Edit: Quick google... This paper [1] shows a 180um fiber + cladding carrying 150W over 1km. According to [2], that diameter equates to a 33awg wire, which has 678 ohm/km resistance and a maximum current of 0.072A. At 1km, the wire would be 678ohms. For maximum power transfer, you want the load to equal the resistance of the wire. Thus you could deliver 0.072A into a 678ohm load - i.e. your maximum power would be 3.5 Watts. The fiber is better by 40x.
[1] https://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=8908271...
To deliver 150W with output voltage of 300V (0.5A) would require an input voltage of 640V at an efficiency of ~45%. Not great. The water could handle a lot higher heat distribution but efficiency is already terrible.
> For maximum power transfer
at a fixed input voltage.
Also, your calculation is for equivalent diameters. If mass was more important then, well, the fiber is 1/3 the density of copper…
Now, the limitation is the insulation of the wire. The insulation will determine how high voltage you can use and when you increase voltage, the insulation requirements grow very quickly.
The fiberoptic can transfer more power, just not for the reason you think it can.
[edit]
Upon further reflection your claim about maximum power transfer being when the load and cable are dissipating the same total power doesn't pass the sniff test, because it would imply that a load connected by the same 33awg cable, but only 1m long could only drive a load of 3.5mW since the load would then only be 678 mohm.
In that case we are selecting the input voltage for maximum power so (Rs, I are fixed, Rl can vary, and for a purely resistive load, Vin is a function of Rl):
Vin=I(Rs+Rl)
Vin = Vs + Vl
Vl/Vs=Rl/Rs
Pl = (I^2*Rl)
Clearly we can always select a Rl (and thus a Vin) that gets the desired Pl at a fixed current. Obviously at some point we are limited by shielding of the wire, and DC/DC conversion at the end-point. We eventually may also be limited by interactions between the medium surrounding the wire and EM fields generated by turning the circuit on and off. But, when you can control the voltage, there's no simple calculation from wire impedance to X maximum watts of load.There's no reason you can't shove many optical fibers into that 5mm diameter. People don't, in short distance applications, to have simpler systems and connectors.
There's big reasons why we don't use power over fiber (the endpoints are very expensive, and the overall system efficiencies are low). None of them have to do with the highest optical power density you could hit.
If the design criteria for cables was "make a good cable", we wouldn't have thousands of different types. There are cables which have different mechanical and environmental properties, different cost constraints, prioritization of different performance characteristics, etc.
> Can you have a copper cable able to deliver power over a long distance but unable to deliver the same power over a shorter distance?
Depending on the particular application, yes. Consider some 12/2 solid Romex. It'll handle 10 watts just fine, for quite a distance, if you want. Won't work well for a 1 meter long phone charging cable, though. Different mechanical requirements.