1

Chapter --IV Thermodynamics I

ENERGY ANALYSIS OF CLOSED SYSTEMS

4.1 MOVING BOUNDARY WORK
One form of mechanical work frequently encountered in practice is associated with
the expansion or compression of a gas in a piston–cylinder device. During this
process, part of the boundary (the inner face of the piston) moves back and forth.
Therefore, the expansion and compression work is often called moving boundary
work, or simply boundary work (Fig. 4–1). Some call it the P dV work.

Consider the gas enclosed in the piston–cylinder device shown in Fig. 4–2. The
initial pressure of the gas is P, the total volume is V, and the cross-sectional area of
the piston is A. If the piston is allowed to move a distance ds in a quasi-equilibrium
manner, the differential work done during this process is Figure 4-1

δ W b=Fds=PAds=P dV

Note in the above equation that P is the absolute pressure, which is always
positive. However, the volume change dV is positive during an expansion
process (volume increasing) and negative during a compression process (volume
decreasing). Thus, the boundary work is positive during an expansion process
and negative during a compression process.

The total boundary work done during the entire process as the piston moves is
obtained by adding all the differential works from the initial state to the final
state:
Figure 4-2
2
W b =∫ P dV
1

The quasi-equilibrium expansion process described is shown on a P-V

diagram in Fig. 4–3. On this diagram, the differential area dA is equal to P
dV, which is the differential work. The total area A under the process
curve 1–2 is obtained by adding these differential areas:
2 2
Area=A=∫ dA=∫ P dV
1 1

A gas can follow several different paths as it expands from state 1 to state
2. In general, each path will have a different area underneath it, and since
this area represents the magnitude of the work, the work done will be
different for each process (Fig. 4–4). This is expected, since work is a path Figure 4-3
Function (i.e., it depends on the path followed as well as the end states).

The cycle shown in Fig. 4–5 produces a net work output because the work done by the system during
the expansion process (area under path A) is greater than the work done on the system during the
compression part of the cycle (area under path B), and the difference between these two is the net
work done during the cycle (the colored area).
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Chapter --IV Thermodynamics I

Figure 4-4 Figure 4-5

Example 4-1

A rigid tank contains air at 500 kPa and 150°C. As a result of heat transfer to the surroundings, the
temperature and pressure inside the tank drop to 65°C and 400 kPa, respectively. Determine the
boundary work done during this process.

Solution

A sketch of the system and the P-V diagram of the process are shown in
Fig. 4–6. The boundary work can be determined to be
2
W b =∫ P dV =0 since dV =0
1

This is expected since a rigid tank has a constant volume and dV = 0 in

this equation. Therefore, there is no boundary work done during this
process. That is, the boundary work done during a constant-volume
process is always zero. This is also evident from the P-V diagram of the
process (the area under the process curve is zero). Figure 4-6

Example 4-2

A piston–cylinder device initially contains 0.4 m3 of air at 100 kPa and 80°C. The air is now
compressed to 0.1 m3 in such a way that the temperature inside the cylinder remains constant.
Determine the work done during this process.

Solution

For an ideal gas at constant temperature T0 (Fig. 4-7)

C
PV =mR T 0∨P=
V

where C is a constant. This leads to

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Chapter --IV Thermodynamics I

Figure 4-7
2 2 2
C dV V2 V2
W b =∫ P dV =∫ dV =C ∫ =C ln =P1 V 1 ln
1 1 V 1 V V1 V1

W b =( 100 kPa ) ( 0.4 m ) ln

3
( 0.1
0.4 )
=−55.5 kJ

The negative sign indicates that this work is done on the system (a work input), which is always the
case for compression processes.

Polytropic Process

During actual expansion and compression processes of gases, pressure

and volume are often related by PV n = C, where n and C are constants.
A process of this kind is called a polytropic process (Fig. 4–8). The
pressure for a polytropic process can be expressed as

P=C V −n

2 2
V −n +1
−V −n +1
P V −P1 V 1
W b =∫ P dV =∫ C V −n dV =¿ C 2 1
= 2 2 ¿
1 1 −n+1 1−n

Figure 4-8

Since C=P1 V n1=P2 V n2. For an ideal gas ( PV =mRT ) , this equation can also be written as

mR ( T 2−T 1 )
W b= n≠1
1−n

For the special case of n = 1 the boundary work becomes

( )
2 2
V2
W b =∫ P dV =∫ C V −1 dV =PV ln
1 1 V1

For an ideal gas this result is equivalent to the isothermal process discussed in the previous example.

