# Specific Heats, Internal Energy and Enthalpy

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```					ME 211 – Addition to Class Notes
Chapter 2 – 4th Ed.

Specific Heats, Internal Energy and Enthalpy

Before the first law of thermodynamics can be applied to systems, ways to calculate
the change in internal energy of the substance enclosed by system boundary must be
determined. For real substances like water the property tables are used to find the
internal energy change. For ideal gases the internal energy is found by knowing the
specific heats. Physics defines the amount of energy needed to raise the temperature of a
unit of mass of a substance one degree as the specific heat at constant volume, CV, for a
constant-volume process and the specific heat at constant pressure, CP, for a constant-
pressure process. Recall that enthalpy, h, is the sum of the internal energy, u, and the
pressure-volume product, Pv.

h  u  Pv
In thermodynamics, the specific heats are defined as

CV 
Fu I

GJ                    CP 
Fh I

GJ
HT K
      v
and
HT K
      P

Additionally, we note, as pointed out in the Textbook { Equations 2-28 and 2-29, the
defining equations }, these equations are property relations and thus are
independent of the type of process. They are valid for any substance undergoing any
process. The terminology of constant pressure and constant volume if only related to the
method in which the respective specific heats are determined. This can be a major
point of confusion and needs a very thorough consideration by the reader.

Simple substance

The thermodynamic state of a simple, homogeneous substance is specified by giving any
two independent, intensive properties. Let's consider the internal energy to be a function
of T and v and the enthalpy to be a function of T and P as follows:

u  u(T , v)               and           h = h(T, P)
The total differential of u is

du 
Fu I dT  Fu I dv

G J GJ     
H K HK
T     v  v                 T

or
Fu I dv

du  C dT  G J
HK
v
v
T

The total differential of h is

dh 
Fh I dT  Fh I dP

GJ GJ      
HT K HP K
     P                    T

or

dh  CP dT 
Fh I dP

GJ
HP K
       T

Using thermodynamic relation theory we could evaluate the remaining partial derivatives
of u and h in terms of functions of P,v, and T. These functions depend upon the equation
of state for the substance. Given the specific heat data and the equation of state for the
substance, we can develop the property tables like the steam tables.
Ideal Gases

For ideal gases, we use thermodynamic function relation theory of Chapter 11 and the
equation of state (Pv= RT) to show that u, h, CV, and CP are functions of temperature
alone. Then for ideal gases,

CV  CV (T )
Fu I  0

GJ
and
Hv K
        T

CP  CP (T )
Fh I  0

GJ
and
HP K
         T

The ideal gas specific heats are written in terms of ordinary differentials as

FI
 GJ
du
CV
HK
dT           ideal gas

FI
 GJ
dh
CP
HK
dT           ideal gas

Figure 2-68, page 102 (4th Ed.) in the text shows how the specific heats vary with
temperature for selected ideal gases.
Figure 2- 68 (Text: 4th Ed.)

The differential changes in internal energy and enthalpy for ideal gases become

du  CV dT
dh  C P dT
The change in internal energy and enthalpy of ideal gases can be expressed as

u  u2  u1                z 1
2
CV (T )dT  CV ,ave (T2  T1 )

h  h2  h1                z1
2
C (T )dT  C
P                       P ,ave (T  T )
2
where CV,ave and CP,ave are average or constant values of the specific heats over the
1

temperature range. We will drop the ave subscript shortly.
2a
P                       2b
1                 2c
T2
T1

V
P-V diagram for several processes for an ideal gas

In the above figure an ideal gas undergoes three different process between the same two
temperatures.
Process 1-2a: Constant Volume
Process 1-2b: P = a + bV, a linear relationship
Process 1-2c: Constant Pressure

These ideal gas processes have the same change in internal energy and enthalpy because
the processes occur between the same temperature limits.

ua  ub  uc  CV (T )dT          z    2

1

ha  hb  hc  CP (T )dT          z1
2

To find u and h we often use average, or constant, values of the specific heats. Some
ways to determined these values are as follows:

1. The best average value (the one that gives the exact results):

See Table A-2c for variable (meaning a function of temperature) specific heat data.

