# What physical property defines how fast a material loses thermal energy? [closed]

I was thinking about the following - let's say, I want to heat a 100 ml of vegetable oil in a 30 cm stainless steel pan (which weighs 1 kg) to 220°C. How should I proceed to reach the desired temperature at the shortest amount of time possible? Should I pre-heat the pan to, say, 300°C and then pour the oil into the pan or should I pour it into the pan while it's room temperature and heat them both? Or will it take the same amount of time regardless of how I proceed? Is there a physical property that defines that sort of thing like, you know, the opposite of thermal conductivity, since while I heat the pan from underneath it still loses energy via radiation and conduction with air. So which of these two materials is better at not transferring the thermal energy - the vegetable oil or the stainless steel?

And one more question - since I already mentioned the diameter and the weight of the pan, what will happen if I swap the 30 cm 1 kg pan with a 26 cm 1 kg pan? How will reducing the surface area of the oil affect the duration of heating the oil i.e. will it start to lose less thermal energy via conductivity?

• Are you starting the clock when you turn on the stove, or when you start pouring oil into the pan? Note that thermal resistivity is a thing, but it's just the reciprocal of thermal conductivity. I believe that the bits that you're missing are the pan's and the oil's heat capacity, and perhaps an appreciation for radiative and conductive heat losses. Sep 12 at 3:21
• In both cases I start the clock when I turn the stove on. Sep 12 at 3:37
• Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer.
– Community Bot
Sep 12 at 4:50

It’s still thermal conductivity. For thermal energy to travel there must be a temperature gradient. This is true for dissipating thermal energy as well. How long it takes to heat the oil will depend on the net effect of gaining thermal energy from the burner and losing thermal energy to the ambient.

To properly answer this question, one needs to consider all interfaces the oil has with its surroundings. This includes the nature of the burner (does it have a constant rate of energy transfer, or is it a constant temperature etc). It includes the shape of the pan (long narrow sides create more surface area to transfer heat to the ambient). It also includes methods of heat transfer from the top of the oil as well (is there any kind of convection going on, can the air be considered to be a constant temperature due to it being a sufficiently ventilated space etc). The nature of the problem requires numerous specifics to form a suitable model.

Wikipedia has a good overview of the methods of heat transfer. https://en.m.wikipedia.org/wiki/Heat_transfer#Convection

As you start the clock when you turn the stove on, there is no real advantage to putting the oil in at a later time. As soon as the pan heats up to a temperature above the ambient it will begin to lose heat to the ambient and whatever it touches. If it doesn’t touch the oil the heat lost will not contribute to heating the oil.

Pre-EDIT: As I go to post this, I see that a better and quite well-constructed answer by @rddr has already been posted (above, no doubt) while I was writing. Nonetheless I will leave this here along with the links below as the work is already done. Among the links is a heat transfer calculator that allows you to input the thermal conductivity as well as the thickness and surface area, so you can experiment with different sized and shaped pans.

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You’re on the right track with thermal conductivity. In fact, thermal conductivity is probably the best answer to your title question as stated. If I am interpreting correctly what you’re trying to get at, another property you my want to consider in trying to think this all through is specific heat.

As you mentioned, material can transfer heat energy in three ways, conduction, convection and radiation.

What you’re describing in the stem is of course many processes occurring over the same time frame. They are essentially all additive, meaning if you could calculate each one and put them all together (with simple addition and subtraction), it would then be fairly straightforward to determine the temperature of the oil at a given time. However, the math/physics to determine the values of all of those components, and as well as how the processes interact with each would be fairly complex. If you want a real life answer for your kitchen, the easiest way will definitely be to test it empirically. If you want to do the math, or at least see the math, I can give you a few links to all the formulas and constants. But keep in mind this is a dynamic environment you’ve described, and it’s therefore not as simple as plugging in some numbers to the formulas. E.g. the thermal conductivity of a fluid such as air is dependent on its pressure and temperature. . . so when the air near the hot pan or stove is heated, its thermal conductivity also changes. Of course it’s also whisked away and quickly replaced with ambient air due to convection.

The Wikipedia page for this kind of thing I typically very good for jumping around and exploring related topics: Specific heat capacity

For the next one, scroll all the way down past the table of thermal conductivities to find a brief summary as well as a calculator that lets you test different thicknesses and surface areas:

Conductive Heat Transfer Calculator
k = thermal conductivity  (W/mK, Btu/(hr ft °F))
s = wall thickness (m, ft)
A = surface area (m2, ft2)
dT = t1 - t2 = temperature difference (oC, oF)
Solids, Liquids and Gases - Thermal Conductivities

This last page includes an Online Air Thermal Conductivity Calculator, where are you can plug in different temperatures and pressures, as well as some nice graphs:

Air - Thermal Conductivity vs. Temperature and Pressure