UPDATED TL;DR: Cast iron has the advantage, because it has a greater capacity to hold heat, and because of its lower conductivity it releases it more slowly, thus its temperature does not fluctuate as much (thus having a more even and steady heat flow)
UPDATE: Answering in terms of the advantage (scroll way down) for the following sections
- Answer to Original Post: Regarding the input/output
- Answer to Original Post: Regarding the advantage
After the some of the comments I feel the following need to be addressed:
- Material properties
- Cooking type (oven or kitchen top)
- Temperature distribution (the full monty)
- Numerical example for temperature distribution
- Why higher Heat Capacity properties are better
- Why lower conductive properties are better
- Answer to Original Post: Regarding the input/output
- Answer to Original Post: Regarding the advantage
Apologies for being so long a post, but I really enjoyed this (even though some seem to disagree).
Material properties
Lets start with the properties, they'll come in handy later
| Material |
Heat Capacity |
Density |
$c_p\cdot \rho$ |
Thermal conductivity |
|
$(c_p)\left[\frac{kJ}{kg K}\right]$ |
$(\rho)\left[\frac{m^3}{ kg}\right]$ |
$\left[\frac{kJ}{m^3 K}\right]$ |
$(k) [\frac{W}{mK}]$ |
| Cast Iron |
0.46 |
7800 |
3.59 |
50~60 |
| Aluminum |
0.887 |
2900 |
2.57 |
150-200 |
| Copper (pure) |
0.386 |
8940 |
3.45 |
380 |
Cooking type
This is drawn mainly the comment from Tigerguy (leading people to heat up cast iron in ovens). It may me realise that some people cook their stakes in the oven instead of the stovetop (I wouldn't do it). Ι would also avoid even more using a skillet in the oven.
| Cooking type |
A |
B |
|
 |
 |
| Where |
Stove top |
in the oven |
| required Utensil |
skillet/pan |
a grill/pan |
| Comments (tongue in cheek) |
better (look at the colors in the picture, although I wouldn't use oil/butter) |
poor |
Obviously if you are talking about in the oven, the temperature distribution is uniform, and to be honest there is not much point to use a pan or a skillet. So from now on I will focus on the type A (on the stove top).
Time to heat up: This goes without saying... I'll never put a steak on a skillet that its not hot. So all my calculations, assume that the skillet is left for a few minutes on the stovetop to heat up and reach a steady state temperature.
The assumption of constant heat flux $q$: This is a tricky one, since most modern ceramic stove tops have some sort of thermal thermostat and switch automatically on/off (correct me if I'm wrong, I've never dissected one). That means that you have a "steady" temperature, and an "average" steady heat flux. On the other hand, gas cookers have constant heat flux, and they can literally burn a cooking utensil if left unattended for too long. Eventually constant heat flux means constant temperature at steady state, but in the case of gas that means burned materials. In any case, I think for the task at hand it's ok to assume constant heat flux $q$.
Temperature distribution
The following image shows a skillet, and we'll be discussing the temperature distribution.
In the image the temperature are the following:
- $T_1$ is the air temperature let's assume 25C
- $T_2$ is the temperature on the top side (cool) of the skillet
- $T_3$ is the temperature on the bottom side (hot) of the skillet on the stove top.
Assuming a constant heat flux, then between zones :
$T_1, T_2$ there is convection (lets assume a convective heat transfer coefficient of $h=100 \left[\frac{W}{m^2K}\right]$). The heat transfer due to convection is:
$$q = h A \Delta T$$
$T_2, T_3$ there is conductivity of the skillet, which is depended on the material used. The heat transfer due to conductivity is:
$$q = \frac{k}{t} A \Delta T$$
where:
Because the heat flux should be constant across the conductive and the convective zone (otherwise we are not on steady state),the following equalities hold :
$$q = \frac{k}{t} A (T_3-T_2) = h A (T_2-T_1) $$
Now assuming a constant heat flux, and a constant cross-section we can simplify the equation as
$$\frac{q}{A} = \frac{k}{t}(T_3-T_2) = h (T_2-T_1) $$
Now since $q/A$ is constant we can calculate
- the temperature $T_2$ from
$$T_2 = \frac{q}{hA} +T_1 $$
- Then temperature $T_3$ from
$$T_3 = \frac{q}{A}\frac{t}{k} + T_2 $$
$$T_3 = \frac{q}{A}\frac{t}{k} + \frac{q}{hA} +T_1 $$
$$T_3 = \frac{q}{A}\left(\frac{t}{k} + \frac{1}{h}\right) +T_1 $$
So both temperatures can be estimated but only $T_3$ is depended on thickness of the material, the heat conductivity and the area, while the top surface is determined only by the air conductivity coefficient (my bad).
