Specific Heat
Macroscopically, specific heat c , is an intensive property
that represents the amount of heat that needs to be added
to 1 kg of material to raise its temperature by 1oK
c
For example :
Q
units :
MdT
KJ
kgo K
The value of c depends on the process of heat addition
For constant volume heat addition:
Q
U  Q  W
Mdu  Q  pdV
M
Hence define specific heat at constant volume as
c
Q
MdT

Mdu du

MdT dT
 cV
du  cV dT
V const
In general cv depends on P and T, so
u 2  u1  1 du  1 cV ( P, T )dT
2
2
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For constant pressure heat addition:
Q
M
U  Q  W
Mdu  Q  pdV
Mdu  Q  p ( Mdv)
Q  M (du  Pdv)
Define new property, enthalpy h
h  u  Pv  dh  du  Pdv  vdP
For constant P process dP=0, so dh = du + Pdv
Substituting into the definition for c
c
Q
MdT

M (du  Pdv) dh

Mdt
dT
 cP
dh  cP dT
P const
In general cP also depends on P and T, so
h2  h1  12 dh  12 c P ( P, T )dT
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Ideal Gas Assumption
For an ideal gas cV and cP only vary with T, not P
u 2  u1  12 cv (T )dT
h2  h1  12 c P (T )dT
Need absolute u and h. Set T2 to T and T1 to Tref
h2  h1  h(T )  h(Tref )  TT c P (T )dT
ref
Take Tref =0 (absolute zero temperature) where h=0
h(T )  0Tc P (T )dT
h (T )  0Tc P (T )dT
for an ideal gas pv  RT
 h  u  pv  u  RT  u  h  RT
u (T )  0Tc P (T )dT  RT
u (T )  0Tc P (T )dT  R T
cp values are tabulated in Table A-20 and Table A-21
gives an empirical relation for (c p / R ) as a function of T
for various gases.
62
Enthalpy and Internal energy as a function of T for air is
given in Table A-22, and for other gases in A-23.
Get cV from cP differentiate h = u + RT with respect to T
dh du

 R  c P (T )  cV (T )  R
dT dT
dh du

 R  c P (T )  cV (T )  R
dT dT
Note :
Since R > 0 and cV > 0  cp > cV
The specific heat ratio k is commonly used, for ideal gas
k (T ) 
cP
cV
since c P  cV  k  1 see Table A - 20
For a substance with molar mass  if you know k at a
specific temperature can get cp and cV
c P (T )  cV (T )  R
c P (T )  cV (T )  R
c P (T )
R
 1
cV (T )
cV (T )
c P (T ) 
cV (T ) 
R
K (T )  1
R
R
K (T )  1
K (T ) R
c P (T ) 
K (T )  1
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Example:
A tank contains 0.042 m3 of oxygen (O2) at 21C and 15 MPa.
Determine the mass of the oxygen, in kg, using
a) the compressibility chart
b) ideal gas model
Comment on the applicability of the ideal gas model
O2
V= 0.042 m3
T= 21 C (294 K)
P= 15 MPa (150 bar)
a) From table A-1for oxygen: molar mass = 32 kg/kmol,
Tc= 126K, Pc=50.5 bar
PR 
P 150 bar
T 294 K

 2.97 ; TR  
 1.91
Pc 50.5 bar
Tc 126 K
From the Generalized Compressibility Chart Z  0.92
ZRT 0.92(8314/32 J/kg  K)(294K)
3
v


0.0047
kg/m
P
150x105 N/m 2
V
0.042 m 3
M 
 8.94 kg
3
v 0.0047 /m /kg
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b) Assuming ideal gas
PV (150x105 N/m 2 )(0.042m 3 )
M 

 8.24 kg
RT
(8314/32 J/kg  K)(294K)
Assuming the mass obtained from the chart is correct, the ideal gas model under predicts the
mass by about 8%, not bad!
Do the same problem with carbon dioxide (CO2):
a) PR 
P 150 bar
T 294 K

 2.03 ; TR  
 0.97
Pc 73.9 bar
Tc 304 K
From the Generalized Compressibility Chart Z  0.3
v = 0.0011 m3/kg  M = 38.2 kg
b) Mideal = 11.34 kg
Assuming the mass obtained from the chart is correct, the
ideal gas model under predicts the mass by about 70%,
bad!
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Example:
A piston-cylinder assembly contains 1 kg of nitrogen gas (N2).
The gas expands from an initial state where T1= 700K and P1= 5
bar to a final state where P2= 2 bar. During the process the
pressure and specific volume are related by Pv1.3= const.
Assuming ideal gas behaviour and neglecting KE and PE effects,
determine the heat transfer during the process, in KJ.
P
1
Q
Pv1.3= const
N2
2
P1= 5 bar P2= 2 bar
T1= 700K
v
U  Q  W
Q  U  W
W  12 pdV 
P2V2  P1V1
1 n
Recall, for an ideal gas PV  MRT
2
1
MR(T2 -T1 )
pdV 
1 n
need T2
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R
n 1
n
P1v1
Pv
P  const 
P  const 
 2 2  1  1 / n   2  1 / n 
T1
T2
T1  P1  T2  P2 
n 1
n
2
P1
P

