Q:

Does a transformer use power when output isn't under load?

I was reading about how AC to DC converters work with a step-down transformer and then a diode bridge to convert the lower, stepped down AC voltage into DC. What I don't understand is since the input AC appears to be connected to the primary coil of the transformer, how does the DC load affect the power used from the AC supply?
Does the DC load somehow feedback and lower the resistance of the primary coil so that more power can be drawn?
When there is no load on the DC side, does power still flow through the AC primary coil, and if so, why doesn't it just melt?

A:

Does the DC load somehow feedback and lower the resistance of the primary coil so that more power can be drawn?

Yes.  It would be simpler to analyze an AC load though.  The diodes are not central to your question:

The impedance of RL is also transformed, so if you have a 10:1 transformer and RL is 2 Ω, the AC source will see the transformer as a 200 Ω resistor ($10^2⋅2$)
As the current in a coil changes, it creates a changing magnetic field.  In the case of a transformer with a load, however, the change in magnetic field creates a current in the secondary, which immediately creates its own changing magnetic field in the opposite direction, cancelling out the primary's field.  People tend to forget that an ideal transformer has no magnetic field while operating.  Any change in either coil's field is immediately cancelled by a change in the other.
The "feedback" is caused by the same effect.  The primary causes the secondary to change, and the secondary causes the primary to change in return.

When there is no load on the DC side, does power still flow through the AC primary coil, and if so, why doesn't it just melt?

With nothing connected to the secondary side, the secondary coil is open circuited and does nothing.  It's just some metal that happens to be nearby.  The circuit is now just an AC source driving the primary coil, which behaves as a lone inductor:

Ideal inductors do not consume any power; they just store energy temporarily in one half of the cycle and return it to the supply on the other half.  Real coils are not made of perfect conductors, though, and have some resistance, so the power consumed by the primary coil will be determined by the resistance of the wire.
Also, it's not quite right to say "power still flow through the AC primary coil".  "Current" is flowing through the primary, and the resistance of the primary to that current causes it to "dissipate energy" (or power) into the room.  "Power" is actually the rate at which energy flows, and energy actually flows through the empty space between the wires, not in the wires themselves.  Once you understand this, a lot of things make much more sense.

A:

A transformer offers resistance to AC current flow due to the magnetic field produced by the current flow. This "AC resistance" is termed "impedance" and is a function of number of turns, core material, air gaap in core , core dimensions and more. 
When there is no load the applied AC voltage will cause "magnetising current" to flow. This will cause some losses due to eddy current losses in the core and copper losses due to resistance in the winding ("I squared R losses" as power = Current^2 x Resistance). 
These losses are relatively small compared to full load power but not trivial at rest. A few percent of full load power would usually be good.
When a DC load is applied it loads the AC secondary circuit which is tightly couple by the core's magnetic fields to the primary winding. So the DC load resistance appears as if it is an AC impedance load on the primary side and input power increases to meet the load. 
If you apply DC (rather than AC) to a transformer winding there is no ongoing magnetic field change, there is no impedance due to the varying magnetic field and current is limited by the resistance which is low compared to the impedance that should be being generated. If the DC supply has enough muscle power the transformer "just melts".

