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From my experience the microwave takes longer to heat two bowls of soup than to heat just one.

But it's not so obvious to me why this should be the case.

Each bowl of soup is a single thermally isolated system, and the same power radiates across each one, so it should take the same amount of time to heat both, no?

Edit: I'm not convinced by the below answers. The answer should depend on the specific mechanism by which a microwave works, because if we ask the same question for a stove instead of a microwave, I would guess that two soups placed on a stove top would take the same amount of time as one soup, as long as the soup bowls have the same geometry and thermal conductivity.

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    $\begingroup$ When encountering thoughts like this, I'll resort to thinking about energy and logic: If it didn't take longer then you've either got free energy to use or the system is using more energy than it needs to. Now think about human motivations and see if either of those outcomes are likely... Doesn't work in every case (there are plenty of wasteful systems out there!), but in general an engineer and/or market forces will have applied pressure to design more efficiently. $\endgroup$
    – Lamar Latrell
    Commented Feb 17, 2018 at 22:16
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    $\begingroup$ Why not try it an find out? All it takes is a thermometer and some bowls of water. $\endgroup$
    – pentane
    Commented Feb 18, 2018 at 0:11
  • $\begingroup$ I think an oven is a better system to compare a microwave to than a stove top. $\endgroup$
    – Omnifarious
    Commented Feb 19, 2018 at 9:01

4 Answers 4

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The interior of a microwave oven functions as sort of resonant cavity, with the food items absorbing some of the microwave energy. So the more food is loaded into the oven, the lower the effective Q value of the resonant cavity, and the lower the equilibrium intensity of the electromagnetic waves inside the microwave. This means that the food absorbs less energy per second.

I seem to recall that, as a rule of thumb, tiny quantities of food (say less than half a cup of water) take a roughly constant time to heat. Large quantities (more than a quart of water) scale almost linearly, because they are absorbing almost all of the power output by the microwave.

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    $\begingroup$ I do not believe it is that simple. The power that a magnetron delivers depends on the load unless there is an isolator between them and high power ferrite isolator being an expensive part is not found in commercial ovens. Microwave sources do not like resonant (reactive) loads (in this case an empty oven) and if they do not blow up then they at least they drop their power. $\endgroup$
    – hyportnex
    Commented Feb 17, 2018 at 22:32
  • $\begingroup$ Yes, I did not mean to assert that the magnetron always has constant output. Maybe I will add sophistication to my answer when I get a minute. $\endgroup$
    – Duncan Harris
    Commented Feb 17, 2018 at 22:49
  • $\begingroup$ @DuncanHarris : Does the answer change if we replace microwave by stove top? $\endgroup$
    – math_lover
    Commented Feb 18, 2018 at 21:02
  • $\begingroup$ @JoshuaBenabou, how would you put two things on the same stovetop element? Stack them on top of each other or have each one off-center? $\endgroup$
    – BowlOfRed
    Commented Feb 19, 2018 at 21:15
  • $\begingroup$ yes placed in separate bowls on the same stovetop element (which we assume is large enough to cover the bottoms of both bowls) $\endgroup$
    – math_lover
    Commented Feb 19, 2018 at 22:00
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After a bit of searching I found the most consistently-used formula that expresses the average microwave power dissipation per unit volume of dialectric material (i.e., delicious soup):

$$Q = \frac{1}{2} \omega ε_0 ε'' |E|^2$$

Where the complex permittivity $ε = ε_{abs}/ε_0 = ε' - i ε''$, and $\omega$ is the angular frequency of the microwaves.

Even though the electric field does not penetrate a constant depth into the soup, let's use the 1-D heat equation to describe the heating process.

$$Q = \frac{m \ c_w \Delta T}{t}$$

Where $c_w$ is the specific heat of water.

Building on what Duncan Harris has said above, when the soup is voluminous enough to absorb all the energy we can set the two heats equal and rearrange to get $t$ on one side:

$$t = \frac{2 \ m \ c_w \Delta T}{\omega ε_0 ε'' |E|^2}$$

Given the same power settings on the microwave, everything but the mass of soup, $m$ is constant, which gives a relation consistent with your observation:

$$t \propto 2 m$$

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Ideally, the microwave would be a perfectly reflecting box, so that the microwaves would bounce around until they were absorbed by the soup. In practice, some energy will be absorbed by the walls before it hits a small item.
In the ideal case, two bowls would take twice as long as one.
Another issue is that, since multiple bowls take longer, they lose more heat by conduction and thermal radiation, so it may take even longer to reach the desired temperature.

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    $\begingroup$ I believe most of the "lost" microwave energy in a modern microwave is actually travelling back up the waveguide pipe and heating the magnetron, not being absorbed by the walls. That is why if you run a microwave empty it burns itself out, but the inside walls never get especially hot. $\endgroup$
    – Duncan Harris
    Commented Feb 17, 2018 at 22:20
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It seems to me that heating two vessels in a microwave will take longer, and not even linearly. The focal point of a microwave is the center. Putting in two bowls will take over twice as long as it would take to heat one placed in the center.

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