressure relative to its critical-point values of 147°C, and 3.39 MPa.2 The system is stationary and thus the kinetic and potential energy changesare zero, KE PE 0 and E U. 3 The pressure remains constantduring the process and thus P 2 P 1 . 4 Nitrogen has constant specific heatsat room temperature.Analysis We take the contents of the cylinder as the system (Fig. 4–31).This is a closed system since no mass crosses the system boundary duringthe process. We observe that a piston–cylinder device typically involves amoving boundary and thus boundary work, W b . Also, heat is lost from thesystem and electrical work W e is done on the system.First, let us determine the electrical work done on the nitrogen:W e VI ¢t 1120 V2 12 A215 60 s2a1 kJ>sb 72 kJ1000 VAThe mass of nitrogen is determined from the ideal-gas relation:m P 1V 1RT 11400 kPa210.5 m 3 210.297 kPa # m 3 >kg # K21300 K2 2.245 kgUnder the stated assumptions and observations, the energy balance on thesystem can be expressed asE in E out ¢E systemNet energy transferby heat, work, and mass⎫ ⎪⎬⎪⎭⎫⎪⎬⎪⎭W e,in Q out W b,out ¢UChange in internal, kinetic,potential, etc., energiesW e,in Q out ¢H m 1h 2 h 1 2 mc p 1T 2 T 1 2since U W b H for a closed system undergoing a quasi-equilibriumexpansion or compression process at constant pressure. From Table A–2a,c p 1.039 kJ/kg · K for nitrogen at room temperature. The only unknownquantity in the previous equation is T 2 , and it is found to be72 kJ 2.8 kJ 12.245 kg211.039 kJ>kg # K21T2 27°C2T 2 56.7°CDiscussion Note that we could also solve this problem by determining theboundary work and the internal energy change rather than the enthalpychange.