Understanding Cv and Cp: Specific Heats Explained

The Intuition: Thermal Inertia

Let's start with a simple observation. Imagine it's a hot, sunny day at the beach. You walk barefoot on the dry sand, and it's scorching hot—you have to run to the water. But when you step into the ocean, the water is still cool and refreshing.

Why is this? Both the sand and the water received the same amount of energy from the sun. Yet, the sand's temperature shot up, while the water's temperature barely changed.

This is because different substances have different "capacities" to absorb heat. You can think of it as a kind of thermal inertia.

• Sand has a low thermal inertia. A little bit of heat causes a big jump in temperature.
• Water has a high thermal inertia. It can soak up a huge amount of heat with only a small change in its temperature.

In thermodynamics, we call this property specific heat. It's a measure of how much energy you need to add to one kilogram of a substance to raise its temperature by one degree Celsius (or one Kelvin, since the size of a Celsius degree and a Kelvin are the same).

The formal definition is:
$c=\frac{Q}{m\Delta T}$
Where:

• $c$ is the specific heat in units like kJ/(kg·°C) or kJ/(kg·K).
• $Q$ is the heat added in kJ.
• $m$ is the mass in kg.
• $\Delta T$ is the change in temperature in °C or K.

But there's a catch, and it's a big one...

Why Are There Two Specific Heats? The "Two Jobs" Analogy

Let's go back to our piston-cylinder device containing a gas. Imagine we want to add heat to raise the temperature of the gas by, say, 10°C. We can do this in two different ways:

Scenario 1: Constant Volume (Rigid Tank)
Imagine we lock the piston in place so it can't move. The volume is fixed. Now, we add heat. Where does all that energy go? Since the gas can't expand and do work, 100% of the heat we add goes into increasing the internal energy of the gas molecules—making them zip around faster. This raises the temperature. In this case, the energy has only one job: raise the internal energy.

Scenario 2: Constant Pressure (Floating Piston)
Now, imagine the piston is free to move up and down, with a constant weight on it (which creates a constant pressure). We start adding heat. As the gas molecules speed up (temperature rises), they push harder on the piston, causing it to move up. The gas expands and does boundary work.

In this scenario, the heat we add has to do two jobs:

1. Job #1: Increase the internal energy of the gas (raise the temperature).
2. Job #2: Provide the energy needed for the gas to do expansion work on the piston.

So, to achieve the exact same 10°C temperature rise, which scenario requires more heat?

It has to be Scenario 2. Because some of the energy is "spent" on doing work, we need to supply more total heat to get the same temperature change.

This is the entire reason we have two different specific heats:

• Specific Heat at Constant Volume ($c_v$): The energy needed to raise 1 kg by 1°C when the volume is held constant (no work is done).
• Specific Heat at Constant Pressure ($c_p$): The energy needed to raise 1 kg by 1°C when the pressure is held constant (energy goes to both internal energy and work).

Because of the "two jobs" principle, it is always true for gases that $c_p>c_v$.

This image perfectly captures the idea:

Recap Break

Let's pause.

• Specific heat is the energy needed to raise the temperature of 1 kg of a substance by 1 degree.
• We have two types because the conditions under which you add the heat matter.
• At constant volume ($c_v$), all heat goes to increasing internal energy.
• At constant pressure ($c_p$), the heat must also cover the energy for expansion work.
• Therefore, for gases, $c_p$ is always greater than $c_v$.

The Formal Definitions: Connecting to the First Law

Now that we have the intuition, let's connect this to the equations we know. Since you're comfortable with calculus, we can be a bit more precise.

Specific Heat at Constant Volume ($c_v$)

Remember the First Law in differential form: $\delta Q-\delta W=dU$.
For a simple compressible system, boundary work is $\delta W=PdV$. So, $\delta Q-PdV=dU$.

Now, consider a constant volume process. If the volume is constant, then $dV=0$, and the work term $\delta W$ becomes zero. The First Law simplifies beautifully to:
$\delta Q=dU(\text{for a constant volume process})$
This says that at constant volume, any heat you add goes directly into changing the internal energy.

By definition, the specific heat at constant volume is $c_v=\frac{\delta Q}{mdT}$ for this process. If we substitute $\delta Q=dU$, and use specific internal energy $u=U/m$, we get:
$c_v={\left(\frac{\partial u}{\partial T}\right)}_v$
This is the formal definition of $c_v$. It's the rate of change of specific internal energy with respect to temperature, while holding volume constant. The partial derivative $\partial$ is used because, in general, internal energy u could depend on both temperature and volume.

The Ideal Gas Shortcut: For an ideal gas, it turns out that internal energy depends only on temperature. This is a foundational property of ideal gases. Since u doesn't depend on volume for an ideal gas, we can drop the "constant volume" constraint and the partial derivative becomes a regular derivative. This leads to a hugely important result:
$du=c_{v}dT(\text{for an ideal gas, ANY process})$
Even though we derived this from a constant-volume process, the final relationship holds for any process (isobaric, isothermal, etc.) as long as it's an ideal gas, because for an ideal gas, internal energy is purely a function of temperature. This allows us to calculate $\Delta u$ for any process just by knowing the temperature change:
$\Delta u=u_2-u_1={\int }_1^2{c}_v(T)dT\approx c_{v,avg}(T_2-T_1)$
If $c_v$ is constant over the temperature range, this simplifies to $\Delta u=c_v\Delta T$.

