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Whenever I put a meal in the microwave which contains cheese, why does the cheese get hot before the rest of the meal is heated through?

3 Answers3

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It is because cheese has a nice combination of water and fat. The water is important since the microwave transfers energy to it by making the water molecules vibrate. On the other hand, oils, in general, have lower specific heat (compared to water). This means that given the same amount of heat, the temperature change is higher for fat than for water. You can see in this table as normally fatty food has greater specific heat. Moreover, oils have higher boiling points so the cheese can reach a temperature above $100\ \mathrm{^\circ C}$.

Edit

Both vegetable and animal oils are made of nonpolar molecules. This means that oils cannot be effectively heated up by dieletric heating (microwave absorption). If we consider the limit case where oil does not absorb microwaves at all, then any combination of water and oil (mixture) outperforms pure oil at the rate of heating up under microwaves. The mixture, in this case, heats up because water is absorbing microwaves and is giving up heat to the oil by thermal conduction. On the other hand to compare the performances of the mixture and pure water we need to take into account the specific heat of both substances. If the specific heat of the mixture is sufficiently smaller than the specific heat of the water, then the former will outperform the latter in heating up under microwaves.

Can we heat up oil in a microwave? Oils' molecule, in general, may have a non-zero dipole moment but it is so small that oil's dielectric loss factor is about a hundredth of the water's one. Recall that the dielectric loss factor roughly expresses the degree to which an externally applied electric field will be converted to heat. It is in general dependent of the frequency of the radiation and for water, it is maximum at $2.45\, \mathrm{GHz}$, the frequency of most microwaves oven. By a simple home experiment, one can easily check that conduction plays a major role. Try to get some containers that respond differently to microwaves, that is, test how the empty containers heat up. Then separate one that does not heat up and one that does heat up. Fill them with the same amount of oil and let them on the microwave oven for the same amount of time. The oil in the microwave interacting container will be much warmer. The explanation is that the oil was mainly heated up by conduction. Note that in a homogeneous mixture of oil and water (like a cheese) this conduction is optimized.

SRS
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Diracology
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The solid structure of cheese helps prevent steam loss

Water absorbs microwaves well, primarily at the boundaries since the water at the boundaries absorbs most of the microwaves before they get deeper into the body of water. If you heat water alone, its boundaries get most of the heat; then, steam can escape, causing much of that absorbed heat to be lost. This results in a powerful cooling effect called evaporative cooling.

Evaporative cooling is an important effect in microwaves. For example, if you get a microwavable dinner, it'll often tell you to cut a slit in the overwrap without actually removing the contents. The slit allows steam to escape a bit so that pressure doesn't cause the bag to pop, but it still keeps in more steam to help retain heat. This reduces evaporative cooling.

Cheese's solid structure should have a similar effect. This is, the water isn't free to just escape as steam, so the heat it captures isn't so easily lost.

Not really about fats and water working together

The currently up-voted answer asserts that cheese and fats work together using their differing levels of microwave absorption and heat capacity to warm up faster than either alone.

Unfortunately this can't be true because it's an entirely first-order mechanism. Fats and water would both heat up at some rate proportional to the microwave absorption divided by their heat capacity, i.e. $$\frac{\text{d}T}{\text{d}t}{\propto}{\frac{\left[\text{absorption ability}\right]}{\left[\text{heat capacity}\right]}}$$ If you combined them without a second order effect, then their combined absorption ability and heat capacity are both a weighted average of the pure values for each, i.e. $${\left.\frac{\text{d}T}{\text{d}t}\right|}_{\text{cheese}}{\propto}{\frac{{x}_{\text{water}}{\left[\text{absorption ability}\right]}_{\text{water}}+{\left(1-{x}_{\text{water}}\right)}{\left[\text{absorption ability}\right]}_{\text{fat}}}{{{x}_{\text{water}}\left[\text{heat capacity}\right]}_{\text{water}}+{\left(1-{x}_{\text{water}}\right)}{\left[\text{heat capacity}\right]}_{\text{fat}}}}$$ Then, let's assume that absorption ability and heat capacity are constant for both fat and water (which isn't really true, but a reasonable simplification). Then, regardless of the actual values for the absorption abilities and heat capacities, there is no combination that can outperform both pure substances. If both pure substances heat up exactly as fast, then their combination should do the same. But if one heats faster, then the more of it the combination has, the faster the combination'll heat. That is, if we optimize ${x}_{\text{water}}$, we'll necessarily find either ${x}_{\text{water}}=0$ (pure fat) or ${x}_{\text{water}}=1$ (pure water) as the optimal solution.

When a combination works like this, there must be a higher-order effect at work. In this case, I'd suspect that the most important higher-order effect is that cheese traps the steam, such that the water molecules that happen to trap the most heat don't just fly away.

Not really about boiling points

Some have pointed out that water boils at ${100}^{\circ}\text{C}$, so the fats might assist by being able to be hotter. As @JirkaHanika pointed out, this isn't really relevant because water isn't bothered by this until it actually hits its boiling point of ${100}^{\circ}\text{C}$.

If you're microwaving your pizza to be that hot, then you're drying it out. This YouTube video shows a guy putting a cup of water in the microwave with his pizza to help keep the crust crispy:

Nat
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I think that the key point in the first answer is oils have higher boiling points so the cheese can reach a temperature above 100 ∘C.

If you heat water, when it reaches 100C it will start boiling and all microwave energy deposited thereafter will be converting water to steam, which promptly escapes.

In cheese, water is emulsified with fat. (I don't know if it's small droplets of water encased in fat or vice versa. I'd guess the former since it's more than 50% fat). In any case, I think it will be possible for the water to become somewhat superheated without turning to steam in this environment, where it's got a very large amount of water surface in contact with fats which can be heated above 100C without boiling. In other words, the mixture with fat may suppress steam-bubble formation and growth.

Also the water in cheese is derived from milk, which means it will contain a very significant amount of water-soluble milk proteins. These long-chain molecules may also serve to stabilize the water at >100C (especially if they have hydrophilic parts and hydrophobic parts, which will tend to bind between the water and fat where the two touch). They may even allow pressure in the water droplets to somewhat exceed ambient atmospheric pressure.

The obvious experiment is to measure the temperature of cheese freshly heated in a microwave, or even during heating. For the former melt it in a well-insulating container (I'd suggest a smallish cheese sample in a hole in an expanded polystyrene block, and a large mug of water in the oven at the same time so most of the microwave energy has somewhere else to go). For the latter you'll need a completely non-metallic thermometer which does not significantly absorb microwaves and which reads well above 100C, which might be an interesting bit of research in itself. My guess is that the cheese will reach a few degrees above 100C.

nigel222
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