Is the rate at which a material heats up dependent on its specific heat capacity? I've recently done an experiment on heating metal bars and analyzing how they expand while heating and contract while cooling.
The experiment consists of running steam through the inside of the metal bars until they've reached the expected length due to dilation, and then cutting the steam flow to see how they contract.
At the beginning of the experiment, we first had to argue whether to choose steel or brass bars to conduct the experiment. I chose brass and argued that as its linear dilation coefficient is larger, the expansion effect would be more noticeable. However, my lab professor told me that while this is true, it's not the main reason why brass is the better option.
That got me thinking on why you would choose brass over steel if it's not for its dilation coefficient and the only thing that came to my mind was arguing that as steel has a larger specific heat capacity then it would take a longer time to heat up and cool down making the experiment longer.
The problem with this is that I've never had a lecture on how fast objects heat up or cool down, so I don't really know if the heat capacity has something to do with the rate at which an object changes temperature.
I'd like someone to let me know if my assumption is correct, and if it's not, I'd appreciate if someone could point out why you would be choosing brass over steel for this particular experiment.
Thanks for the help in advance.
 A: The characteristic time $t$ for a diffusive process (in this case, conductive heat transfer) to progress substantially toward completion is $t\sim \frac{L^2}{D}$, where $L$ is a characteristic length and $D$ is the relevant diffusivity—here, the thermal diffusivity $\frac{k}{\rho c_P}$, which scales proportionally with the thermal conductivity $k$ and inversely proportionally with the density $\rho$ and (constant-pressure) specific heat $c_P$.
Thus, the specific heat (i.e., the amount of heating per unit mass required to obtain a certain temperature increase, if all you're doing is heating the material) does play a role, although not the sole role.
For a given geometry, therefore, you'll see the fastest absolute expansion from a material with large $\alpha D$, where $\alpha$ is the thermal expansion coefficient. This is about 0.0003 mm²/s·°C for steel and 0.0006 mm²/s·°C for brass.
A: This answer is incorrect. See the last edit.
Alright so in order to learn how fast the temperature of a material rises as time goes on we need to do some crash-course in thermodynamics.
So there is this quantity called internal energy $U$ that is the sum of a kinetic energy and potential energy of the particles of a system. We can think of $U$ as a function of the chemical composition of a system, its pressure, volume and temperature. So in a system composed of substances $A$, $B$ and $C$, for instance, $U = U(n_A,n_B,n_C,p,V,T)$ where $n_A$ is the number of moles of substance $A$.
$$
\mathrm{d}U=
 \left(\frac{\partial U}{\partial p}\right)_{n_A,n_B,n_C,V,T}\mathrm{d}p
+\left(\frac{\partial U}{\partial V}\right)_{n_A,n_B,n_C,p,T}\mathrm{d}V
+\left(\frac{\partial U}{\partial T}\right)_{n_A,n_B,n_C,V,p}\mathrm{d}T
+\sum_{i=A,B,C}\left(\frac{\partial U}{\partial n_i}\right)_{n_{j\neq i},p,V,T}\mathrm{d}n_i
$$
The subscripted quantities are kept constant for the computation of the partial derivatives. This is a very complicated thing. Buuuuut, we won't need to know all of that. Your brass bar doesn't undergo any relevant chemical changes during the experiment and neither does the steam you're blowing inside it. So we can approximate by saying we're trying to compute the variation in internal energy at constant composition.
$$
\mathrm{d}U_n=
 \left(\frac{\partial U}{\partial p}\right)_{n,V,T}\mathrm{d}p
+\left(\frac{\partial U}{\partial V}\right)_{n,p,T}\mathrm{d}V
+\left(\frac{\partial U}{\partial T}\right)_{n,V,p}\mathrm{d}T
$$
Where the subscript in $\mathrm{d}U_n$ means the total differential of the internal energy with the composition constant.