Example 4-3

A piston–cylinder device contains 0.05 m3 of a gas initially at 200 kPa. At this state, a linear spring
that has a spring constant of 150 kN/m is touching the piston but exerting no force on it. Now heat is
transferred to the gas, causing the piston to rise and to compress the spring until the volume inside
the cylinder doubles. If the cross-sectional area of the piston is 0.25 m 2, determine (a) the final
pressure inside the cylinder, (b) the total work done by the gas, and (c) the fraction of this work done
against the spring to compress it.

Solution
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Chapter --IV Thermodynamics I

A sketch of the system and the P-V diagram of the process are shown in Fig. 4–9.
(a) The enclosed volume at the final state is

V 2=2 V 1=( 2 ) ( 0.05 m3 )=0.1 m3

Then the displacement of the piston (and of the

spring) becomes
3
V ( 0.1−0.05 ) m
x= = =0.2 m
A 0.25m2

The force applied by the linear spring at the final state is

Figure 4-9

F=kx =( 150 kN /m )( 0.2 m )=30 kN

The additional pressure applied by the spring on the gas at this state is

F 30 kN
P= = =120 kPa
A 0.25 m2

Without the spring, the pressure of the gas would remain constant at 200 kPa while the piston is
rising. But under the effect of the spring, the pressure rises linearly from 200 kPa to320 kPa at the
final state.

b) An easy way of finding the work done is to plot the process on a P-V diagram and find the area
under the process curve. From Fig. 4–9 the area under the process curve (a trapezoid) is determined
to be
( 200+320 ) kPa
W =area=
2
[ ( 0.1−0.05 ) m3 ]=13 kJ

Note that the work is done by the system.

(c) The work represented by the rectangular area (region I) is done against the piston and the
atmosphere, and the work represented by the triangular area (region II) is done against the spring.
Thus,
1
W spring = [ (320−200 ) kPa ] ( 0.05 m3 )=3 kJ
2

4.2 ENERGY BALANCE FOR CLOSED SYSTEMS

Energy balance for any system undergoing any kind of process was expressed as

( )( )
Net energy transfer by change∈internal ,kinetic
heat , work ∧mass = potential , etc . energies
E¿ −Eout ∆ E system
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Chapter --IV Thermodynamics I

or, in the rate form, as

( )( )
Rate of net energy transfer by Ra te of change ∈internal, kinetic
heat , work ∧mass = potential , etc . energies
Ė¿ − Ėout d Ėsystem /dt

For constant rates, the total quantities during a time interval Δt are related to the quantities per unit
time as
Q=Q̇ ∆t W =Ẇ ∆ t ∆ E=( dE/dt ) ∆ t ( kJ )

The energy balance can be expressed on a per unit mass basis as

e ¿ −e out =∆ esystem ( kJ /kg )

For a closed system undergoing a cycle, the initial and final states are identical,
and thus ΔEsystem = E2 - E1 = 0. Then the energy balance for a cycle simplifies to
Ein - Eout = 0 or Ein - Eout. Noting that a closed system does not involve any mass
flow across its boundaries, the energy balance for a cycle can be expressed in
terms of heat and work interactions
Figure 4-10
W net ,out =Qnet ,∈¿∨Ẇ =Q̇ ¿ net ,out net ,∈¿ ( for acycle ) ¿

That is, the net work output during a cycle is equal to net heat input (Fig. 4-10).

it is common practice to use the classical thermodynamics sign convention and to assume heat to be
transferred into the system (heat input) in the amount of Q and work to be done by the system (work
output) in the amount of W, and then to solve the problem. The energy balance relation in that case
for a closed system becomes

Qnet ,∈¿−W net ,out =∆ Esystem ∨Q−W=∆ E ¿

Where Q=Q net ,∈¿=Q −Q ¿ is the net heat input and W =W net ,out =W out −W ¿ is the net work output.
¿ out

Obtaining a negative quantity for Q or W simply means that the assumed direction for that quantity
is wrong and should be reversed.

