Why does canola oil heat up in the microwave?

Question

Introduction: I read on Wikipedia's list of common misconceptions that microwaves work not by emitting the resonant frequency of water, but as a result of dielectric heating. As I understand it, this process heats a substance by emitting a constantly changing electric field, which makes the polar molecules in the substance attempt to align themselves with the field, thus introducing more molecular motion, that is, thermal energy. This makes me think that the heat introduced to the substance should be directly proportional to the polarity of the molecule. I then conducted a brief experiment:

Method: I heated 25g of canola oil and 25g of water in plastic cups in a 1250 W microwave oven for 15 seconds each, measuring the temperatures before and after. I couldn't find the frequency of the waves emitted from the microwave, though I'm pretty sure it's the standard 2.45 GHz. If the frequency is necessary to know for sure, I think I could go back and partially melt a chocolate bar in it, finding the wavelength, and use the speed of light to find the frequency. (http://www.planet-science.com/categories/over-11s/physics-is-fun!/2012/01/measure-the-speed-of-light-using-chocolate.aspx)

Data:

	Water	Canola oil
mass (g)	25	25
time in microwave (s)	15	15
initial temperature (C)	25	24
final temperature (C)	51	33
heat deposited (J)	2718	472

Results:

Using estimated specific heat for canola oil (from https://www.sciencedirect.com/topics/neuroscience/canola-oil and https://doi.org/10.1080/10942910701586273) of 2.1 J/gK and 4.182 J/gK for water, it can be found that the change in energy of the water is:

$q = mc(\Delta T)=(25~\mathrm{g})(4.182~\mathrm{J/gK})(26~\mathrm{K}) = 2718 ~\mathrm{J}$

And the change of energy in the oil is:

$q = mc(\Delta T)=(25~\mathrm{g})(2.1~\mathrm{J/gK})(9~\mathrm{K}) = 472~\mathrm{J}$

So there was about six times more energy given to the water than the oil.

Discussion: This seems odd to me. First, if dielectric heating relies on polarity, why is the canola oil heating at all? Second, why is it heating as much as it is? I would think that the hydrogen bonds in the water are far more than six times as strong as the London dispersion forces within the oil. Is it because the oil is diluted with a polar substance? Is it because the lowered polarity makes it easier to move the molecules, and therefore impart heat? What should be expected to happen if a completely nonpolar material is microwaved?

Answer 1

This makes me think that the heat introduced to the substance should be directly proportional to the polarity of the molecule.

In addition to the good answer from Gert, there's a problem in this step. The microwave oven is a metal box. The purpose of the metal box is to reflect whatever microwaves aren't absorbed by the food (either because they "missed the target" or because they passed through). This reflection isn't 100% efficient, but it's actually pretty close, so for the sake of discussion we can pretend that it is. What that means is that basically all of the power emitted by the magnetron goes into the food. If you put a sample in there with a lower dielectric loss it will absorb less energy, and heat up less, "on the first pass", but that just means that more of the energy will be available to bounce off of the walls and take another try, and another, and another — the field density will increase until it reaches the point where absorbed power equals input power.

Obviously there are practical limits to this (an empty microwave oven would either have to deliver all of its heat to itself, or else automatically turn off or reduce power), but for reasonable samples of food-like stuff it's close enough to true to seriously mess with the concept of your experiment.

Answer 2

First, if dielectric heating relies on polarity, why is the canola oil heating at all? Second, why is it heating as much as it is?

Firstly, Canola oil is a vegetable oil, more specifically a triglyceride. Vegetable oils contain polarity, coming from the glyceride ester bonds.

Secondly, the temperature increase observed in those nice experiments you conducted, depends not only on the microwave energy converted to heat but also on the amount of mass and the specific heat capacity $c_p$.

Adiabatic heating of an amount of substance $m$ is given by:

$$\Delta Q=m c_p \Delta T$$

where $\Delta Q$ is the amount of heat and $\Delta T$ the temperature increase$^\dagger$.

Water has characteristically a high $c_p$, including compared to vegetable oil, so it needs more heat energy to reach the same temperature (all other things being equal).

For water, $c_p=4.18\mathrm{kJ/kgK}$ and for sunflower oil, $c_p=2.24\mathrm{kJ/kgK}$

$^\dagger$ $c_p$ is assumed temperature-invariant over small temperature intervals.

Answer 3

First of all, I'm impressed by the work you did with background reading, experimental investigation and hypothesising! Great job!

One issue here is that the concept of "polarity" is almost always used somewhat loosely in most contexts in chemistry, even though it has a rigorous physicochemical interpretation (the presence or absence of an electric dipole moment in a molecule). There is no sharp cut-off between liquids classified as "polar" or "non-polar", it's a somewhat arbitrary boundary. Proof of this is that there are several different scales which attempt to quantify this notion of "chemical polarity", and they don't agree perfectly with each other (see, for example, this table with six distinct measures of polarity).

Furthermore, in truth no material can be completely "non-polar". Even molecules containing only bonds between the same atoms (e.g. nitrogen gas or liquid bromine) or where any dipole moments between atoms cancel out due to symmetry (e.g., carbon tetrachloride) will generate "instantaneous" dipoles due to quantum mechanical uncertainty - at any single point in time, electrons might preferentially slosh towards one side of the molecule, even if on average the molecule has no net dipole. These instantaneous dipoles will interact with the microwave radiation.

The combination of the above points should already make it clear that whatever substance you put in your microwave, you should expect it to absorb some amount of energy. Now trying to quantify how much would be absorbed is very tricky business, whether it's based purely on theoretical calculations, or based on experimental parameters such as "polarity". This kind of problem needs extensive treatment using rather hefty electrodynamics calculations. I'm afraid your extrapolation that solvent A absorbing six times more energy than B means that "solvent A has six times more polarity than solvent B" is too crude.

