Let us assume that the rod is rotating in a vacuum. We assume that the rod is slender, i.e. its diameter is much smaller than its length $L$. The moment of inertia of such a rod rotating about one of its ends is $I(L)=mL^2/3$. Due to temperature increase $\Delta T$ the length of the rod increases by $\delta x$. The new moment of inertia is $I(L+\delta x)=m(L+\delta x)^2/3$. Let $\omega$ and $\omega'$ be the initial and final angular velocity of the rod. Then since no torque is applied to the rod due to heating, angular momentum is conserved:
\begin{align}
I(L+\delta x)\omega' &=I(L)\omega\\
(L+\delta x)^2\omega' &=L^2\omega\\
\omega' &=\frac{1}{(1+\delta x/L)^2}\omega\\
&\approx\left(1-2\frac{\delta x}{L}\right)\omega\quad\textrm{for }\frac{\delta x}{L}\ll 1
\end{align}
Fractional change in kinetic energy of the rod is:
\begin{align}
\eta&=\frac{I(L+\delta x)\omega'^2}{I(L)\omega^2}-1\\
&=\frac{\omega'}{\omega}-1\\
&\approx -2\frac{\delta x}{L}=-2k\Delta T\quad\textrm{for }\frac{\delta x}{L}\ll 1
\end{align}
See that we have not used energy conservation equation so far. This becomes:
Change in K.E. of the rod+Change in internal energy of the rod+Change in some other energy of the rod = Heat input. Energy conservation only tells you that there should be at least one other type of energy involved, but not what it is. That is up to you to decide depending on specifics of the problem. In a real rod one possible type would be strain energy stored inside the rod due to its elongation. Of course there could be other sources of energy storage too: for example, if there are attracting magnets attached to the ends of the rod, then the rod must do work in elongating against this attraction.