Example 4-4

A piston–cylinder device contains 25

g of saturated water vapor that is
maintained at a constant pressure of
300 kPa. A resistance heater within
the cylinder is turned on and passes a
current of 0.2 A for 5 min from a 120-
V source. At the same time, a heat loss
of 3.7 kJ occurs. (a) Show that

Figure 4-11
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Chapter --IV Thermodynamics I

for a closed system the boundary work Wb and the change in internal energy ΔU in the first-law
relation can be combined into one term, ΔH, for a constant pressure process. (b) Determine the final
temperature of the steam.

Solution

The tank is stationary and thus the kinetic and potential energy changes are zero, ΔKE = ΔPE = 0.
Therefore, ΔE = ΔU and internal energy is the only form of energy of the system that may change
during this process.

We take the contents of the cylinder, including the resistance wires, as the system (Fig. 4–11). This is
a closed system since 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 W b. The
pressure remains constant during the process and thus P2 = P1. Also, heat is lost from the system and
electrical work We is done on the system.

(a) We take the direction of heat transfer Q to be to the system and the work W to be done by the
system. We also express the work as the sum of boundary and other forms of work (such as electrical
and shaft). Then the energy balance can be expressed as

E¿ −E out =E system

Q−W =∆U + ∆ KE+∆ PE

Q−W other −W b =U 2−U 1

For a constant-pressure process, the boundary work is given as W b =P0 ( V 2−V 1 ). Substituting this
into the preceding relation gives

Q−W other −P0 ( V 2−V 1 ) =U 2−U 1

However,

P0=P 2=P1 → Q−W other =( U 2 + P2 V 2 ) −( U 1 + P1 V 1 )

Also H=U + PV , and thus

Q−W other =H 2−H 1 ( kJ )

(b) The only other form of work in this case is the electrical work, which can be determined from

W e =VI ∆ t=( 120 V )( 0.2 A ) ( 300 s )=7.2 kJ

State I :P 1=300 kPa

sat . vapour }
h 1=h g @300 kPa=2724.9 kJ /kg
 7
Chapter --IV Thermodynamics I

we can use the general energy balance relation with the simplification that the boundary work is
considered automatically by replacing ΔU by ΔH for a constant-pressure expansion or compression
process:
W e ,∈¿−Q −W =∆ U ¿ out b

W e ,∈¿−Q out =∆ H =m ( h2−h1) ( since P=constant ) ¿

7.2 kJ −3.7 kJ =( 0.025 kg ) ( h2−2724.9 kJ /kg ) ⇒ h2=2864.9 kJ /kg

Now the final state is completely specified since we know both the pressure and the enthalpy. The
temperature at this state is

h2=2864.9 kJ /kg
2
}
State 2 : P1=300 kPa T =200℃ ( Table A−6 )

Therefore, the steam will be at 200°C at the end of this process.

Example 4-5

A rigid tank is divided into two equal parts by a

partition. Initially, one side of the tank contains
5 kg of water at 200 kPa and 25°C, and the
other side is evacuated. The partition is then
removed, and the water expands into the entire
tank. The water is allowed to exchange heat
with its surroundings until the temperature in
the tank returns to the initial value of 25°C.
Determine (a) the volume of the tank, (b) the
final pressure, and (c) the heat transfer for this process. Figure 4-12

Solution

We take the contents of the tank, including the evacuated space, as the system (Fig. 4–12). This is a
closed system since no mass crosses the system boundary during the process. We observe that the
water fills the entire tank when the partition is removed (possibly as a liquid–vapor mixture).

(a) Initially the water in the tank exists as a compressed liquid since its pressure (200 kPa) is greater
than the saturation pressure at 25°C (3.1698 kPa). Approximating the compressed liquid as a
saturated liquid at the given temperature, we find

v ≅ v f @ 25 ℃=0.001003 m3 /kg ≅ 0.001m3 /kg ( Table A−4 )

Then the initial volume of the water is

V 1=mv 1= (5 kg ) ( 0.001 m3 /kg )=0.005m3

The total volume of the tank is twice this amount:

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Chapter --IV Thermodynamics I

V tank =( 2 ) ( 0.005 m3 ) =0.01 m3

b) At the final state, the specific volume of the water is

V 2 0.01 m3 3
v 2= = =0.002 m /kg
m 5 kg

which is twice the initial value of the specific volume. This result is expected since the volume
doubles while the amount of mass remains constant.
3 3
At 25℃ : v f =0.001003 m / kg∧¿ v g =43.34 m /kg ( Table A−4 ) ¿

