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

where T 0 is the constant temperature of the system and Qis the heat transfer
for the internally reversible process. Equation 7–6 is particularly useful for
determining the entropy changes of thermal energy reservoirs that can
absorb or supply heat indefinitely at a constant temperature.
Notice that the entropy change of a system during an internally reversible
isothermal process can be positive or negative, depending on the direction
of heat transfer. Heat transfer to a system increases the entropy of a system,
whereas heat transfer from a system decreases it. In fact, losing heat is the
only way the entropy of a system can be decreased.

Chapter 7 | 335

EXAMPLE 7–1 Entropy Change during an Isothermal Process

A piston–cylinder device contains a liquid–vapor mixture of water at 300 K.

During a constant-pressure process, 750 kJ of heat is transferred to the

water. As a result, part of the liquid in the cylinder vaporizes. Determine the

entropy change of the water during this process.

Solution Heat is transferred to a liquid–vapor mixture of water in a piston–

cylinder device at constant pressure. The entropy change of water is to be

determined.

Assumptions No irreversibilities occur within the system boundaries during

the process.

Analysis We take the entire water(liquid 
vapor) in the cylinder as the

system (Fig. 7–4). This is a closed systemsince no mass crosses the system

boundary during the process. We note that the temperature of the system

remains constant at 300 K during this process since the temperature of a

pure substance remains constant at the saturation value during a phase-

change process at constant pressure.

The system undergoes an internally reversible, isothermal process, and

thus its entropy change can be determined directly from Eq. 7–6 to be

Discussion Note that the entropy change of the system is positive, as

expected, since heat transfer is tothe system.

¢Ssys,isothermal

Q

Tsys



750 kJ

300 K

2.5 kJ/K

7–2 ■ THE INCREASE OF ENTROPY PRINCIPLE

Consider a cycle that is made up of two processes: process 1-2, which is
arbitrary (reversible or irreversible), and process 2-1, which is internally
reversible, as shown in Figure 7–5. From the Clausius inequality,

or



2

1

¬

dQ

T



1

2

a

dQ

T

b

int rev

 0

¬

dQ

T

 0

T

T = 300 K = const.

∆Ssys = Q = 2.5 kJK

Q = 750 kJ

FIGURE 7–4

Schematic for Example 7–1.

SEE TUTORIAL CH. 7, SEC. 2 ON THE DVD.

INTERACTIVE

TUTORIAL