| 3C-3 : | Liquid Heat Capacities and Specific Heats | 2 pts |
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| The heat capacity of liquid heptane at various temperatures is tabulated in the NIST Webbook. Determine the constant volume specific heat of liquid heptane at 5°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 variable can be independently specified. | |||||||||
| Given : | T | 5 |
°C | P = | 1 |
atm | |||
| Find : | CV | ??? |
J/kg K | °Free = | ??? |
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| Assumption: | - The mixture is at equilibrium | ||||||||
| Solution : | 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°C. Use the isobaric option for a range of temperatures including 5°C or use the isothermal option including a pressure of 1 atm. Selecting the correct units makes this task easier. Use temperature in °C 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 |
-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: | °Free = 2 + C - P | |||||||
| °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 | |||||||
| °Free = 2 + 1 - 1 = | 2 | ||||||||
| Note: | We only need 2 intensive properties, such as: | ||||||||
| to completely determine the state of the system. | |||||||||
Example Problem with Complete Solution
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