Since vf < v2 < vg, the water is a saturated liquid–vapor mixture at the final state, and thus the
pressure is the saturation pressure at 25°C:

P2=P sat@ 25 ℃=3.1698 kPa ( Table A−4 )

(c) Under stated assumptions and observations, the energy balance on the system can be expressed
as
Q¿ =∆ U =m ( u 2−u1 )

Notice that even though the water is expanding during this process, the system chosen involves fixed
boundaries only (the dashed lines) and therefore the moving boundary work is zero. Then W = 0
since the system does not involve any other forms of work. Initially,

u1 ≅ u f @ 25℃ =104.38 kJ /kg (Table A−4 )

The quality at the final state is determined from the specific volume information:

v 2−v f 0.002−0.001 −5
x 2= = =2.3 ×10
v fg 43.34−0.001
then,
u2=u f + x 2 u fg

¿ 104.38 kJ /kg+ ( 2.3 ×10 ) ( 2304.3 kJ /kg )=104.88 kJ /kg

−5

Substituting yields,

Q¿ =( 5 kg ) [ ( 104.88−104.38 ) kJ /kg ]=0.25 kJ

The positive sign indicates that the assumed direction is correct, and heat is transferred to the water.

4.3 SPECIFIC HEATS

We know from experience that it takes different amounts of energy to raise the temperature of
identical masses of different substances by one degree. For example, we need about 4.5 kJ of energy
to raise the temperature of 1 kg of iron from 20 to 30°C, whereas it takes about 9 times this energy
(41.8 kJ to be exact) to raise the temperature of 1 kg of liquid water by the same amount. Therefore,
 9
Chapter --IV Thermodynamics I

it is desirable to have a property that will enable us to compare the energy storage capabilities of
various substances. This property is the specific heat.
The specific heat is defined as the energy required to raise the temperature of a unit mass of a
substance by one degree. In general, this energy depends on how the process is executed. In
thermodynamics, we are interested in two kinds of specific heats: specific heat at constant volume
cv and specific heat at constant pressure cp.

Physically, the specific heat at constant volume c v can be viewed as the energy required to raise the
temperature of the unit mass of a substance by one degree as the volume is maintained constant. The
energy required to do the same as the pressure is maintained constant is the specific heat at constant
pressure cp.

c v dT =du at constant volume

or
c v= ( ∂∂ Tu ) v

Similarly, an expression for the specific heat at constant pressure cp can be obtained by considering a
constant-pressure expansion or compression process. It yields

c P= ( ∂∂hT ) P

A common unit for specific heats is kJ/kg · °C or kJ/kg · K. Notice that these two units are identical
since ΔT(°C) = ΔT(K), and 1°C change in temperature is equivalent to a change of 1 K. The specific
heats are sometimes given on a molar basis. They are then denoted by c v ∧c P and have the unit
kJ/kmol · °C or kJ/kmol · K.

4.4 INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF IDEAL

GASES

Using the definition of enthalpy and the equation of state of an ideal gas, we have

}
h=u+ P v h=u+ RT
P v=RT

Since u and h depend only on temperature for an ideal gas, the specific heats c v and cp also depend, at
most, on temperature only. Therefore, at a given temperature, u, h, c v, and cp of an ideal gas have
fixed values regardless of the specific volume or pressure.

Thus, for ideal gases, the partial derivatives can be replaced by ordinary derivatives. Then the
differential changes in the internal energy and enthalpy of an ideal gas can be expressed as

du=c v ( T ) dT
and
dh=c P ( T ) dT
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Chapter --IV Thermodynamics I

The change in internal energy or enthalpy for an ideal gas during a process from state 1 to state 2 is
determined by integrating these equations:
2
∆ u=u2−u 1=∫ c v (T ) dT ( kJ /kg )
1
and
2
∆ h=h2 −h1=∫ c P (T ) dT ( kJ /kg )
1

To carry out these integrations, we need to have relations for cv and cp as functions of temperature.
The specific heat functions in the above equations can be replaced by the constant average specific
heat values. Then the integrations in these equations can be performed, yielding

u2−u 1=c v , avg ( T 2−T 1 )

and
h2 −h1=c P ,avg ( T 2−T 1 )

The specific heat values for some common gases are listed as a function of
temperature in Table A–2b. The average specific heats c p,avg and cv,avg are
evaluated from this table at the average temperature (T1 + T2)/2, as shown
in Fig. 4–13. If the final temperature T2 is not known, the specific heats
may be evaluated at T1 or at the anticipated average temperature. Then T 2
can be determined by using these specific heat values. The value of T 2 can
be refined, if necessary, by evaluating the specific heats at the new average
temperature.
Another way of determining the average specific heats is to evaluate them
at T1 and T2 and then take their average.

