The Thermophysical
Properties of Fluid Systems page in the NIST WebBook lists the heat capacity of liquid heptane at various temperatures. 
Use this resource to determine the constant volume specific heat of liquid heptane at 7^{o}C and 1
atm. How many intensive
properties of liquid heptane
can be independently specified? 


Read : 
Looking up the heat
capacity in the NIST WebBook
is straightforward. 

Gibbs
Phase Rule will tell us how many intensive variables can be independently specified. 










Diagram: 
A diagram is not
necessary for this problem. 










Given: 
T 
5 
^{o}C 



P 
1 
atm 










Find: 
C_{V} 
??? 
J/kg K 



^{o}Free = 
??? 











Assumptions: 
 The mixture is at
equilibrium 















Equations
/ Data / Solve: 


















First, we determine
the constant volume specific heat at of liquid heptane. 











From the NIST WebBook, we can obtain C_{p} and C_{v} for heptane at 1 atm
and 5 ^{o}C.
Use the isobaric
option for a range of temperatures including 5^{o}C or use the isothermal option including a pressure of 1
atm.
Selecting the correct units makes this task easier. Use temperature in degrees Celsius and pressure in atmospheres. 











From the NIST WebBook, I obtained the
following data : 















Temp.
(C) 
Pressure (atm) 
Density (kg/m^{3}) 
Volume (m^{3}/kg) 
Internal Energy
(kJ/kg) 
Enthalpy (kJ/kg) 
Entropy (J/g*K) 
Cv (J/g*K) 
Cp (J/g*K) 

2 
1 
698.87 
0.0014309 
226.32 
226.18 
0.70175 
1.6984 
2.159 

3 
1 
698.04 
0.0014326 
224.16 
224.01 
0.69391 
1.7016 
2.1624 

4 
1 
697.2 
0.0014343 
222.00 
221.85 
0.68609 
1.7048 
2.1657 

5 
1 
696.37 
0.001436 
219.83 
219.68 
0.67828 
1.7080 
2.1690 

6 
1 
695.54 
0.0014377 
217.66 
217.51 
0.67049 
1.7112 
2.1724 

7 
1 
694.7 
0.0014395 
215.48 
215.34 
0.66272 
1.7144 
2.1758 

8 
1 
693.87 
0.0014412 
213.31 
213.16 
0.65496 
1.7177 
2.1792 











C_{P}
= 
2169 
J/kg K 



C_{V}
= 
1708 
J/kg K 











Degrees of Freedom: 


Gibbs
Phase Rule is: 
^{o}Free = 2 + C  P 












^{o}Free = 
Degrees of freedom or
the number of intensive properties that can be independently specified 

^{} 









C = 
Number of chemical
species within the system 





C = 
1 
species 

















P = 
Number of phases 







P = 
1 
liquid phase 
















^{o}Free = 2 + 1  1 = 
2 

















Note: 









We only need 2 intensive
properties, such as: 





to completely determine
the state of the system. 















Verify: 
The equilibrium
assumption cannot be verified from the data available in this problem. 










Answers : 
C_{V} 
1708 
J/kg K 



^{o}Free 
2 










