Your microwave creates a standing electromagnetic wave inside itself but it doesn't consume much energy to create this wave, so if the oven is empty it will only consume a small amount of power - a perfect microwave oven would consume no power when empty. Power is only consumed when you put something in the oven that absorbs energy from the standing wave. The oven needs to take energy from the mains to replace the energy absorbed by whatever's inside it.
The rate at which something inside the oven absorbs energy depends on what you put in. For example if you put in a cup of cooking oil it will absorb energy slowly and heat up slowly (I wouldn't try this at home as traces of water in the oil can boil explosively!!).
The 750W rating of the oven doesn't mean it pours 750W into whatever is inside it; it means that's the maximum amount of power it can pour in. The actual power absorbed will typically be less than this and possibly much less.
It would be an interesting experiment to try heating two (or more) mugs of coffee at the same time. I would bet you'll find the total energy absorbed increases as you put more mugs in, up to the 750W limit.
Later:
Prompted by Anna I have done the experiment. I used two identical coffee mugs containing 400g of water each. I first heated just one mug and measured the temperature rise every ten seconds from about 10C up to about 25C - I didn't want to go higher because you have to start worrying about heat loss to the mug and air.
I then replaced the water and heated the two mugs together, spaced as widely apart as possible, and again graphed the temperature as a function of time again up to about 25C. The results? Well my oven is rated at 600W and with one mug I measured the rate of temperature rise to be 481W (plus or minus a few percent). With two mugs the rate of temperature rise was 530W.
Now I'll just post the results up to the arxiv :-)
Nice intuition! When you say your equation wasn't accurate, do you mean that it didn't accurately fit experimentally obtained time-temperature measurements? That's a little surprising, because I'd expect a similar relationship.
If the insulation is good enough, we might assume that the temperature inside the box and throughout the beaker is uniform (because heat can transfer much more easily within the box than through the sides). In heat transfer analysis, this is called a lumped-capacitance assumption. It lets us ascribe a single temperature $T$ to the inside of the box over time $t$: $T(t)$.
I agree with you that the rate of heat loss can be expected to slow down over time. Certain relevant mechanisms of heat transfer, namely, conduction and convection, depend linearly on the difference between two temperatures. Therefore, we might express the rate of heat loss $q$ from the box (in units of power, or energy per unit time) as $$q=C_1\left[T(t)-T_\infty\right]$$
where $C_1$ a constant representing the material properties and geometry of the insulation and the nature of the dominant heat transfer mechanism and $T_\infty$ is the ambient temperature or room temperature.
(You might even decompose $C_1$ into $C_1=\frac{C_2 C_3 A}{d}$ where $C_2$ is solely the strength of the dominant heat transfer mechanism, $C_3$ is a solely a material property such as the thermal conductivity of the insulation, $A$ is the surface area of the box, and $d$ is its thickness. Here, it makes intuitive sense that the rate of heat transfer scales up with increased surface area and is inversely related to the insulation thickness.)
The rate of heat loss is going to affect the remaining energy in the box (including the beaker) over time. That is, we could write in energy terms that $$C_4 \Delta T(t)=-q\Delta t$$
or
$$C_4 \frac{\Delta T(t)}{\Delta t}=-q=-C_1\left[T(t)-T_\infty\right]$$ where $C_4$ represents some type of thermal capacitance of the beaker and water—that is, a constant that connects their temperature to energy—and $\Delta T(t)$ is the temperature change over some time interval $\Delta t$.
(You might even decompose $C_4$ into $C_4=C_5 C_6+C_7C_8$ where $C_5$ and $C_6$ are the heat capacity and mass, respectively, of the water and and $C_7$ and $C_8$ are the heat capacity and mass, respectively, of the beaker. Maybe this last term is negligible; I don't know. Note that it's essential, though, for the units to agree here and throughout this exercise.)
In differential form, this is $$C_4\frac{dT(t)}{dt}=-C_1\left[T(t)-T_\infty\right]$$ and the solution to this differential equation is $$T(t)=T_\infty+(T_0-T_\infty)\exp(-C_1t/C_4)$$
where $T_0$ is the initial temperature. This equation has a form very similar to the one you postulated. Understanding its derivation may be of use as you continue to characterize and optimize your system.
Best Answer
What do you mean by heat loss number? Depending on your definition, it might be necessary to know the cubic volume of the room.
1) Specific Heat Capacity of air is approx. 1.006 kJ/kgC
2) Air density (assuming dry air) is approx 1.2574 kg/m3
3) Mass = density x volume
4) Temperature differential is 0.3C
Gives us the following equation: heat capacity x mass x temperature differential = heat in KJ
For the second part of your question, you simply need to figure out the heat loss over 8 hours and supply that much heat to the air.
Som practical issues that will complicate things:
1) Outside air temperature. The greater the difference, the faster the heat transfer (into or out of the room).
2) Solar heating of room's exterior walls.
3) Calculating the amount of heat your system actually adds to the room air temperature.
4) Mixing the air to ensure it is evenly heated.