Buuuuut, $p$, $V$ and $T$ are really all related and if you know two of them, you know the value of the other is determined even if you can't compute it. So if we decide that we'll let $V$ and $T$ vary freely then the pressure exerted by the system against the walls of the recipient is not a degree of freedom. In general, that pressure is not a very relevant variable to the problem when the system is a solid, so we'll work with $V$ and $T$. The total variation in the internal energy of a system with constant composition is:
$$
\mathrm{d}U_n=
\left(\frac{\partial U}{\partial V}\right)_{n,T}\mathrm{d}V
+\left(\frac{\partial U}{\partial T}\right)_{n,V}\mathrm{d}T
$$
We'll drop the $n$ subscript because we know enough about the transformation taking place in your experiment to know that the composition is constant. So we can write the variation in internal energy in the $(V,T)$ space of states as:
$$
\mathrm{d}U=
\left(\frac{\partial U}{\partial V}\right)_T\mathrm{d}V
+\left(\frac{\partial U}{\partial T}\right)_V\mathrm{d}T
$$
By dividing both sides by $dt$ we get:
$$
\frac{\mathrm{d}U}{dt}=
\left(\frac{\partial U}{\partial V}\right)_T\frac{\mathrm{d}V}{dt}
+\left(\frac{\partial U}{\partial T}\right)_V\frac{\mathrm{d}T}{dt}
$$
But we already determined that $V,p$ and $T$ are related so $V(T,p)$. We can thus write:
$$
\frac{\mathrm{d}V}{dt} = \left(\frac{\partial V}{\partial T}\right)_p\frac{\mathrm{d}T}{dt}
$$
which means that:
$$
\frac{\mathrm{d}U}{dt}=
\left(\frac{\partial U}{\partial V}\right)_T
\left(\frac{\partial V}{\partial T}\right)_p
\frac{\mathrm{d}T}{dt}
+\left(\frac{\partial U}{\partial T}\right)_V\frac{\mathrm{d}T}{dt}
$$
The term $\left(\frac{\partial U}{\partial V}\right)_T$ is the internal pressure $\pi$ of the brass alloy you have been using. $\pi$ is a tabulated value, I believe, and it is generally a function of temperature so you might have to check several values at the table. The term $\left(\frac{\partial V}{\partial T}\right)_p$ is $\alpha V$ where $\alpha$ is the thermal expansion coefficient. The term $\left(\frac{\partial U}{\partial T}\right)_V$ is the $C_v$, which is the heat capacity of the substance at constant volume. You can use the heat capacity at constant volume even though the volume of  your substance is changing because the internal energy is a state function. So you end up with this
$$
\frac{\mathrm{d}U}{dt}=
\pi
V\alpha
\frac{\mathrm{d}T}{dt}
+C_V\frac{\mathrm{d}T}{dt}
$$
Which we can simplify to:
$$
\dot U=
(\pi
V\alpha
+C_V)\dot T
$$
Where the dots denote time-derivatives.
$$
\frac{1}{\pi \alpha V+C_V}\dot U=
\dot T
$$
And if we're working with specific heat capacities:
$$
\frac{1}{\pi \alpha V + m c_V}\dot U=
\dot T
$$
This equation is general for any system with constant composition, which your bars certainly are. The term $\dot U$ is the power of your heating system. The amount of energy it dumps into your metal bar per unit of time. Presumably, it will be the same for both bars so there is no reason to simplify this further.
$$
\dot T=\frac{\dot U}{\pi \alpha V + m c_V}
$$
Presuming you'll be heating your systems using the same power source, it turns out that how fast the temperature of your bar changes is inversely proportional to both the $C_V$ and the $\alpha$.
So while it's nice that the brass will have the most pronounced expansion, making it easier to measure, that also means it would take longer to heat up, but if it's $C_V$ is much lower than that of steel, the effect is compensated and then you may be able to have your cake and eat it too as brass will both have a larger expansion and will heat up faster. It all depends on the internal pressure $\pi$.
Conceptually, the internal pressure is involved with how tightly bound are the particles in the system and it captures how difficult it is to increase the system's volume. The harder it is to increase a system's volume, the more its internal energy (and thus its temperature) decreases as its volume increases because the expansion is done against the mutual attraction of the particles that make up the material.
I didn't find values of internal pressure for the solids you're looking for though the Table III of this paper from 1970 provides a table for pure solid metals and there we have that iron, copper and zinc and $540\ \mathrm{kbar}$, $340\ \mathrm{kbar}$ and $156\ \mathrm{kbar}$ of internal pressure, respectively.
Now Brass is a general term for a range of copper-zinc alloys. This site provides the density of a generic 69% Cu 29% Zn 1% Mg brass alloy, you might want to approximate $\pi_{brass} = 0.70 \pi_\mathrm{Cu} + 0.30 \pi_\mathrm{Zn}$ and use the iron one for steel since carbon makes up such a small proportion. However, be warned, the internal pressure is a measure of how the particles of a system interact with one another, so I presume it might change quite a lot depending of the material so the pressure of steel might be very different from that of iron, dooming the computation.
The same site where we can find the density of the 1% Mg brass also gives the density of high-carbon steel, which comes at 1% Carbon.
The paper also mentions that the value of internal pressure for the metals studied is nearly constant from room temperature to $1000\ \mathrm{K}$ so if you want to use those values as approximations for steel and brass you can consider them as constants through the entire experiment.
Also, I expect your bars to have the same size, which means the same volume $V$ but different masses, which you can calculate using the densities for brass and steel.
