1. Specific Heats Chapter 4

2. Importance We know it takes more energy to warm up some materials than others
For example, it takes about 10 times as much energy to warm up 1 kg of water, as it does to warm up the same mass of iron.
it is desirable to have a property that will enable us to compare the energy storage capabilities for different substances.

3. Specific Heat (heat capacity)
Definition: It is the energy required to raise the temperature of a unit mass by one degree
Units
kJ/(kg 0C) or kJ/(kg K)
cal/(g 0C) or cal/(g K)
Btu/(lbm 0F) or Btu/(lbm R)
Simple mathematical definition:
If DT = 1 degree and m = 1 kg the DE = C

4. Constant volume and Constant pressure specific heats, Cv and Cp
Cv can be viewed as energy required to raise the temperature of a unit mass by one degree as the volume is maintained constant.
Cp can be viewed as energy required to raise the temperature of a unit mass by one degree as the pressure is maintained constant.
Cp > Cv

5. Mathematical form of Cv
Consider constant volume system. Heat it from T1 to T2.
E=U+KE +PE
DE = DU
dE = dU
But dE= mCvdT
du = CvdT

6. We’ll worry about the math later, but…
Mathematical form of Cp
h includes the internal energy and the work required to expand the system boundaries

7. Observations
Cp is always bigger than Cv. This is because it takes more energy to warm up a constant pressure system due to the system boundaries expansion.
That is you need to provide the energy to
increase the internal energy
do the work required to move the system boundary

8. Observations (continued)
Both are expressed in terms of u or h, and T, which are properties and thus Cv and Cp are also properties (get them from a table).
Because they are properties, they are independent of the type of process!! They are valid for any substance undergoing any process. The constant volume in Cv this is how the values of Cv are determined.

9. In general Cp, Cv are f(T,P)
cv and cp are expressed in terms of other properties
cv and cp must be properties themselves.
a substance depend on the state that, in general, is specified by two independent, intensive properties.
the energy required to raise the temperature of a substance by one degree is different at different temperatures and pressure

10. Cv dependence on T for an ideal gas.
Recall that :
Joule found experimentally that the internal energy of an ideal gas is a function of temperature only
Hence, Cv is at most function of T for an ideal gas.
The partial derivative becomes ordinary derivative for an ideal gas. .

11. Joule’s Observations
Joule found that after opening the valve ( pressure and volume of air changed)
Water temperature Still the same
Since no work done then
the internal energy of air did not change
Even the V and P changed
 u of the air must be f(T) only

12. Cp dependence on T for an ideal gas.
Recall that
Hence, Cp is at most function of T for an ideal gas.
The partial derivative becomes ordinary derivative for an ideal gas.

13. Internal energy and Enthalpy as functions of Cv and Cp for an ideal gas.

14. To carry out these integrations, we need Cv and Cp as functions of T.
Analytical expressions are available in Table A-2c.
In this table, Cp is given as
Cp = a + bT + cT2 + dT3
The constants a, b, c, and d are tabulated for various gases.

15. Method 1
This is inconvenient!! Only do it if you really need to be very accurate!!
Isn’t there an easier way?

16. Method 2
These integrations of Eqs and 4-26 were tabulated in Table A-17 page 910.
T0=0 K (u, h assigned =0) was chosen to be an arbitrary reference.
This choice has no effect on
The u and h data are given in KJ/kg for air.
Other gases in KJ/Kmol ( table A18-27).

17. Method 3
The variation of specific heat for gases with complex molecules are higher and increase with temperature.
The variation of specific heats is smooth and can be approximated as linear over small temperature interval (a few hundred degrees or less)

18. Method 3
Assume Cp and Cv is constant over a short temperature range (a few hundred degrees or less).
The constant specific heats are evaluated at the average temperature (T1+T2)/2.
u2-u1=Cv,av(T2-T1)
h2-h1=Cp,av(T2-T1)
Table A2

19. Three Ways to Calculate Du

20. Important Observation
The previous relations are not restricted to any kind of process.
The presence of constant volume specific heat in an equation should not lead to the concept that this equation is valid only for constant volume process.
The constant volume or constant pressure part of the name defines only how they are measured for each substance (see figure).
Once we have Cv or Cp as function of T, we can perform the integration for any process.
Recall, Cp and Cv is F(T) only for an ideal gas
Du = Cv DT for any ideal gas undergoing any process.

21. Specific heat relations of an Ideal Gas
Cp is modeled in the Appendix A-2(c) as a function of temperature – so you could calculate dh, but what if you want to calculate du?
You’d need Cv. There is no corresponding Cv table !!
Recall that h=u+RT
dh = du + RdT
CpdT=CvdT+ RdT
Cp=Cv + R
Cp-Cv = R

22. Specific Heat Ratio
k also varies with temperature, but this variation is very mild then k = constant.
k = 1.4 for diatomic gases (like air)
k = for noble gases (Monatomic gases)
Diatomic gases: having two atoms in the molecule N2 ;
Monatomic gases: consisting of one atom Ar

23. Example 4-7
Air at 300 K and 200 kPa is heated at constant pressure to 600 K. Determine the change in the internal energy of air per unit mass using:
Data from air tables (Table A-17)
The functional form of the specific heat (Table A-2c)
The average specific heat value (Table A-2b).

26. EXAMPLE 4–9 Heating of a Gas by a Resistance Heater
A piston–cylinder device initially contains 0.5 m3 of nitrogen gas at 400 kPa and 27°C.
An electric heater within the device is turned on and is allowed to pass a current of 2 A for 5 min from a 120-V source.
Nitrogen expands at constant pressure, and a heat loss of 2800 J occurs during the process.
Determine the final temperature
of nitrogen.

28. Specific heats of Solids and Liquids
The specific volume of incompressible substances remain constant during a process.
The Cv and Cp values of incompressible substances are identical and are denoted by C.
Cp = Cv = C

29. Internal energy of Solids and Liquids
C in Table A3

30. Enthalpy of Solids Small for solids
But dv is 0 if the system is incompressible
Small for solids

31. Enthalpy of Liquids
We have two cases: Constant pressure process, as in heaters Constant temperature process, as in pumps

32. Enthalpy of Liquids for constant temperature
Consider state 2 to be the compressed liquid state at a given T and P and state 1 to be the saturated liquid state at the same temperature, the enthalpy of the compressed liquid can be expressed as
This is an improvement over the assumption that the enthalpy of the compressed liquid could be taken as hf at the given temperature
BUT The improvement often is very small

33. Example 4-11

34. Example 4-11….

35. Example 4-12

36. Example 4-11

39. Take the system as the hand and the affected portion of the face ( the is a closed system)