Bringing things back down to earth a bit, let's look at your examples. While vegetable oils are often considered overall "non-polar", and while they do have some properties similar to other solvents which are generally considered "non-polar", it's important to note that they are largely composed of a mixture of fatty acids. These molecules have a carboxylic acid functionality at the end, which is often considered a "polar" functional group (in the sense that it generates a significant local dipole), and its presence may account for the fact that vegetable oils absorb as much microwave radiation as they do. In fact, it's even worth pointing out that there are other substances in vegetable oils, including water. Even though water will typically represent less than 1% of the weight of a sample of vegetable oil, it can absorb disproportionately more microwave radiation.

Finally, I'd like to share an example of the concepts involved in your question playing a role in chemistry. There is extensive work in synthetic chemistry using microwave radiation as energy input for a reaction (as opposed to, say, putting a flask on top of a heated surface). For a while there was some back-and-forth discussion whether microwave heating had any "special" effects, but it is now largely understood that in the vast majority of cases, all the microwave radiation really does is transfer heat energy to the solvent. In that case, it becomes important to study how well different solvents heat up in a microwave reactor. However, this is usually directly measured by simply running some tests and comparing the different curves of reactor temperature versus time. As you can see, the least polar solvent in that graph, 1,4-dioxane, heats up much more slowly than the rest. However, it does heat up, at least up to a point where it loses as much heat to the surroundings as it can absorb from the microwave radiation.

Answer 4

Interesting query. Let us make an assumption. The container of water and oil did not contribute to heating (i.e., it did not absorb microwaves).

The fundamental requirement for microwave absorption is that the molecule must have a permanent dipole moment. Second, the incoming radiation must have energy(ies) corresponding to the rotational energy levels of the molecules.

After absorption of (resonant) microwave radiation, the net effect is that the molecules begin to rotate. The greater the dipole moment of the molecule, the stronger is its microwave absorption. Microwave spectroscopists usually prefer small molecules for their studies (< 300 daltons). The main components of vegetable oils of canola oil are triglycerides of long-chain fatty acids. Their molecular weight is pretty considerable from "microwave standards."

Once you have large molecules, two things happen:

1. Their rotational frequency becomes slow; imagine a tiger vs. an elephant. Larger molecules rotate slowly.

2. They have hundreds to thousands of rotational lines, implying that microwave radiation of a particular single resonant frequency will have minimal heating effects.

Therefore, your observations are not unexpected. Although the above arguments are for gas-phase molecules (microwave spectroscopy is, after all, a gas phase technique), the arguments can be extended to liquids with a grain of salt. All foods have water, and the microwave frequency of, say, 2.45 GHz corresponds to water absorption lines. Note that I am unable to locate the bandwidth of 2.45 GHz frequency of a standard microwave. The question is, how broad is the incoming radiation from a microwave oven?

Some of the rotational lines of canola oil components (mainly triglycerides of oleic and other fatty acids) must be resonant with the microwave source and hence the minor heating effects you observe in your experiment.

Answer 5

Saying that microwave ovens do not heat water by operating at a specific resonance frequency of water but rather "dielectric heating" is about as meaningful as saying that chlorophyll does not get its green color by absorbing red and blue light but rather by "electromagnetic illumination".

Of course is the fact, that water gets heated efficiently at the common microwave oven frequencies, a result of the fact that water has some resonances more or less near those frequencies. That is pretty trivial. If it did not have resonances there, it would only be heated a little. And every resonance has some linewidth, so it is not essential to hit one resonance exactly, especially if multiple resonances get excited simultaneously.

The actual question is what mechanism causes these resonances. And for water it is the (quantum mechanical) rotation modes that cause it to respond to the microwave oven's radiation. They are determined by the water molecule's dipole moment and its moments of inertia. However, in liquid water, the individual molecules do not move independently, but they are subject to intense interactions (contrary to water vapour). These interactions cause relaxation ("damping") that broadens the resonance lines extremely. Hence, if a water molecule in the vapour phase had a rotational resonance in the ~20 GHz range, this would normally not allow substantial absorption at ~2 Ghz, but due to linewidth broadening/relaxation in the liquid phase, such absorption far off the center of a resonance becomes possible. The absorption coefficient is low, but significant. Actually this is desired, because for higher absorption (closer to the free resonance), the microwaves would get absorbed already at the surface, and the core of the meal would stay cold.

But the fact that water has resonances in that broad range, does not preclude that other substances couldn't have resonances also in that range. Remember that moment of inertia determines the resistance to rotational acceleration. According to quantum mechanics, the higher the moment of inertia, the lower the rotational resonance frequencies:

$$E=\frac{h^2}{8 \pi^2 I}J(J+1)$$

where $I$ is the moment of inertia around the respective axis, and $J$ is the angular momentum quantum number. If the moment of inertia is big (as e.g. for a long fatty acid molecule), the transition frequencies get smaller. This might compensate a smaller excitability due to smaller electric dipole moment.

Apart from these basic facts of electromagnetic absorption, what Gert has written about heat capacity also applies. The result is probably a combination of both, absorption coefficient and heat capacity. Possibly also heat conduction, because the higher the absorption, the more the matter is only heated at the surface, and it may take some time for the heat to distribute (a problem that anyone ever having heated some food in the microwave oven has faced). So, you'd have to make sure that your measuring an almost equilibrium state by waiting some time, but not too long, because that might already cool down your samples.

Bottom line: you can't just predict absorptivity of a material by handwaving arguments like "but the dipoles... non-polar... etc.". If ever, you'd have to do detailed quantum numerical computations, that take into account relaxation effects.