Though we can make all of these approximations an arrive at a final ratio between $\frac{dT}{dt}$ of steel by that of the brass to crudely estimate which one heats faster, even if we presume that our shoddy computation is correct, I doubt your professor was doing all of that in his head as he spoke to you.
As you can see there are many properties of the system that influences how quickly its temperature change because the process depends on various ways that energy can flow into and out of the system. If it was indeed the $C_V$ your professor was talking about, then I'm guessing there might be a rule of thumb to somehow disregard $\pi$ in certain situations or maybe the product $\pi \alpha$ might be negligible in solids or whatever. I couldn't find no such thing, and if you can't find that as well then I guess your professor had something else in his mind when he told you brass was the better choice but not for its thermal expansion coefficient.
Nevertheless, I hope to at least have answered the question about how the rate at which the temperature of a system rises depends on its specific heat capacity =D.
#technically-answered
Suggestion by Chemomechanic
Chemomecanic suggested we work with $C_p$, the heat capacity at constant pressure, rather than work with this problematic internal pressure. In order to get $C_p$ in our equation for $\frac{dT}{dt}$ we need to work with a different state function, the enthalpy ($H$), because $C_p$ is defined as the partial pressure of enthalpy with respect to temperature at constant pressure. Though enthalpy could be defined as $H(V,T)$, the partial derivatives you would get from this would hardly correspond to tabulated quantities, so they would be of little value. It is more common to define $H(p,T)$ which is an equivalent formulation, since $V(p,T)$. So with this we have that the total differential of a system with constant composition is given as:
$$
\mathrm{d}H =
\left(\frac{\partial H}{\partial p}\right)_{T}\mathrm{d}p
+\left(\frac{\partial H}{\partial T}\right)_{p}\mathrm{d}T
$$
Dividing both sides by $\mathrm{d}t$ gives us:
$$
\dot H =
\left(\frac{\partial H}{\partial p}\right)_{T}\frac{\mathrm{d}p}{\mathrm{d}t}
+\left(\frac{\partial H}{\partial T}\right)_{p}\dot T
$$
The term $\frac{\mathrm{d}p}{\mathrm{d}t}$ can once more be broken down further by knowing that $p(V,T)$ such that:
$$
\frac{\mathrm{d}p}{\mathrm{d}t}= \left( \frac{\partial p}{\partial T} \right)_V \frac{\mathrm{dT}}{\mathrm{dt}}
$$
We then substitute it back into the formula to get:
$$
\dot H =
\left(\frac{\partial H}{\partial p}\right)_{T}\left( \frac{\partial p}{\partial T} \right)_V \frac{\mathrm{dT}}{\mathrm{dt}}
+\left(\frac{\partial H}{\partial T}\right)_{p}\dot T
$$
So the term $\left(\frac{\partial H}{\partial T}\right)_{p}$ is the constant-pressure heat capacity $C_p$. But the other terms are a mess. The first one  $\left(\frac{\partial H}{\partial p}\right)_{T}$ is often expressed for gases as a product of $C_p$ multiplied by the Joule-Thomson coefficient, but I don't think that makes any sense for solids. And again you have a term that describes how the enthalpy of the system lowers dow with respect to the change in the pressure that it exerts against the atmosphere.
I suspect that, for solids, the pressure that the system exerts against the atmosphere is just the pressure of not collapsing in on itself, os it is involved with the structure of the material. That rate of change then has to do with how difficult it is to expand the material, which has to be related with how strongly are the inter-particle interactions that maintain the material structure of the system since any expansion would be done against the binding energy of the internal structure. It's just a more complicated way to express the $\left(\frac{\partial U}{\partial V}\right)_{T}$ term of the main body of the answer. I think I read once that this measure is called free volume, let's call it $\gamma$.
The term $\left( \frac{\partial p}{\partial T} \right)_V$ is a measure of how much the pressure that the material exerts change as a function of its temperature. It's the pressure equivalent of the thermal expansion coefficient, I don't know its name but let's call it $\beta$ just to keep the greek-letter motif. With this we have that:
$$
\dot H =
\gamma \beta \frac{\mathrm{dT}}{\mathrm{dt}}
+C_p\dot T
$$
From here it is easy to see that if we have a heat source with power $\dot H$ then:
$$
\dot T=\frac{\dot H}{\gamma \beta + m c_p}
$$
which has the same structure as  the other one. This is expected since the terms we are trying to get rid of are related to the strength of the chemical interactions in the material structure of the system, which is a big deal for solids.
New Suggestion by Chemomechanic
So in a new comment Chemomechanic said that:

All those $dP$ terms can be taken as zero—again, the experiment is conducted at constant atmospheric pressure, and the bars are unstressed.

which once more I decided to address here at the answer because it might be something many people are thinking about and I wanted to elaborate further than a comment would allow.
