Where does the excess energy emitted by a microwave go? If there is nothing in the microwave, where does the excess radiation go?
Why doesn't the radiation accumulate and blow it up?
Should I cook two pieces of Canadian Bacon twice as long as I cook one?
 A: The power going into an empty microwave will accumulate as an EM field inside the cavity up to the point where the input power equals the rate of leakage.  Energy inside the cavity will leak out through the little holes in the metal screen in the door, through ohmic losses in the sided of the microwave, feedback into the power source, etc.   The rate of power leakage from the cavity is given by $P_{leak} = 2\pi E f / Q$ where $E$ is the energy stored in the cavity, $f$ is the resonant frequency of the cavity and Q is the "quality factor".  In equilibrium then, $E = QP_{input}/{2\pi f}$.  For an resonant frequency of around 2.5 GHz, input power of 1kW, Q of say 100 (guess), the accumulated energy is only about a milli-joule.  Of course, the materials making up the microwave probably aren't designed to dissipate 1kW of power over an extended time, so eventually the microwave may explode or something due to overheating of components.
A: The microwave would blow up, eventually. If you read the user manual, some warning will placed about not operating the microwave when it's empty. Being a vegetarian, I have no idea about the bacon, but I can guess that it won't behave in a different way than any other material placed in the microwave. In small amount (single or double dish (?)) the energy absorbency is roughly linear and the heating time should be doubled.
