Usage of General Specific Heats Cp0 and Cv0

In summary, the second equation in the solution for the question states that the transformation of the heat energy (hi) into the thermal energy (Cp*Ti) is valid only when the temperature range of interest is considered. Additionally, the equations must be valid at constant volume and pressure, and the mass added between the states must not change the results.
  • #1
messier992
8
0
Homework Statement
What conditions for usage of the general specific heats?
Relevant Equations
u2-u1=~Cv0(T2-T1)
h2-h1=~Cp0(T2-T1)
Cv=du/DT at constant volume
Cp=dh/dT at constant Pressure
Q-W=me*he-mi*hi+m2*u
1. I am rather confused about the usage of Cp0 and Cv0 in the solutions to the question below:
242473

In the solution to the depicted question, the solution involves setting up the equation as follows:
Q-W=me*he-mi*hi+m2*ue-m1*u1
-W=-mi*Cp*Ti+m2CvT2-m1CvT1
I don't follow for the second step. How can the transformation of hi, u1, and u2 be valid? I thought the relation between Cp0 and Cv0 only held over an interval due to the nature of the derivation of the equations:
Cv=δu/δT at constant volume => ∫Cvδu=∫δT => u2-u1=~Cv0(T2-T1)
Cp=δh/δT at constant Pressure = ∫Cvδu=∫δT => h2-h1=~Cp0(T2-T1)2. Additionally, what are the conditions for the usage of these equations?
u2-u1=~Cv0(T2-T1)
h2-h1=~Cp0(T2-T1)
I thought they were only valid at constant volume and pressure? In the case above, the pressure/volume changes linearly, are the relations still valid then?
Lastly, mass is also added between states 1 and 2, in the case above. I understand the equation relates specific terms, per unit mass, but does this addition of mass pose a problem?
 
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  • #2
messier992 said:
1. I am rather confused about the usage of Cp0 and Cv0 in the solutions to the question below:
View attachment 242473
In the solution to the depicted question, the solution involves setting up the equation as follows:
Q-W=me*he-mi*hi+m2*ue-m1*u1
-W=-mi*Cp*Ti+m2CvT2-m1CvT1
I don't follow for the second step. How can the transformation of hi, u1, and u2 be valid? I thought the relation between Cp0 and Cv0 only held over an interval due to the nature of the derivation of the equations:
Cv=δu/δT at constant volume => ∫Cvδu=∫δT => u2-u1=~Cv0(T2-T1)
Cp=δh/δT at constant Pressure = ∫Cvδu=∫δT => h2-h1=~Cp0(T2-T1)
The assumption is made that, over the temperature range of interest, the heat capacities are constant. If the answer had employed a reference temperature other than absolute zero (say, 27 C, for example), the 2nd equation would have come out the same in the end, but the derivation would have been much more acceptable to you

2. Additionally, what are the conditions for the usage of these equations?
u2-u1=~Cv0(T2-T1)
h2-h1=~Cp0(T2-T1)
I thought they were only valid at constant volume and pressure? In the case above, the pressure/volume changes linearly, are the relations still valid then?
For the special case of an ideal gas, the thermodynamic state functions u and h are functions only of temperature, independent of pressure and volume, so it doesn't matter if the pressure and volume changed.
Lastly, mass is also added between states 1 and 2, in the case above. I understand the equation relates specific terms, per unit mass, but does this addition of mass pose a problem?
Well, you need to determine the change in mass ##m_i=m_2-m_1##, so you would write $$m_2=m_1+m_i$$
 
  • #3
Thanks for the reply! I am still confused about how hi can transform into Cp*Ti. I know there is a relationship between Δh = Cp(ΔT), but do not know if there is one for h=CpT. Or are there more complicated immediate steps in between the aforementioned Steps 1 & 2 that were not in the solution?
 
  • #4
messier992 said:
Thanks for the reply! I am still confused about how hi can transform into Cp*Ti. I know there is a relationship between Δh = Cp(ΔT), but do not know if there is one for h=CpT. Or are there more complicated immediate steps in between the aforementioned Steps 1 & 2 that were not in the solution?
If we let ##T_R## represent some arbitrary datum absolute temperature at which the internal energy per unit mass is ##u_R##, then in the open system (control volume) version of the 1 st law applied to this problem, then we would write $$u=u_R+C_{v0}(T-T_R)$$ and $$h=u+Pv=u+RT=u_R+C_{v0}(T-T_R)+RT=C_{p0}(T-T_R)+u_R+RT_R$$The energy balance would be written as:$$m_2u_2-m_1u_1=m_ih_i+Q-W$$If we then substitute the equations for u and h in terms of the reference temperature ##T_R##, we obtain:$$m_2C_{v0}T_2-m_1C_{v0}T_1+(m_2-m_1)(u_R-C_{v0}T_R)=m_iC_{p0}T_i+m_i(u_R-C_{p0}T_R+RT_R)+Q-W$$But, since ##m_2-m_1=m_i##, the terms involving the reference state temperature ##T_R## and reference state internal energy ##u_R## drop out, and we are left with:$$m_2C_{v0}T_2-m_1C_{v0}T_1=m_iC_{p0}T_i+Q-W$$
 
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Related to Usage of General Specific Heats Cp0 and Cv0

1. What are general specific heats Cp0 and Cv0?

General specific heats Cp0 and Cv0 are thermodynamic properties that describe the amount of heat required to raise the temperature of a substance by one degree. Cp0 is the specific heat at constant pressure, while Cv0 is the specific heat at constant volume.

2. How are Cp0 and Cv0 different from each other?

Cp0 and Cv0 differ in the conditions under which they are measured. Cp0 is measured under constant pressure, meaning that the substance can expand or contract as heat is added or removed. Cv0, on the other hand, is measured under constant volume, meaning that the substance cannot change in size as heat is added or removed.

3. What is the significance of Cp0 and Cv0 in thermodynamics?

Cp0 and Cv0 are important in thermodynamics because they help us understand how substances behave when heat is added or removed. They are used in calculations to determine the amount of energy needed to change the temperature of a substance, and they also provide information about the specific properties of a substance.

4. How are Cp0 and Cv0 related to each other?

Cp0 and Cv0 are related through the equation Cp0 = Cv0 + R, where R is the gas constant. This means that Cp0 is always greater than Cv0, since the constant pressure condition allows for more expansion and thus more heat absorption compared to the constant volume condition.

5. How do Cp0 and Cv0 vary with temperature and pressure?

Cp0 and Cv0 can vary with temperature and pressure, as they are dependent on the properties of the substance. In general, both values tend to increase with increasing temperature, but they can also vary depending on the state of the substance (solid, liquid, gas). Additionally, pressure can affect the values of Cp0 and Cv0, but this effect is more significant for gases compared to liquids and solids.

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