Figure 4-13

Specific Heat Relations of Ideal Gases

A special relationship between cp and cv for ideal gases is

c P=c v + R ( kJ /kg . K )

When the specific heats are given on a molar basis, R in the above equation should be replaced by
the universal gas constant Ru

c P=c v + Ru

At this point, we introduce another ideal-gas property called the specific heat ratio k, defined as

cP
k=
cv
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Chapter --IV Thermodynamics I

The specific ratio also varies with temperature, but this variation is very mild. For monatomic gases,
its value is essentially constant at 1.667. Many diatomic gases, including air, have a specific heat
ratio of about 1.4 at room temperature.

Example 4-6

Air at 300 K and 200 kPa is heated at constant pressure to 600 K. Determine the change in internal
energy of air per unit mass, using (a) data from the air table (Table A–17), (b) the functional form of
the specific heat (Table A–2c), and (c) the average specific heat value (Table A–2b).

Solution

The internal energy change Δu of ideal gases depends on the initial and final temperatures only, and
not on the type of process. Thus, the following solution is valid for any kind of process.

(a) One way of determining the change in internal energy of air is to read the u values at T 1 and T2
from Table A–17 and take the difference:

u1=u @300 K =214.07 kJ /kg

u2=u @ 600 K =434.78 kJ /kg

Thus,
∆ u=u2−u 1=( 434.78−214.07 ) kJ /kg=220.71kJ /kg

(b) The c P ( T ) of air is given in Table A–2c in the form of a third-degree polynomial expressed as

2 3
c P ( T )=a+bT + c T + d T

where a = 28.11, b = 0.1967 x 10-2, c = 0.4802 x 10-5, and d = -1.966 x 10-9.

2 3
c v ( T ) =c P −Ru=( a−Ru ) + bT + c T + d T

2 T2

∆ u=∫ c v ( T ) dT =∫ [ ( a−R u )+ bT + c T 2+ d T 3 ] dT
1 T1

Performing the integration and substituting the values, we obtain

∆ u=6447 kJ /kmol

The change in the internal energy on a unit-mass basis is determined by dividing this value by the
molar mass of air (Table A–1):

∆ u 6447 kJ /kmol
∆ u= = =222.5 kJ /k g
M 28.97 kg /kmol

which differs from the tabulated value by 0.8 percent.

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Chapter --IV Thermodynamics I

(c) The average value of the constant-volume specific heat c v,avg is determined from Table A–2b at the
average temperature of (T1 + T2)/2 = 450 K to be

c v , avg=c v, @ 450 =0.733 kJ /kg . K

Thus,
∆ u=c v , avg ( T 2−T 1 )=( 0.733 kJ /kg . K ) [ ( 600−300 ) K ] =220 kJ / kg

Example 4-7

A piston–cylinder device initially contains 0.5 m3 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

The system is stationary and thus the

kinetic and potential energy changes
are zero, ΔKE = ΔPE = 0 and ΔE =
ΔU. The pressure remains constant
during the process and thus P2 = P1.
Nitrogen has constant specific heats at
room temperature.
We take the contents of the cylinder as
the system (Fig. 4–14). This is a closed
system since no mass crosses the Figure 4-14

system boundary during the process. We observe that a piston–cylinder device typically involves a
moving boundary and thus boundary work, W b. Also, heat is lost from the system and electrical work
We is done on the system.

