When do we use $ΔU = mcΔT$?

Question

So we usually use $Q = mcΔT$, and in a specific task they used $ΔQ = ΔU + ΔW$ and wrote ΔQ = mcΔT + pΔV. So basically ΔU = mcΔT. I am a bit confused on what formula to use for ΔU when? I have been using 3/2 nRΔT for the past tasks, but sometimes see ncvΔT and mcΔT! Can someone explain when do I use what for ΔU?

Answer 1

In classical thermodynamics, we usually start with defining the exact differential of the property we want to examine (because we are interested mostly in differences) rather than exact quantities.

The mathematical definition for an exact differential is:

$${\displaystyle d\omega =\sum _{i=1}^{n}{\frac {\partial f}{\partial x_{i}}}\,dx_{i}.}$$

So if you are interested in the internal energy $dU$, you would write the exact differential like so:

$$dU(T,V)=\frac{\partial U}{\partial T}\,dT + \frac{\partial U}{\partial V} \,dV\tag1$$

That is let's say the most general definition for internal energy (and as you can see not a very useful one).

We than define the specific heat capacity at constant volume as:

$$c_V = \frac{1}{m} \left({\frac{\partial U}{\partial T}}\right)_V$$

and you see that equation $(1)$ reduces to the more familiar form:

$$dU=mc_VdT + \frac{\partial U}{\partial V} \,dV$$

Now, for the second summand, we use the fact that for the ideal gas the internal energy is only a function of temperature. That fact is usually known as Joule's second law.

So you would obtain that

$$dU=m\,c_V\, dT$$

So, it is important to remember that this equation is only valid for an ideal gas.

Now if you have the mass of the ideal gas you would use the equation with $m$ in it.

If you have the amount of substance of the ideal gas (in moles) you would use:

$$dU=nC_{MV}dT$$

where $n$ is the amount of substance in moles and $C_{MV}$ is the molar heat capacity in $\frac{J}{K\cdot mol}$. So that $dU$ will always have the unit of $J$ as expected.

Now for the last equation, kinetic theory of gasses tells us that all monatomic gases store energy in kinetic energy and that it is equal to:

$${\displaystyle E={\frac {3}{2}}nRT}$$

equivalently the molar heat capacity of all monoatomic gasses is the same and equal to:

$$C_{VM}=\frac{3}{2}R$$

where $R$ is the ideal gas constant.

So, by substituting you get that:

$$dU=\frac{3}{2}R\space n\space dT$$

You use the two equations with $m$ and $n$ when you have the mass and the amount of substance of the ideal gas respectively. The last equation you can use if the task asks you to use it specifically to maybe obtain a more precise solution.

Answer 2

There is no "$c$", you must always specify under what conditions you measure the heat capacity (OK, sometimes people are lazy and only the context tells you what subscript is operational).

With a slight notational modification, you wrote $$\delta Q = mc_vdT + \tilde pdV \tag{1},$$ but you did not define what you meant by the symbols. Also, for the equation to make sense one assumes that $\tilde p=\tilde p(T,V)$ and $c_v=c_v(T,V).$

If your thermodynamic system consists of the simplest material, one that is subject to a single mechanical interaction, here pressure-volume, and the process it undergoes is a reversible one, then, in Eq. (1) the symbols $dT, dV$ are infinitesimal steps separating two neighboring equilibrium states of the system.

As written, Eq. (1) assumes that the reversibly absorbed heat $\delta Q$ is really a function of the state characterized by its temperature $T$ and volume $V.$ This actually is true for any of the simplest two-variable thermodynamic systems. It is also true in a more general case but mathematically the situation is much more complicated for more than two variables.

To move from a state $T,V$ to one at $T+dT, V+dV$ the system has absorbed $\delta Q$ heat and worked on the environment by expanding its volume in the amount of $dV$.

The quantity $c_v$ is the mass specific heat capacity at constant volume, that is if we fix $V$, $\delta V=0,$ and then $\delta Q = mc_v dT$ representing pure heat absorption without performing work, the basic calorimetric measurement.

Eq. 1 is called the doctrine of specific and latent heats for a good reason: when the system absorbs heat while it changes temperature is due to its $mc_v$ a material constitutive characteristic, but the system can also absorb (or reject) heat $\delta Q = \tilde pdV$ even if its temperature does not change, i.e., $dT=0.$ It is a latent heat because it can be absorbed without a sensible temperature change.

Energy conservation is a different story. You are correct in writing that $dU=\delta Q+\delta W$ where $dU$ is the increment of internal energy, $\delta Q$ and $\delta W$ are the absorbed heat and work increments. If quasi-static pressure-volume work is the only non-thermal interaction then $dU=\delta Q - pdV,$ and again for a constant volume, $\delta V=0,$ pure thermal interaction $dU=\delta Q=mc_vdT,$ and this holds for any two-variable $T,V$ system, $m$ being kept a constant, with $c_v=c_v(T,V)$ at constant $V.$

But note that from $\delta Q = dU+pdV = mc_vdT+\tilde p(T,V)dV$ and $dU(T,V)=\frac{\partial U}{\partial T}dT + \frac{\partial U}{\partial V}dV$ we have $mc_v=\frac{\partial U}{\partial T}|_V \tag{2}$ $\tilde p(T,V) = p(T,V)+\frac{\partial U}{\partial V}|_T \tag{3a}$ It can also be shown that, in general, $\tilde p(T,V) = T\frac{\partial p}{\partial T}|_T\tag{3b}.$

Ideal gases are defined by their caloric equation of state $U=\hat k \frac{m}{M}RT,$ where $\hat k(\text {monatomic gas}) = \frac{3}{2}$, $\hat k(\text {diatomic gas})=\frac{5}{2},$ and thermal equation of state $pV=\frac{m}{M}RT.$ Therefore $dU=\hat k \frac{m}{M}RdT = mc_vdT,$ and then $c_v=\hat k \frac{R}{M}.$

Since $U=U(T)$ independently of $V,$ we also have that for an ideal gas $\tilde p(T,V) = p(T,V).$