I think there are 2 main sources of confusion:
First, because of gravity, extending your arms feels like work. We're only interested in the radial movement, though, and in this direction, the skater's arms are pulled by the centrifugal force (in the long tradition of spherical cows in vacuum, we could replace the figure skater with two beads on a spinning rod).
Second, the idea of rotational energy as kinetic energy. The relevant work variable is (as already mentioned) the radial extension of the skater's arms, and as far as that's concerned, rotational energy plays the part of potential energy.
Think of the skater pulling in her arms as compressing a spring, and extending the arms as its release.
Going by either the bead or spring model, the rotational energy gets converted into kinetic energy of the arms, accelerated by the centrifugal force in direction of the radial work variable and ultimately dissipating via vibrations when the arms abruptly reach maximal extension.
Of course, if the skater doesn't let her arms be accelerated and slowly extends them instead, the energy dissipates right away, which might be the more realistic approach.
As you can see from this wiki link, the relation between torque and angular acceleration is derived in this way:
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
$$\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}$$
where $L$ is the angular momentum vector and $t$ is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
$$\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}$$
For rotation about a fixed axis,
$$\mathbf{L} = I\boldsymbol{\omega}$$
where $I$ is the moment of inertia and $\omega$ is the angular velocity. It follows that
$$\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha}$$
where $α$ is the angular acceleration of the body, measured in $rad/s^2$.
Your answer specifically lies in the following paragraph:
This equation has the limitation that the torque equation describes
the instantaneous axis of rotation or center of mass for any type of
motion – whether pure translation, pure rotation, or mixed motion. $I$
= Moment of inertia about the point which the torque is written (either instantaneous axis of rotation or center of mass only). If
body is in translatory equilibrium then the torque equation is the
same about all points in the plane of motion.
Few points I would like to emphasise on:
- In general, in the formula, $\mathbf{L} = I\boldsymbol{\omega}$, $I$ is the moment of inertia tensor and depending on the axis of rotation and our assumption of co-ordinate axes, it becomes a scalar or a vector and so on.
- In $\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha}$, $I$ is a constant, hence it is taken out of the term $\frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t}$ to form the term $I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t}$.But here $I$ is not constant. So you cannot use that formula.
Best Answer
The energy comes from the ice-skater's muscles; they have to work to pull their arms in.
There is no external work done on the skater - the energy is converted from the chemical potential energy stored in the skater's body to kinetic energy.