In a microwave oven, why do smaller portions heat up faster? Sorry for brevity, but what is the exact physics explanation of why smaller quantities placed inside a microwave oven heat up faster than when you place a larger quantity of a similar material inside?
 A: Microwaves do not fill the microwave cabinet uniformly.  They form a "standing wave pattern" with high-intensity and low-intensity regions.  When something is placed in the cabinet, it distorts the standing wave pattern, so the pattern of hot and less-hot portions can change.  The reason most microwave ovens include a rotating turntable is to attempt to "blur out" the exposure of different parts of the thing being cooked so the heating is more uniform.
A: I'm assuming you are comparing a microwave oven to a conventional oven.
In short, the reason is microwave energy penetrates the food deeper than a conventional oven. As a result, a small portion will heat up faster throughout in a microwave oven than a conventional oven where the electromagnetic energy absorbed is primarily infrared.
On the other hand, though the temperature of the small portion is more uniform for a microwave oven, the temperature on the surface of the small portion is greater for the infrared radiation. That's because most of the IR is absorbed on the outer layers of the portion.
Hope this helps.
A: The magnetron injects microwave radiation at a certain rate. Ignoring losses, that radiation bounces around the walls until it’s absorbed by the food. If you put two burritos in there instead of one, on average there will be fewer bounces before absorption. That means that with two burritos, the average intensity of the radiation impinging on any point is less—some of photons, if you want to think of it that way, that would have been hitting the spot aren’t there because they’ve already been absorbed.
This is quite different from a regular oven—as long as there is enough power to keep the air temperature at the desired setting, it doesn’t much matter how many burritos you put in there, as long as there’s air space between them. They are heated by conduction from the air, which is unaffected by neighbors, and blackbody radiation from the surroundings which is only affected a bit.
A: Let's say you have a glass of water.  It should be rather obvious that the amount of energy required to heat that water is positively correlated with the amount of water.  Otherwise you would be arguing it's just as easy to heat a pot of water one degree as it is to heat an ocean one degree.
So how does a oven or microwave heat something?  It applies energy over some amount of time.  But the energy output of the microwave (running at max) is fixed so the only other variable that can influence the temperature rise of the object inside is time.
Note: see @Gert answer for the math.
A: The energy sent into a microwave oven's chamber deposits almost entirely in the moisture content in the chamber.
The walls of the chamber are either metal, or have holes too small for the microwave radiation to exit out of.  The magnetron and other components can absorb the microwave radiation, but if there is a non-trivial amount of moisture in the cooking chamber it gets almost all of it.
The food has most of the moisture, so most of the energy pumped into the microwave cooking chamber goes into the food.  If there is more food, less energy is deposited per unit of food.
The microwaves penetrate reasonably deeply into food, heating not only the surface but some of the interior.

In comparison, a conventional electric oven heats the elements up as black body radiators.  Most of this energy is in the form of infrared light, which deposits on food, the walls of the oven, and even in the air in the oven reasonably well.
Heat transfer to the food occurs from air-food physical interaction, food-pan physical interaction, and infrared light radiating from the heating elements and the walls of the chamber.
This doesn't penetrate deeply into the food, unlike microwave radiation.  The interior of the food is mostly heated through conduction from the heated surface, instead of directly from infrared photons.
This means if you attempt to heat up food fast in an oven, the surface burns while the inside stays relatively cold.  You can see this effect when you toast bread under a broiler; the top (near the broiler) can be charred while the bottom of the piece of bread stays cool.
So you heat the oven up to a temperature that won't burn the food's surface before the interior is cooked or warmed.
Two things go on here.
First, for many cooking tasks, much of the heat energy radiates from the oven outside of the oven, instead of into the food.  Energy is spent keeping the cooking chamber warm, not directly at depositing energy into the food.
Second, your cooking rate limitation is on the transfer of heat from the surface of the food to the interior.  Unless your oven is crowded, this isn't limited by adding more food.
Now, adding more food can slow an oven cooking process down.
If you take a small turkey and cook it, it will cook faster than a large one; here, the "more food" forms an insulating layer around the inside of the turkey, slowing the cooking process.
But if you have a large oven, and place an additional sausage reasonably far away from an existing one, the two sausages (a) don't cool the oven significantly, and (b) don't interfere with the other sausage's interior getting heated.
The same 2nd sausages in a microwave oven would absorb MW photons that the other sausage would have absorbed, slowing the warming process.
A: Let $P$ be the power (in $\mathrm{Watt}$) the microwave delivers then a simple heating model can be stated as follows.
The heat energy $q$ needed to heat an object is:
$$q=\varepsilon mc_p\Delta T$$
where $\varepsilon$ is an efficiency factor (for food stuffs with a high moisture/water content $\varepsilon \approx 0.9 - 1$). $m$ is the mass of the object, $c_p$ its specific heat capacity and $\Delta T$ its rise in temperature on heating.
Because:
$$P=\frac{\mathrm{d}Q}{\rm{d}t}$$
It can then be shown with minimal calculus that the heating time $\Delta t$ is given by:
$$\Delta t=\frac{\varepsilon mc_p\Delta T}{P}$$
So all other things being equal (same composition, same $\Delta T$), objects of smaller mass $m$ require smaller heating times.
A: There is a nice answer by Gert, which explains you in classical terms that all other things being equal, objects of smaller mass require smaller heating times. Now I feel like I need to add about the underlying mechanism.
Now microwaves are calibrated so that they correspond to the vibrations of water molecules, thus they can effectively transfer energy to the water molecules (kinetic energy) rotational and vibrational freedom.

This induces polar molecules in the food to rotate and produce thermal energy in a process known as dielectric heating. Microwave ovens heat foods quickly and efficiently because excitation is fairly uniform in the outer 25–38 mm (1–1.5 inches) of a homogeneous, high water content food item.

https://en.wikipedia.org/wiki/Microwave_oven
The emphasis is on the outer layers. The answer to your question is that the underlying mechanism is penetration depth. Microwaves, as they enter the food, they start interacting with the outer layers of water molecules, and transfer their energy to the molecules. Now whenever they do transfer, and photons get absorbed, photons cease to exist (they transform their energies to the molecules), and the microwave loses intensity. It is the number of photons per unit area that decreases, thus the microwave loses intensity as it penetrates deeper and the inner layers get less energy. Thus, contrary to popular belief, a bigger piece of food needs more energy to heat up even with the microwave technology. And so the answer to your question is that a smaller piece heats up faster because the microwaves need to penetrate less layers and reach the center at higher intensity faster.
