EGR 334 Thermodynamics
Chapter 3: Section 9-10
Lecture 08:
Specific Heat Capacity
Quiz Today?
Today’s main concepts:
Introduce Specific Heats, cv and cP.
Understand when it is acceptable to apply specific heat.
Calculate changes of energy using specific heats
For liquids and solids, the saturated properties values may be
used to approximate properties values for supercooled property
values, if the model can be treated as incompressible.
• For incompressible liquids and solids, cv ≈ cp
• the specific heat ratio is defined as k = cp/cv
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Reading Assignment:
• Read Chap 3: Sections 11
Homework Assignment:
From Chap 3: 49, 55,68, 78
3
Think back to those simple, happy, carefree days of your
youth, back when you were in Physics lab….
You performed experiments to
try and figure out how the
world around you worked.
… it requires 1 calorie
of heat to raise the
temperature of
1 gram of water
by 1 deg C.
Q  cm  T
In one experiment you dropped hot
aluminum pellets into a container of
water and measured the change of
temperature the water underwent.
Using that information you found
a property of aluminum called
specific heat capacity.
Sec 3.9 : Specific Heat
4
In Thermodynamics, two common forms
of specific heat capacity are used.
 u 
cv  

 T v
Constant Volume
process
-----
cp
and
 h 


 T  p
Constant Pressure
process
Used to calculate changes of energy between states
Defined for pure, simple compressible substances
May be used only under certain “special conditions”
Usually applied to “ideal gas” model
When it can be applied: (if cv and cp are often treated as constants)
u 2  u 1  c v (T 2  T1 )
and
h 2  h1  c p (T 2  T1 )
Sec 3.9 : Specific Heat
5
Specific Heats
isobars
T2
the internal energy change or enthalpy
change when heat is added at constant
pressure or constant volume.
T
At constant volume
U  f T , V
T1

v
This is an exact differential
 U 
 U 
 U 
dU  
 dT  
 dV  c V dT  
 dV
  T V
 V T
 V T
U 
where 
  cV
 U 
  T V
At constant volume, dV=0; thus, 
 dV  0
 V T
Finally:
dU  cV dT   U 

T2
T1
cV dT
Sec 3.9 : Specific Heat
6
isobars
At Constant Pressure H  f T , P 
T2
This is an exact differential
T
 H 
 H 
dH  
 dT  
 dP
 T  P
 P T
 H 
 c P dT  
 dP
 P T
At constant pressure, dP=0; thus,
Finally,
T1
v
 H 
  cP
where 
 T  P
 H 

 dP  0
 P T
dH  c P dT   H 

T2
T1
c P dT
Sec 3.9 : Specific Heat
7
Specific Heat Ratio:
k 
cP
cV
For air (at 68oF (20oC) and 14.7 psia (1 atm)):
k a ir 
1 .0 1 kJ / kg  K
0 .7 1 8 kJ / kg  K
0 .2 4 B tu / lb m  F
o

0 .1 7 B tu / lb m  F
o
 1 .4
Fig. 3.9 Shows that cv and cp for water/steam
vary with temperature and pressure.
In practice, specific heats will be used as constants which are looked up on
tables based on standard values of temperature and pressure.
Example of Table:
Gases - Specific Heats and Individual Gas Constants
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Evaluating Properties of Liquids and Solids:
While Appendix A does have a table for super-cooled
water, for many other liquids, a super-cooled table is not
available.
What to do?
For liquids and solids, it is acceptable practice to
approximate
v (T , p )  v f (T )
and
u (T , p )  u f (T )
h (T , p )  h f (T )
if a super cooled table is not available.
Sec 3.10.1 : Approximations for Liquids using Saturated Data
For a liquid, there is little change in v, u, h, s at different pressure
and fixed T. Therefore,
v f  T , P   v f  T
u f  T , V
hf
 uf
 T , P   h f

 T 
 T 
Evaluate liquids at the
saturated state
Since these properties are essentially only a function of T and not P,
we call them Incompressible.
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Example:
What is the enthalpy for Refrigerant 22 at
T = 10 deg F. and p = 40 psi.
(Refer to Table A-7E)
Temperature
pressure
T=10 deg C
*
spec. vol.
h (T , p )  h f (T )
*
o
p = 40 psi
spec. vol.
at T=10 deg….hf=13.33 Btu/lbm
Sec 3.10.2 : Incompressible Substance Model
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Incompressible Substance: Includes any substance whose properties
do not change with pressure.
For liquids and solids: v f , s  co n stan t
Thus c
V
u f ,s  u T

h f ,s  h T

dU
 U 


c


V
dT
  T V
But, h T , P   u T   PV
So now what? Take partial with respect to T
 H 
 U 
  PV 
 U 

 
 
 
 0
 T  P
 T  P  T  P
 T  P
Thus,   H 
dU
 c P  cv  c

 
dT
 T  P
Sec 3.10.2 : Incompressible Substance Model
14
Use the heat capacity to calculate the change in internal energy.
 u  u 2  u1 

T2
T1
c T dT
 h  h 2  h1  u 2  u 1  v  P2  P1  

T2
T1
c T dT  v  P2  P1 
If c = constant then that means,
v  P2  P1  is small and can usually be dropped
Therefore,
 h  h 2  h1 

T2
T1
c T dT  c T 2  T1 
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Table of specific heat for incompressible materials.
See the course website for the complete tables of specific
heats for both compressible and incompressible materials.
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End of Slides for Lecture 08