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32 minutes 44 seconds

Here's a simplified (i.e., probably not the best) way of looking at it assuming continuous, constant movement of the hands of the clock. One nice, simplifying thing is that the hour hand and minute hand start at the same point, let's say $(0,0)$.

Let's use $(t,\theta)$ to denote the internal angle of the circle $\theta$ swept out over time $t$, where $t$ is measured in minutes. For the hour hand, we start at $(0,0)$ and end at $(60,\frac{1}{12}\cdot2\pi)$ over the course of an hour. We can let the function $\theta_h(t)=\frac{\pi t}{360}$ describe such motion. Similarly, for the minute hand, we start at $(0,0)$ and end at $(60,2\pi)$ over the course of an hour. We can let the function $\theta_m(t)=\frac{\pi t}{30}$ model this motion.

Hence, we are essentially looking for a $t$ for which $\theta_m-\theta_h=\frac{\pi}{2}$ as well as a $t$ for which $\theta_m-\theta_h=\frac{3\pi}{2}$. Or, more helpfully, we are trying to find a $t$ for which $$ \frac{\pi t}{30}-\frac{\pi t}{360}=\frac{\pi}{2},\frac{3\pi}{2}, $$ and doing some basic algebra shows that what we get are solutions when $t=\frac{180}{11}, \frac{540}{11}$; hence, the difference in minutes between these two instances is $$ \frac{540}{11}-\frac{180}{11}=\frac{360}{11}=32\frac{8}{11}\;\text{minutes} $$

or, more simply, 32 minutes and 44 seconds.

Side note: The hour and minute hands will be at right angles at (roughly) 12:16:22pm and 12:49:05pm.

As said above, this is probably not the cleanest way of approaching it, but maybe it will be somewhat clearer.