Then you can see that for constant $q$, and thickness $t$ you'd have a greater temperature difference between the cool and the hot side of the skillet.
Numerical Example
This section is mainly added to reply to @mart comment, regarding the fact that the cook sets the temperature of the skillet. However, it is very crucial to know what is the actual maximum achievable temperature of the skillet, because then the cook has more options.
For this numerical example I will assume:
- heat flux $q=2[kW]$
- skillet diameter $d=10''= 25.6[mm]$
- thickness of material $t=10[mm]$ (a thick one)
| Material |
Max $T_2$ $[^oC]$ |
Max $T_3$ |
Heat capacity compared to cast iron |
| Cast Iron |
~420 |
427.5 |
1 |
| Aluminum |
~420 |
421.8 |
0.956 |
| Copper |
~420 |
420.7 |
0.713 |
So compared to the other two, cast iron will have a higher temperature on side $T_3$ (the hot side of the skillet, but the biggest advantage over aluminimium is that it can hold approximately 1/0.713=40% more heat energy (or aluminium can hold approximately 28% less energy than cast iron).
why greater thermal capacity is better
The overall heat energy in the material for a given temperature difference is greater for the cast iron because
$$ C_{p,steel} \rho_{steel} \gt C_{p,Cu} \rho_{Cu} \gt C_{p,Al} \rho_{Al}$$
$$ 3.588 \left[\frac{kJ}{m^3 K}\right] \gt
3.45\left[\frac{kJ}{m^3 K}\right] \gt2.572\left[\frac{kJ}{m^3 K}\right]$$
This means than when the steak and the skillet come into contact:
there is more readily available heat energy to be transmitted to the steak.
Additionally, for the same heat energy that is imparted, the temperature of the cast iron skillet compared to the aluminium will lower less (greater thermal inertia).
why lower thermal conductivity is better
Although lower thermal conductivity is not as detrimental, (and in terms of heating time and response is worse), it allows for slightly greater temperatures on the hot side (for a constant flux). (If constant $T_3$ is assumed then you will have greater $T_2$). The added temperature has a (small) beneficial effect on the heat capacity (because the average temperature is greater).
Answer to Original Post: Regarding the input/output
Especially since there's no way the heat output from the system is greater than the heat input.
You need to think the heat capacity of the skillet as a battery/heat energy buffer. Although you are right that there is continuous heat input, the moment you put a room temperature steak, there is a very sudden temperature difference (due to the higher conductivity and heat capacity of of the steak). That creates momentarily an increase in the heat flux, which is away from the previous steady state.
Initially the temperature of the stake increases, as heat is transferred from the skillet, until there is an equilibrium. During that time the temperature of the skillet drops. That is why you will notice a dip in the temperature.
The more thermal inertia (i.e. heat capacity of the skillet the better). As we've seen an similarly shaped and dimensioned cast iron has much greater thermal capacity and therefore, the temperature drop will be smaller.
Answer to Original Post: Regarding the advantage
"Advantage" in this context means the ability to transfer heat to the food at an even and steady rate. (the common fear is that plopping a cold steak in the pan will instantly cool down the pan and make it unable to keep cooking, or change the rate at which to food cooks)
If by "advantage you mean even and steady rate, then the cast iron since (again) it has higher thermal capacity and lower thermal conductivity, it will transfer heat energy at a slower rate, and for a longer time. This gives time to the system to reach equilibrium, beyond which the heat flux is steady.
Final thoughts
I enjoyed writing this a LOT. It was a thought exercise that I meant to do, but never actually had the chance. Going through it (and with the driving help of some comments), I managed to clear up some my thoughts better and set the things straight (i.e. conductivity doesn't contribute that much). If you feel, I am wrong somewhere or if you need some further clarification, don't hesitate to put a comment below.
Also sincere apologies to my fellow husband, for not agreeing with him and having an opposite opinion. (You don't have to show this to your wife). Us -married men- should stick together :-), but when science (or cooking) is involved... cast iron wins.
