Constant pressure specific heats when temperature changes

In summary, the conversation is about calculating the constant pressure specific heat for an ideal gas undergoing a process where the temperature is changing. The equation used to calculate Cp0 is dependent on the temperature, so the question is whether it is valid and more accurate to use the corresponding Cp for each temperature or just the same Cp for the different temperatures. It is concluded that using enthalpies instead of constant specific heats is more accurate, unless there is an equation for the average specific heat as a function of the two end point temperatures.
  • #1
Imolopa
21
1

Homework Statement


Im trying to understand what would be the correct approach for calculating the constant pressure specific heat for an ideal gas undergoing a process where the temperature is changing.

The reason I am asking is because the equation used to calculate Cp0 is dependent on the temperature. So when you want to calculate the work or/and heat exchange for a process where the temperature is changing over time, would you still use this equation and in that case what temperatures would you input? Maybe sum up the start temperature with the end temperature and divide the result by 2? That last method doesn't seem like it would be a very accurate way of doing it, therefore I believe there must some other way that is commonly used?

Homework Equations



Cp0 = C0 + C1θ + C2θ2 + C3θ3

where θ = T /1000 and Ci = constants (gas dependent) found in tables
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  • #2
Imolopa said:

Homework Statement


Im trying to understand what would be the correct approach for calculating the constant pressure specific heat for an ideal gas undergoing a process where the temperature is changing.

The reason I am asking is because the equation used to calculate Cp0 is dependent on the temperature. So when you want to calculate the work or/and heat exchange for a process where the temperature is changing over time, would you still use this equation and in that case what temperatures would you input? Maybe sum up the start temperature with the end temperature and divide the result by 2? That last method doesn't seem like it would be a very accurate way of doing it, therefore I believe there must some other way that is commonly used?

Homework Equations



Cp0 = C0 + C1θ + C2θ2 + C3θ3

where θ = T /1000 and Ci = constants (gas dependent) found in tables
[/B]
You integrate to get the enthalpy change.
 
  • #3
Chestermiller said:
You integrate to get the enthalpy change.

Yeah so with enthalpy values found in tables for the start and end temperature would basically give us what we need in other words the equation: 1Q2 = m(h2-h1)

Combining with the relation:

1Q2= m(u2-u1 ) + 1W2

where 1W2 = mCv(T2-1) = mP(v2-v1)

So solving for Cv in short.
 
  • #4
Imolopa said:
Yeah so with enthalpy values found in tables for the start and end temperature would basically give us what we need in other words the equation: 1Q2 = m(h2-h1)
Yes.
Combining with the relation:

1Q2= m(u2-u1 ) + 1W2

where 1W2 = mCv(T2-1) = mP(v2-v1)
This last equation is incorrect.
 
  • #5
ah yeah the correct ones would be, thank you!:
1Q2 = m(u2-u1 ) + 1W2 = mCv(T2-T1) + 1W2

where 1W2 = mP(v2-v1)
 
  • #6
Imolopa said:
ah yeah the correct ones would be, thank you!:
1Q2 = m(u2-u1 ) + 1W2 = mCv(T2-T1) + 1W2

where 1W2 = mP(v2-v1)
Are you trying to determine Cv? The equation you gave certainly doesn't determine ##\Delta U## correctly.
 
  • #7
Chestermiller said:
Are you trying to determine Cv? The equation you gave certainly doesn't determine ##\Delta U## correctly.

Yes that is right I want to determine Cv.
 
  • #8
Imolopa said:
Yes that is right I want to determine Cv.
Well, ##C_v(\theta)=C_v(\theta)-R##, where R is the universal gas constant and these are the molar heat capacities..
 
  • #9
Chestermiller said:
Well, ##C_v(\theta)=C_v(\theta)-R##, where R is the universal gas constant and these are the molar heat capacities..

Ok so conclusionwise either choose a temperature and use in combination with above formula to find Cp which is inputted in the last one to get Cv or calculate using enthalpy with h2 and h1.
 
  • #10
Imolopa said:
Ok so conclusionwise either choose a temperature and use in combination with above formula to find Cp which is inputted in the last one to get Cv or calculate using enthalpy with h2 and h1.
I have no idea what you're saying.
 
  • #11
Let's get back to my initial question, and now let's assume that we don't have the values for enthalpy h available.

Given the following relation that is approximately true:
h2 - h1 = Cp(T2-T1)

Now given that Cp is a function of the temperature T by the formula:
Cp0 = C0 + C1θ + C2θ2 + C3θ3
where θ = Ti /1000 and Cj = constants (gas dependent) found in tablesSo with the above in mind my question is if it is valid, and more accurate to actually rewrite the top formula to become:
h2 - h1 = Cp(T2) T2 - Cp(T1) T1
In short words use the corresponding Cp for each temperature, rather than just the same Cp for the different temperatures. If not how can one find a Cp that is accurate enough taking into account the different temperatures in the process?
 
  • #12
Your proposed equation is not accurate unless Cp is constant. Even if Cp varies linearly with temperature, it will give the wrong answer. Here are two more accurate versions that are both exact if Cp varies linearly with temperature:
$$\Delta h=\frac{C_p(T_1)+C_p(T_2)}{2}(T_2-T_1)$$
$$\Delta h=C_p|_{\frac{(T_1+T_2)}{2}}(T_2-T_1)$$
 
  • #13
Thank you!
So to conclude we can say that it is more accurate to use enthalpies instead of constant specific heats, right?
 
  • #14
Imolopa said:
Thank you!
So to conclude we can say that it is more accurate to use enthalpies instead of constant specific heats, right?
Sure, unless you have an equation for the average specific heat as a function of the two end point temperatures.
 

Related to Constant pressure specific heats when temperature changes

1. What is the definition of constant pressure specific heat?

Constant pressure specific heat is a thermodynamic property that measures the amount of heat required to raise the temperature of a unit mass of a substance by one degree at constant pressure. It is denoted by the symbol Cp and is expressed in units of J/kg•K (joules per kilogram per Kelvin).

2. How does constant pressure specific heat differ from constant volume specific heat?

Constant pressure specific heat (Cp) is measured at a constant pressure, while constant volume specific heat (Cv) is measured at a constant volume. This means that Cp takes into account the work done by the substance as it expands against a constant pressure, while Cv does not account for any work done. As a result, Cp is generally higher than Cv for most substances.

3. How does the specific heat of a substance change with temperature?

The specific heat of a substance generally increases with temperature. This is because as the temperature increases, the average kinetic energy of the particles in the substance also increases, making it more difficult to raise the temperature. This means that more heat is required to cause the same increase in temperature, resulting in a higher specific heat.

4. How does the specific heat of a substance affect its thermal energy storage capacity?

The specific heat of a substance is directly related to its thermal energy storage capacity. A substance with a higher specific heat can store more thermal energy per unit mass compared to a substance with a lower specific heat. This means that substances with higher specific heats are able to absorb and release more heat energy without experiencing significant changes in temperature.

5. What factors can affect the constant pressure specific heat of a substance?

The constant pressure specific heat of a substance can be affected by various factors such as temperature, pressure, and phase changes. Additionally, the molecular structure and composition of the substance can also impact its specific heat. For example, substances with more complex molecular structures tend to have higher specific heats compared to simpler substances.

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