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? |
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Read : |
Looking up the heat
capacity in the NIST WebBook
is straightforward. |
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Gibbs
Phase Rule will tell us how many intensive variables can be independently specified. |
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Diagram: |
A diagram is not
necessary for this problem. |
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Given: |
T |
5 |
oC |
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P |
1 |
atm |
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Find: |
CV |
??? |
J/kg K |
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oFree = |
??? |
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Assumptions: |
- The mixture is at
equilibrium |
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Equations
/ Data / Solve: |
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First, we determine
the constant volume specific heat at of liquid heptane. |
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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. |
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From the NIST WebBook, I obtained the
following data : |
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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) |
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2 |
1 |
698.87 |
0.0014309 |
-226.32 |
-226.18 |
-0.70175 |
1.6984 |
2.159 |
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3 |
1 |
698.04 |
0.0014326 |
-224.16 |
-224.01 |
-0.69391 |
1.7016 |
2.1624 |
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4 |
1 |
697.2 |
0.0014343 |
-222.00 |
-221.85 |
-0.68609 |
1.7048 |
2.1657 |
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5 |
1 |
696.37 |
0.001436 |
-219.83 |
-219.68 |
-0.67828 |
1.7080 |
2.1690 |
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6 |
1 |
695.54 |
0.0014377 |
-217.66 |
-217.51 |
-0.67049 |
1.7112 |
2.1724 |
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7 |
1 |
694.7 |
0.0014395 |
-215.48 |
-215.34 |
-0.66272 |
1.7144 |
2.1758 |
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8 |
1 |
693.87 |
0.0014412 |
-213.31 |
-213.16 |
-0.65496 |
1.7177 |
2.1792 |
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CP
= |
2169 |
J/kg K |
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CV
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1708 |
J/kg K |
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Degrees of Freedom: |
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Gibbs
Phase Rule is: |
oFree = 2 + C - P |
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oFree = |
Degrees of freedom or
the number of intensive properties that can be independently specified |
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C = |
Number of chemical
species within the system |
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C = |
1 |
species |
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P = |
Number of phases |
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P = |
1 |
liquid phase |
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oFree = 2 + 1 - 1 = |
2 |
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Note: |
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We only need 2 intensive
properties, such as: |
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to completely determine
the state of the system. |
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Verify: |
The equilibrium
assumption cannot be verified from the data available in this problem. |
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Answers : |
CV |
1708 |
J/kg K |
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oFree |
2 |
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