T1
T2
T2  P2 
   
T1  P1 
n 1
n
v
 
v
n
1
n
2



n 1
n
T2  v1 
   
T1  v2 
n 1
0.3
1.3
 2 bar 
T2  
 (700K)  567K
 5 bar 
MR(T2 -T1 )
W
1 n
1 kg(8.314/28 kJ/kg  K)(567 - 700)K

 132kJ
1  1.3
Note: work is positive  work done by the system
The molar internal energy for nitrogen from Table A-23:
u (700K)  14,784 kJ/kmol
u (567K)  11,858 kJ/kmol
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u u 
U  M (u 2  u1 )  M  2 1 
  
 11,858- 14,784 kJ/kmol 
 1 kg 
  104.5 kJ
28 kg/kmol


Q= U + W = (-104.5 kJ) + (132 kJ)= +27.5 kJ
Note: heat transfer is positive heat transferred into the
system
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Constant Specific Heat Assumption
This assumption can be made if the specific heat does not
vary much in the temperature range of interest, use
average values c~V and c~P such that:
u 2  u1  TT cV (T )dT  c~V TT dT  c~V (T2  T1 )
h  h  T c (T )dT  c~ T dT  c~ (T  T )
2
2
1
1
2
2
1
T1
2
p
P T1
P
2
1
the average value is taken to be
c (T )  cV (T2 )
c~V  V 1
2
c (T )  c P (T2 )
c~P  P 1
2
Example:
Recalculate U in last example assuming constant
specific heat
The specific heats for nitrogen from Table A-20 for the two temperatures are
cV (700K)  0.801kJ/kg  K
cV (567K)  0.771kJ/kg  K
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The average value over the temperature range is thus
(0.801  0.771)
~
cV 
 0.786 kJ/kg  K
2
U  M (u  u )  Mc~ (T  T )
2
1
V
2
1
 1 kg (0.786kJ/kg  K)(567 - 700)K
 - 104.5 kJ
The value for U obtained using Table A-23 which takes
into account the change in cV with temperature is exactly
the same!!
If tables are not available can approximate cV by using the
specific heat ratio K, which in the temperature range of
300K to 1000K, is taken to be constant (see Table A-20):
Diatomic molecules
 K= 7/5= 1.4
(N2, O2, H2, CO, “Air”,…)
Monatomic molecules
(Ar, He, Ne,…)
cV 
 K=5/3=1.67
R
8.314/28kJ/kg  K

 0.742 kJ/kg  K
K 1
1.4 - 1
compared to 0.786 kJ/kgK calculated above
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Incompressible Assumption
A substance is incompressible if there is negligible
change in specific volume with change in pressure,
e.g., liquids and solids
The specific internal energy of an incompressible
substance does not vary much with pressure, therefore the
specific heat only depends on temperature
cV (T ) 
du(T )
dT v const
By definition enthalpy varies both with P and T
h(P,T) = u(T) + Pv
For incompressible v is constant, differentiating with
respect to T while maintaining P constant yields,
dh(T )
du(T )

dT P const
dT v const
cP = cv = c
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u 2  u1  12 c(T )dT
Also, Δh  Δ(u  Pv )
h2  h1  (u 2  u1 )  ( P2 v2  P1v1 )
h2  h1  12 c(T )dT  v( P2  P1 )
Two special cases:
i) Constant pressure heat addition into an incompressible
substance (P= const.)
h2  h1  12 c(T )dT
ii) Isothermal heat addition into an incompressible
substance (T= const.)
h2  h1  v( P2  P1 )
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Approximating enthalpy in the compressed liquid
region of steam tables
Want to get enthalpy at (P*, T*) without using
compressible liquid tables
Liquid at
(P*,T*)
Liquid at
(P*,T*)
Saturation line
P* isobar
T*
P* isobar
T1=T*
P1(= Psat(T*)) isobar
Saturation state (P1,T1)
Consider the isothermal process (incompressible) from
(P1,T1)  (P*,T*) we now know h*- h1=v(P*- P1)
h*  h1  v( P *  P1 )
h( P*, T *)  h f sat (T *)  v f sat (T *)( P *  Psat (T *))
This gives a more accurate result than just taking
h( P*, T *)  h f sat (T *) as proposed earlier for u(P*,T*)
The last term is a correction which is normally small
because vfsat is very small
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