theonlysaneone wrote:
I know how to figure out the weight in grams of the metal for the acid part. Assuming they are all perfect cylinders, you square the cross-sectional radius and multiply it by pi, then multiply that by the length of the cylinder. This gives you the volume of the cylinder in cubic millimeters. Since the density of the rod is in cubic centimeters, you divide your last answer by 1000 then multiply by the given density. This will give you the number of grams of metal you have.
As for how to turn that into a number of seconds, I have no earthly idea.
If we suppose that the acid eats into the bar a constant *depth* per second:
Let k = depth per second = dh/dt = dr/dt
r = original radius
h = original height
d = density
Original volume: pi * r^2 * h
End volume: pi * (r-k*t)^2 * (h-k*t)
Material dissolved: d * pi * (r^2 * h - (r-k*t)^2 * (h-k*t) )
...and solve for t. I'm not sure if I want to touch this one - I already did the one for the gravatics...
I suppose that TNT *could* have done something insane and accounted for kinetics. Then dh/dt = dr/dt would be something insane, involving the K_a_ and volume and concentration of acid in the vat. But I don't think they're quite that evil, or realistic... If you
add an alkali metal into water, you produce a base, not an acid.
But it would explain why the Conundum Lenny is "on vacation"...