Please read all the other topics where I explain why the cable wont melt.
Few Watts are dissipated on the cable and will depends on the average crest factor of the audio signal.
I have given up on trying to convince TNT, being that he is quite an obstinate fellow.
Assume 600A RMS draw for one second, with 1.5' of 2/0 wire.
If I remember correctly it was about 24W of dissipation.
Okay, with that information, I can tell you about how much it will heat for a given amount of time.
First find the volume of the cable, multiply it by the density, then apply the specific heat for the given dissipation upon the mass.
r^2(pi) R is in cm, and so is the length
(((.9266)/2)^2)pi(45.72) = 30.830509 cubic centimeters
30.830509*8.94 g·cm = 275.62475 grams
0.385 J/g•K is the specific heat of copper.
275.62475*.385 = 106.115529
It takes 106.115529J to raise a sample of 275.6g of copper 1 degree kelvin.
24 / 106.115529 = 0.226168594
With 24J per second of heat being dissipated upon 275.6g of copper will raise the sample 0.226168594 Kelvin per second, or 0.407103 degrees Fahrenheit per second. I don't think that's flashing, or melting the insulation instantly. Though, I am not saying the voltage drop is not horrible. xD It will drop about 0.04676V No, I did not type that up on the keyboard, but that's what is what I get from ohms law.
If anyone catches something wrong, please point it out. //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif
If this is not enough to convince you, I will go on.
Edit: Crap, it was 28W.
28 / 106.115529 = 0.26386336
With 28J per second of heat being dissipated upon 275.6g of copper will raise the sample 0.26386336 Kelvin per second, or 0.474954 degrees Fahrenheit per second. That's a little bit faster, but still is not "OMFG MY WIRE MELTED/VAPORIZED/FLASH/etc"