Microsoft Word - Cengel and Boles TOC _2-03-05_.doc

186 | Thermodynamics

EXAMPLE 4 –9 Heating of a Gas by a Resistance Heater

A piston–cylinder device initially contains 0.5 m^3 of nitrogen gas at 400 kPa

and 27°C. An electric heater within the device is turned on and is allowed to

pass a current of 2 A for 5 min from a 120-V source. Nitrogen expands at

constant 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 electric

resistance heater. Nitrogen expands at constant pressure while some heat is

lost. The final temperature of nitrogen is to be determined.

Assumptions 1 Nitrogen is an ideal gas since it is at a high temperature and

low 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 changes

are zero, KE PE 0 and EU. 3 The pressure remains constant

during the process and thus P 2 P 1. 4 Nitrogen has constant specific heats

at room temperature.

Analysis We take the contents of the cylinder as the system(Fig. 4–31).

This is a closed systemsince no mass crosses the system boundary during

the process. We observe that a piston–cylinder device typically involves a

moving boundary and thus boundary work, Wb. Also, heat is lost from the

system and electrical work Weis done on the system.

First, let us determine the electrical work done on the nitrogen:

The mass of nitrogen is determined from the ideal-gas relation:

Under the stated assumptions and observations, the energy balance on the

system can be expressed as

Net energy transfer Change in internal, kinetic,

by heat, work, and mass potential, etc., energies

since UWbHfor a closed system undergoing a quasi-equilibrium

expansion or compression process at constant pressure. From Table A–2a,

cp1.039 kJ/kg · K for nitrogen at room temperature. The only unknown

quantity in the previous equation is T 2 , and it is found to be

Discussion Note that we could also solve this problem by determining the

boundary work and the internal energy change rather than the enthalpy

change.

T 2 56.7°C

72 kJ
2.8 kJ 1 2.245 kg 21 1.039 kJ>kg#K 21 T 2
27°C 2

We,in
Qout¢Hm 1 h 2 
h 12 mcp 1 T 2 
T 12

We,in
Qout
Wb,out¢U

Ein
Eout¬

¬

¢Esystem

m

P 1 V 1

RT 1



1 400 kPa 21 0.5 m^32

1 0.297 kPa#m^3 >kg#K 21 300 K 2

2.245 kg

WeVI ¢t 1 120 V 21 2 A 215 60 s2a

1 kJ>s

1000 VA

b72 kJ

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