Specific Heat at Constant Pressure ($c_p$)

Let's do the same for a constant pressure process. The First Law is still $\delta Q-PdV=dU$.
This time, work is being done. Rearranging gives: $\delta Q=dU+PdV$.

This looks familiar! Remember our property enthalpy ($H$)? We defined it as $H=U+PV$.
In differential form, $dH=dU+PdV+VdP$.
For a constant pressure process, $dP=0$, so the last term vanishes:
$dH=dU+PdV(\text{for a constant pressure process})$
Comparing our two equations, we see that for a constant pressure process:
$\delta Q=dH$
This is why enthalpy is so useful! Enthalpy represents the heat transferred during a constant pressure process.

By definition, $c_p=\frac{\delta Q}{mdT}$ for this process. Using specific enthalpy $h=H/m$ and substituting $\delta Q=dH$, we get the formal definition:
$c_p={\left(\frac{\partial h}{\partial T}\right)}_p$
This is the rate of change of specific enthalpy with respect to temperature at constant pressure.

The Ideal Gas Shortcut (again): Just like internal energy, for an ideal gas, enthalpy also depends only on temperature. So, for any process involving an ideal gas:
$dh=c_{p}dT(\text{for an ideal gas, ANY process})$
And the change in enthalpy is:
$\Delta h=h_2-h_1={\int }_1^2{c}_p(T)dT\approx c_{p,avg}(T_2-T_1)$
If $c_p$ is constant, this is just $\Delta h=c_p\Delta T$.

To see all of this derived from the First Law with clear visuals, let's watch a video.

The Relationship Between $c_p$ and $c_v$ for an Ideal Gas

We know $c_p$ is greater than $c_v$. But by how much? We can find an exact relationship for an ideal gas.

1. Start with the definition of enthalpy: $h=u+Pv$.
2. For an ideal gas, we know the equation of state: $Pv=RT$, where R is the specific gas constant.
3. Substitute this into the enthalpy equation: $h=u+RT$.
4. Now, let's differentiate this whole equation with respect to temperature, $T$:
   $\frac{dh}{dT}=\frac{du}{dT}+\frac{d(RT)}{dT}$
5. Since $R$ is a constant, this becomes:
   $\frac{dh}{dT}=\frac{du}{dT}+R$
6. We just defined these derivatives! For an ideal gas, $dh/dT=c_p$ and $du/dT=c_v$.
   $c_p=c_v+R\text{or}{c}_p-c_v=R$

This is a fundamental and elegant result. It proves that for an ideal gas, the specific heat at constant pressure is larger than the specific heat at constant volume by exactly the amount of the gas constant, $R$. That constant R represents the extra energy per degree that goes into doing expansion work in the constant pressure case.

Worked Numerical Problems

Let's put this into practice. For air, let's use the following average values:
$c_p=1.005\text{ kJ/(kg⋅K)}$
$c_v=0.718\text{ kJ/(kg⋅K)}$
$R=0.287\text{ kJ/(kg⋅K)}$
(Note that $c_p-c_v=1.005-0.718=0.287=R$, as expected!)

Problem 1: Constant Volume Heating
A 2 m³ rigid tank contains air at 100 kPa and 22°C. The tank is heated until the pressure is 300 kPa. Determine the heat transfer.

Problem 2: Constant Pressure Heating
A piston-cylinder device contains 0.5 kg of air initially at 300 K. It is heated at a constant pressure until its volume doubles. Find (a) the heat added, (b) the work done, and (c) the change in internal energy.

Conclusion

Excellent work today! We've added two of the most useful properties in all of thermodynamics to our toolkit.

Key Takeaways:

• Specific Heat ($c$) is a measure of the energy required to raise the temperature of a unit mass of a substance by one degree.
• We need two different specific heats for gases because the work done during heating can be different.
• Specific Heat at Constant Volume ($c_v$) applies when no work is done. For an ideal gas, this lets us find the change in internal energy: $\Delta u=c_v\Delta T$.
• Specific Heat at Constant Pressure ($c_p$) applies when the gas expands and does work. For an ideal gas, this lets us find the change in enthalpy: $\Delta h=c_p\Delta T$.
• For any gas, $c_p>c_v$. For an ideal gas, their relationship is simple and powerful: $c_p-c_v=R$.

Preview of the Next Lesson

We now have the tools to calculate changes in $u$ and $h$ for ideal gases. This is a game-changer. In our next lesson, we will use this knowledge to fully analyze the four fundamental thermodynamic processes:

1. Constant Volume (Isochoric)
2. Constant Pressure (Isobaric)
3. Constant Temperature (Isothermal)
4. No Heat Transfer (Adiabatic)

We will learn how to calculate both the work ($W$) and heat ($Q$) for each of these processes, and see how they look on a P-V diagram. This will give us a complete picture of how energy moves and transforms in these basic but essential scenarios.