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

W e =VI ∆ t=( 120 V )( 2 A ) ( 300 s )=72 kJ

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

P1 V 1 ( 400 kPa ) ( 0.5 m3 )

m= = =2.245 kg
R T 1 ( 0.297 kPa . m 3 /kg . K ) (300 K )

Under the stated assumptions and observations, the energy balance on the system can be expressed
as
W e ,∈¿−Q −W =∆ U ¿ out b, out

W e ,∈¿−Q out =∆ H =m ( h2−h1)=m c P ( T 2 −T 1) ¿

 13
Chapter --IV Thermodynamics I

From Table A–2a, cp = 1.039 kJ/kg · K for nitrogen at room temperature. The only unknown quantity
in the previous equation is T2, and it is found to be

72 kJ −2.8 kJ =( 2.245 kg )( 1.039 kJ / kg . K ) ( T 2−27 ℃ ) ⇒T 2 =56.7 ℃

Example 4-8

Piston–cylinder device initially contains

air at 150 kPa and 27°C. At this state, the
piston is resting on a pair of stops, as
shown in Fig. 4–15, and the enclosed
volume is 400 L. The mass of the piston is
such that a 350-kPa pressure is required
to move it. The air is now heated until its
volume has doubled. Determine (a) the
final temperature, (b) the work done by
the air, and (c) the total heat transferred
to the air.

Figure 4-15
Solution

We take the contents of the cylinder as the system (Fig. 4–15). This is a closed system since 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, the boundary work is done by the
system, and heat is transferred to the system.

(a) The final temperature can be determined easily by using the ideal-gas relation between states 1
and 3 in the following form:

P 1 V 1 P 3 V 3 ( 150 kPa ) ( V 1 ) ( 350 kPa ) ( 2 V 1 )

= ⇒ = ⇒ T 3=1400 K
T1 T3 300 K T3

(b)The work done could be determined by integration, but for this case it is much easier to find it
from the area under the process curve on a P-V diagram, shown in Fig. 4–15:

A=( V 2−V 1 ) P2=( 0.4 m 3) ( 350 kPa ) =140 m3 . kPa

Therefore, W 13=140 kJ

The work is done by the system (to raise the piston and to push the atmospheric air out of the way),
and thus it is work output

(c) Under the stated assumptions and observations, the energy balance on the system between the
initial and final states (process 1–3) can be expressed as
 14
Chapter --IV Thermodynamics I

Q¿ −W b , out =∆U =m ( u3−u1 )

The mass of the system can be determined from the ideal-gas relation:

P1 V 1 (150 kPa ) ( 0.4 m3 )

m= = =0.697 kg
R T 1 ( 0.287 kPa . m3 /kg . K ) (300 K )

The internal energies are determined from the air table (Table A–17) to be

u1=u @300 K =214.07 kJ /kg

u2=u @1400 K =1113.52 kJ /kg

Thus,
Q¿ =140 kJ + ( 0.697 kg ) [ ( 1113.52−214.07 ) kJ /kg ] =767 kJ

4.5 INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF SOLIDS

AND LIQUIDS

It can be mathematically shown that the constant-volume and constant-pressure specific heats are
identical for incompressible substances. Therefore, for solids and liquids, the subscripts on c p and cv
can be dropped, and both specific heats can be represented by a single symbol c. That is,.

c v =c P=c

Specific heat values for several common liquids and solids are given in Table A–3.

Internal Energy Changes

Like those of ideal gases, the specific heats of incompressible substances depend on temperature
only. Thus, the partial differentials in the defining equation of cv can be replaced by ordinary
differentials, which yield
du=c v dT =c ( T ) dT

The change in internal energy between states 1 and 2 is then obtained by integration:
2
∆ u=u2−u 1=∫ c v (T ) dT ( kJ /kg )
1

For small temperature intervals, a c value at the average temperature can be used and treated as a
constant, yielding
∆ u ≅C avg ( T 2−T 1) ( kJ /kg )
Enthalpy Changes

∆ h=∆u+ v ∆ P ≅ c avg ∆ T + v ∆ P ( kJ / kg )
 15
Chapter --IV Thermodynamics I

For solids, the term v ΔP is insignificant and thus ∆ h=∆u ≅ c avg ∆ T . For liquids, two special cases
are commonly encountered:
1. Constant-pressure processes, as in heaters (ΔP = 0): ∆ h=∆u ≅ c avg ∆ T .
2. Constant-temperature processes, as in pumps (ΔT = 0): Δh = v ΔP.

For a process between states 1 and 2, the last relation can be expressed as h 2 - h1 = v(P2 - P1). By
taking state 2 to be the compressed liquid state at a given T and P and state 1 to be the saturated
liquid state at the same temperature, the enthalpy of the compressed liquid can be expressed as

h@ P ,T ≅ hf @ T + v h @ T ( P−P sat@ T )

However, the contribution of the last term is often very small, and is neglected.

