Well, let's look at the relevant part of the fundamental equation of thermodynamics:
$$\mathrm dE = T\, \mathrm dS + (\text{terms irrelevant for this question})$$
Here $\delta Q = T\,\mathrm dS$ is the heat energy that goes into or out of the system. Obviously, if $T=0$, the $\delta Q=0$. In other words, you can dump entropy in it ($\mathrm dS>0$) without heating it ($\delta Q=0$).
This implies that you could use it to build a perpetual motion machine of the second kind (PM2): You extract heat ("entropic energy") from the environment, dump its entropy into the zero-temperature object, and put the energy into work.
Indeed, if you look at the Carnot efficiency,
$$\eta_C = \frac{T_H - T_C}{T_H}$$
which is the maximal efficiency of a heat engine, and insert $T_C = 0$, you get $\eta_C=1$, that is, perfect efficiency, aka PM2.
However there's of course a caveat: The Carnot efficiency can only be reached for infinitely slow processes. However, there's also a formula for the efficiency at maximum power output, the Curzon-Ahlborn efficiency:
$$\eta_{CA} = 1 - \sqrt{\frac{T_C}{T_H}}$$
Now if you insert $T_C=0$, you again get $\eta_{CA} = 1$. That is, with absolute zero temperature you can actually achieve an efficiency of $1$ at maximum power. That is, you can build a PM2 that actually outputs energy!
Also note that $T\,\mathrm dS=0$ also means that reversible processes cannot heat up a zero-temperature object, so it would stay at zero. Of course that equation doesn't say anything about irreversible processes (it only applies strictly to reversible ones), so an irreversible heating might still be possible.
Now if you dig deeper, I'd expect to sooner or later find some contradiction. After all, there's a reason why thermodynamics says that $T=0$ cannot be achieved.