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I did simple observation when I was heating water using stainless steel container and aluminum container. Indeed they are not exactly same but quite comparable. What I discovered was that, with the same amount of water I put inside them, I found like below:

  • Using the aluminum container, the water get boiled faster.
  • The aluminum container get cooled faster counted since the boiled water get boiled and the flame is turned off or I lifted the container off from the stove.
  • When the water get boiled and the water is poured out, the aluminum container is get cooled faster compared to the stainless steel with the same treatment.
  • After the water get boiled and poured out, the stainless steel container is hotter than the aluminum (for this, I didn't do scientific measurement with measurement tool, just based on my finger sense).

I can say that stainless steel container is like it has inertia to the thermal. The container and the water inside was slower get heated/boiled but that container also slower get cooled. I am not very sure what kind of stainless steel was it, but probably it was 316. I searched in internet, its thermal conductivity is 16.5W/m.K, while thermal conductivity of aluminum is 247W/m.K, 15 times than the stainless steel's. There is also mentioned that stainless steel has among the lowest thermal conductivity of any metal.

What I am asking is, what property the stainless steel has so it slower get heated but also slower get cooled as it has thermal inertia? Is that due to that thermal conductivity?

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2 Answers 2

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I'll add another perspective, -- although probably DKNGuyen answer is what you are after ( I upvoted it). The reason I am also providing another answer is because you are also considering the rate (how fast the material changes temperature). From my view both properties a) heat capacity $c_p$ of the material, and b) heat conductivity $k$ of the material play a role.

Heat capacity was covered in DKNguyen answer thoroughly, so I'll just say that it basically shows how much heat energy in Joules is required to raise the temperature of the material. The Specific Heat Capacity is in $\frac{J}{kg\cdot K}$ and translates to how much energy energy is required to raise by 1 degree Kelvin 1 kg of the material.

However, because you are also looking for the rate of something getting hotter and colder, the thermal conductivity also plays a role. Thermal conductivity shows (simply put) how easy heat flows through a material. If there is no conductivity then there is an energy buildup, which leads to higher temperature at one end and no change at the other. So you could have a material with low specific heat capacity that is a good insulator (low conductivity) that would not allow the liquid inside the container to get hot.

E.g. wood has a heat capacity about double of aluminium (1.76 compared to .9 kj/kgK), but its density is almost 1/3 of the aluminium (aluminium has 2700 kg/$m^3$ while most wood floats so it under 1000 kg/$m^3$). So by volume, the heat capacity is less. However, you'd never be able to boil water inside a wooden container, because wood is a poor conductor of heat.

So, if you are after boiling the liquid inside then the property of thermal diffusivity is probably the most relevant:

$$ D = \frac{k}{c_p\cdot \rho}$$

where:

  • D is the thermal diffusivity
  • k is the conductivity coefficient
  • $c_p$ is the specific heat capacity.
  • $\rho$ is the density

Basically, Diffusivity would show how fast would the temperature rise at one end of a bar that is heated from the other end. Higher values would indicate faster temperature rise.

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  • $\begingroup$ What's the technical difference between diffusivity and conductivity? I guess it takes into account how quickly the heat travels between adjacent areas but also how much heat it takes to heat up adjacent areas of the material before moving on? $\endgroup$
    – DKNguyen
    Feb 17, 2023 at 16:50
  • $\begingroup$ Very interesting. I had no idea what was that heat capacity and thermal diffusivity. Regarding the heat capacity, as you explained, is "how much energy is required to raise by 1 degree Kelvin 1 kg of the material." There is no volume is considered? Or, is that description like saying, Heat capacity of 'this' material is xJ/kg.K? What if I want to say "heat capacity of stainless steel 316 is xJ/kg.K?" $\endgroup$ Feb 17, 2023 at 17:28
  • $\begingroup$ @DKNguyen the diffusivity has to do with how easily the heat is "diffused" into the material. Assuming the same conductivity, a material with higher capacity would be slower to heat up. Conversely, assuming same conductivity, a material with higher conductivity will allow heat to travel faster through it. Diffusivity is a derived quantity (not a basic material property like heat capacity and conductivity). $\endgroup$
    – NMech
    Feb 17, 2023 at 17:57
  • $\begingroup$ @AirCraftLover with specific heat capacity the heat capacity is with respect to the mass. If you want to consider it with respect to volume you only need to multiply specific heat capacity to the density. $\endgroup$
    – NMech
    Feb 17, 2023 at 17:59
  • $\begingroup$ @AirCraftLover You could define specific heat capacity relative to volume of material present, but convention does not because it is less convenient for a number of reasons. On the other hand it makes no sense to define thermal conductivity with respect to mass since it is geometry dependent. Therefore you must take this difference into account when thinking about the two. $\endgroup$
    – DKNguyen
    Feb 17, 2023 at 18:19
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The "inertia" is called specific heat capacity.

Thermal conductivity is not that relevant in your scenario since the pot has been heated for a long time by the time the water comes to a boil. Everything has recahed the same temperature by then. Also, the overwhelming amount of water provides a short cut around the thermal resistance of the pots largely bypassing time differences it takes for heat to travel through the pot material.

To test thermal conductivity you would need to feel how long it takes for the top of an empty pot to reach the same temperature as the heated bottom. Or how long it takes for a stir stick to burn your hand (heat capacity doesn't matter too much here since the water 's heat capacity overwhelms that and the heating element is powerful enough to heat the water so heat capacity in the stick doesn't have much impact on how fast the stick heats up).

Seeing how long it takes for the pot to cool after emptying it is a good indication of heat capacity. That gives clear indications of heat capacity. Then you need to account for mass to see the specific heat capacity (talked about later).

Seeing how long the water boils and cools while sitting in a pot is a somewhat decent way to test thermal conductivity. Better if you keep the lid closed.


Back to specific heat capacity:

You need to account for the mass and volume of those pots. Steel is 3x denser than aluminum so if your pots are similar mass then the aluminum, which already has higher thermal conductivity per cross section area, will have even more due to wall thickness and thus more cross section in the walls.

But if your pots are have similar volume of material (similar wall thickness) then the steel has 3x mass which more than makes up for how its specific heat capacity is only half that of aluminum. You can see in that example steel will have much more heat capacity if the pots have similar dimensions even though aluminum has much more specific heat capacity. Specific heat capacity is given in terms of mass and although aluminum has double the specific heat capacity of steel, steel is so much denser it can make up for that if you keep material cross sections the same.

Your aluminum pots probably don't have walls 3x thicker than the steel pots or weigh the same which would make them fall into the second scenario.

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  • $\begingroup$ What formula to relate to that density? The volume of the two containers are almost same. $\endgroup$ Feb 17, 2023 at 14:35
  • $\begingroup$ See edits. Not volume the container can hold. Volume of material in the container. $\endgroup$
    – DKNguyen
    Feb 17, 2023 at 15:07

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