So it turns out that the various $dp$ terms do not refer to the atmospheric pressure, which does remain constant throughout the experiment, but to the internal pressure of the system.
Any system that wants to maintain a finite volume against the constant weight of the atmosphere needs a form of internal pressure to maintain mechanical equilibrium with the atmospheric pressure so as to keep the volume stationary. For a gas inside a balloon that pressure is the same type as that of the atmosphere: molecules of ar bombarding the inside surface area of the balloon, but for a solid something equivalent must happen, other wise the solid would be compressed into nothing. That something is related to the molecular structure of the solid and the chemical bonds that hold the thing together.
When one heats the thing up, we provide internal energy for the system. Since we're heating the system, the kinetic energy of the system will increased. This increased kinetic energy corresponds to an increased internal pressure. The mechanical imbalance between this new internal pressure and the external pressure is what drives the increase in volume until the new equilbrium is reached. Something like what is shown below.

The thing is that the $dp$ sucks a bit of the $dU$ acquired from the $dq$ because increasing the internal pressure is something done against the chemical bonds that keep everything together. In order to increase the internal pressure by $dp$, work must be done against these internal chemical influences, so there is a small $-dU$ associated with it. The complex interplay between internal energy and the internal pressure is given by this term: $\left( \frac{\partial H}{\partial p} \right)_T$, which we called $\gamma$ in the comment of the first suggestion.
Hopefully by now you should see how that is related to $\pi = \left( \frac{\partial U}{\partial V} \right)_T$, which is related to that $dV$ we see in red in the above figure and is entwined with $\gamma$, its pressure equivalent.  So although the external pressure remains constant, that simply means that we have a non-adiabatic isobaric expansion, but the dynamics of this mechanical balance between the forces holding the material together and the external pressure is what controls the thermal expansion, and these forces seem like a big deal for solids, which are mostly characterized by the type of chemical bonds holding them together.
Deciding that $dp=0$ really is a possible approximation depending on what application you have in mind but is the same as saying that $\alpha=0$, which is to say that there is no relevant thermal expansion. This strikes me as an odd approximation to make in order to design an experiment to measure thermal expansion.
So maybe that was what your professor had in mind. Maybe the magnitude of $\gamma \beta p$ might be negligible compared to $C_p$ and that is the answer, but since data for these things is hard to come by, I can't give a definitive answer.
Last Edit: Suggestion by Chemomechanic
So it turns out that Chemomechanic convinced me that my approach was misguided. There were interpretation mistakes along the way. One can arrive at the correct formula by taking $U$ as a function of pressure and temperature $U(p,T)$ like so:
$$
\mathrm{d}U =
\left(\frac{\partial U}{\partial p}\right)_{T}\mathrm{d}p
+\left(\frac{\partial U}{\partial T}\right)_{p}\mathrm{d}T
$$
under the assumption that the system is in equilibrium with the atmosphere and that the pressure is constant we have that the $dp$ term disappears leaving just:
$$
\mathrm{d}U =
\left(\frac{\partial U}{\partial T}\right)_{p}\mathrm{d}T
$$
If we assume that the only work done or received by the system is expansion work, then $\mathrm{d}U=T\mathrm{d}S - p\mathrm{d}V$ so:
$$
\mathrm{d}U =
\left(\frac{\partial [T\mathrm{d}S - p\mathrm{d}]}{\partial T}\right)_{p}\mathrm{d}T
$$
which can be further broken down into:
$$
\mathrm{d}U =
 T\left( \frac{\partial S}{\partial T} \right)_p \mathrm{d}T
-p\left( \frac{\partial V}{\partial T}\right)_p \mathrm{d}T 
$$
Since $\left( \frac{\partial V}{\partial T}\right)_p = V\alpha$ and $T\left( \frac{\partial V}{\partial T}\right)_p= C_p$ then the above equation can be written as:
$$
\mathrm{d}U = (C_p-pV\alpha)\mathrm{d}T
$$
Dividing both sides by $\mathrm{d}t$ and assuming that $C_p$ and $\alpha$ remain constant in the temperature range of the experiment we get that:
$$
\dot U=  (C_p-pV\alpha) \dot T
$$
and thus that:
$$
\dot T = \frac{\dot U}{C_p-pV\alpha}
$$
Chemomechanic metioned that $pV\alpha$ is much smaller than $C_p$ so the above equation can be approximated as
$$
\dot T \approx \frac{\dot U}{C_p} 
$$
Which implies that the rate with which a material increases its temperature in the experiment described is roughly directly proportional to the power output of your heating system and inversely proportional to the heat capacity of the material being heated.
This, I think, is the most correct and satisfying answer. Thanks to @Chemomechanic for pointing the mistakes!