Cv ,ave 
z1
2
CV (T )dT
T2  T1
and          CP ,ave 
z  2
CP (T )dT
1
T2  T1
2. Good average values are:

CV (T2 )  CV (T1 )                                       CP (T2 )  CP (T1 )
Cv ,ave                                  and       CP ,ave 
2                                                         2
and

Cv ,ave  CV (Tave )        and       CP ,ave  CP (Tave )
where
T2  T1
Tave 
2
3. Sometimes adequate (and most often used) values are the ones evaluated at 300 K:

Cv ,ave  CV (300 K )             and     CP ,ave  CP (300 K )

Let's take a second look at the definition of u and h for ideal gases. Just consider the
enthalpy for now.

h  h2  h1  CP (T )dT   z1
2

Let's perform the integral relative to a reference state where
h = href at T = Tref.

h  h2  h1              z  Tref

T1
CP (T )dT          z Tref
T2
CP (T )dT 

or
h  h2  h1               z T2

Tref
CP (T )dT           z T1

Tref
CP (T )dT 

 (h2  href )  (h1  href )

At any temperature, we can calculate the enthalpy relative to the reference state as

h  href         z T

Tref
CP (T )dT 

or
h  href         z T

Tref
CP (T )dT 

Similarly, the internal energy change relative to the reference state is

u  uref         z  T

Tref
CV (T )dT 

or
u  uref         z  T

Tref
CV (T )dT 

These last two relations form the basis of the air tables (Table A-17 on a mass basis) and
the other ideal gas tables (Tables A-18 through A-25 on a mole basis). When you review
Table A-17, you will find h and u as functions of T in K. Since the parameters Pr, vr, and
so apply to air in a particular process, called isentropic, you should ignore these
parameters until we study Chapter 6. The reference state for these tables is defined as

uref  0 at Tref  0 K
href  0 at Tref  0 K
We may note in the analysis to follow the “ave” notation is dropped. In many
applications for ideal gases, the values of the specific heats at 300 K given in Table A-2
Relation between CP and CV for ideal gases

Using the definition of enthalpy (h = u + Pv) and writing the differential of enthalpy,
the relationship between the specific heats for ideal gases is

h  u  Pv
dh  du  d ( RT )
CP dT  CV dT  RdT
CP  CV  R

where R is the particular gas constant. The specific heat ratio, k, (fluids texts often use
 instead of k) is defined as

CP
k
CV

Extra Problem

Show that

kR                                R
CP                   and         CV 
k 1                             k 1
Example 2-4

Two kilograms of air are heated from 300 to 500 K. Find the change in enthalpy by
assuming
a. Empirical specific heat data.
b. Air tables.
c. Specific heat at the average temperature.
d. Use the 300 K value for the specific heat.

a. Table A-2c gives the molar specific heat at constant pressure for air as

kJ
CP = 2811 + 01967 x10-2 T + 0.4802 x10-5 T 2 - 1966 x10-9 T 3
.     .                                  .
kmol - K
The enthalpy change per unit mole is

h  h2  h1            z1
2
CP (T )dT

=   z 500 K

300 K
(2811 + 01967 x10-2 T + 0.4802 x10-5 T 2
.     .
- 1966 x10-9 T 3 )dT
.
01967 x10-2 2 0.4802 x10-5 3
.
 (2811T +
.                   T +          T
2             3
1966 x10-9 4 500K
.
-            T ) 300K
4
kJ
 5909.49
kmol
kJ
5909.49
h           kmol  203.9 kJ
h     
M            kg           kg
28.97
kmol
kJ
H  mh  (2 kg )(203.9                        )  407.98 kJ
kg

b. Using the air tables, Table A-17, at T1 = 300K, h1 = 300.19 kJ/kg and at T2 = 500K,
h2 = 503.02 kJ/kg

kJ
H  mh  (2 kg )(503.02  30019)
.       405.66 kJ
kg
The results of parts a and b would be identical if Table A-17 had been based on the same
specific heat function listed in Table A-2.

c. Let’s use a constant specific heat at the average temperature.

Tave = (300 + 500)K/2 = 400 K. At Tave , Table A-2 gives CP = 1.013 kJ/(kg-K).
For CP = constant,
h  h2  h1  CP ,ave (T2  T1 )
kJ
 1013
.           (500  300) K
kg  K
kJ
 202.6
kg
kJ
H  mh  (2 kg )(202.6)                          405.2 kJ
kg

d. Using the 300 K value from Table A-2a, CP = 1.005 kJ/kg- K.

For CP = constant,
h  h2  h1  CP (T2  T1 )
kJ
 1005
.           (500  300) K
kg  K
kJ
 2010
.
kg
kJ
H  mh  (2 kg )(2010)
.                          402.0 kJ
kg

Extra Problem

Find the change in internal energy for air between 300 K and 500 K.

Discussion:
What Methods?

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