Example 4-9

Determine the enthalpy of liquid water at 100°C and 15 MPa (a) by using compressed liquid tables,
(b) by approximating it as a saturated liquid, and (c) by using the correction equation.

Solution

At 100°C, the saturation pressure of water is 101.42 kPa, and since P > P sat, the water exists as a
compressed liquid at the specified state.

(a) From compressed liquid tables, we read

}
P=15 MPa h=430.39 kJ /kg (Table A−7 )
T =100 ℃

This is the exact value.

(b) Approximating the compressed liquid as a saturated liquid at 100°C, as is commonly done, we
obtain
h ≅ hf @ 100℃ =419.17 kJ /kg

This value is in error by about 2.6 percent.

(c) From the enthalpy equation

h@ P ,T ≅ hf @ T + v h @ T ( P−P sat@ T )

¿ ( 419.17 kJ /kg ) + ( 0.001 m3 /kg ) [ ( 15,000−101.42 ) kPa ]=434.07 kJ /kg

Example 4-10

A 50-kg iron block at 80°C is dropped into an insulated tank that contains
0.5 m3 of liquid water at 25°C. Determine the temperature when thermal
equilibrium is reached.

Solution
 16
Chapter --IV Thermodynamics I

We take the entire contents of the tank as the system (Fig. 4–16). This is a closed system since no
mass crosses the system boundary during the process. We observe that the volume of a rigid tank is
constant, and thus there is no boundary work. The energy balance on the system can be expressed as
Figure 4-16

∆ U =0

The total internal energy U is an extensive property, and therefore it can be expressed as the sum of
the internal energies of the parts of the system. Then the total internal energy change of the system
becomes

∆ U sys=∆U iron+ ∆ U water =0

[ mc ( T 2−T 1 ) ]iron+ [ mc ( T 2−T 1) ]water =0

The specific volume of liquid water at or about room temperature can be taken to be 0.001 m 3/kg.
Then the mass of the water is

V 0.5 m3
m= = =500 kg
v 0.001m3 /kg

Sspecific heats of iron and liquid water are determined from Table A–3 to be c iron = 0.45 kJ/kg · °C
and cwater = 4.18 kJ/kg · °C. Substituting these values into the energy equation, we obtain

( 50 kg )( 0.45 kJ / kg . ℃ ) ( T 2−80 ℃ ) + ( 500 kg )( 4.18 kJ / kg .℃ ) ( T 2−25 ℃ )=0

T 2=25.6 ℃

Therefore, when thermal equilibrium is established, both the water and iron will be at 25.6°C. The
small rise in water temperature is due to its large mass and large specific heat.

Example 4-11

If you ever slapped someone or got slapped yourself, you probably remember the burning sensation.
Imagine you had the unfortunate occasion of being slapped by an angry person, which caused the
temperature of the affected area of your face to rise by 1.8°C (ouch!). Assuming the slapping hand
has a mass of 1.2 kg and about 0.150 kg of the tissue on the face and the hand is affected by the
incident, estimate the velocity of the hand just before impact. Take the specific heat of the tissue to be
3.8 kJ/kg · °C.

Solution

We analyze this incident in a professional manner without involving any emotions. First, we identify
the system, draw a sketch of it, and state our observations about the specifics of the problem. We
take the hand and the affected portion of the face as the system. This is a closed system since it
involves a fixed amount of mass (no mass transfer). We observe that the kinetic energy of the hand
decreases during the process, as evidenced by a decrease in velocity from initial value to zero, while
the internal energy of the affected area increases, as evidenced by an increase in the temperature.
 17
Chapter --IV Thermodynamics I

There seems to be no significant energy transfer between the system and its surroundings during this
process.

Under the stated assumptions and observations, the energy balance on the system can be expressed
as
E¿ −E out =∆ E system

0=∆ U affected tissue + ∆ KE hand

0=( mc ∆ T )affected tissue + [ m ( 0−V 2 ) /2 ]hand

That is, the decrease in the kinetic energy of the hand must be equal to the increase in the internal
energy of the affected area. Solving for the velocity and substituting the given quantities, the impact
velocity of the hand is determined to be

V hand =
√ 2 ( mc ∆ T )affected tissue
mhand

¿
√ 2 ( 0.15 kg )( 3.8 kJ /kg . ℃ ) ( 1.8℃ )
1.2 kg
=41.4 m/ s ( ¿ 149 kh/h )