# Example Problem with Complete Solution

3C-3 : Liquid Heat Capacities and Specific Heats 2 pts
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 7oC 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 oC P 1 atm
Find: CV ??? J/kg K oFree = ???
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 Cp and Cv for heptane at 1 atm and 5 oC.  Use the isobaric option for a range of temperatures including 5oC 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/m3) Volume (m3/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
CP = 2169 J/kg K CV = 1708 J/kg K
Degrees of Freedom: Gibbs Phase Rule is:  oFree = 2 + C - P
oFree =  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
oFree = 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 : CV 1708 J/kg K oFree 2 