Puzzler's corner

Puzzle 1:
Let there be a regular pentagon P with a side length of 20 millimeters, inside which is inscribed a regular star sharing the vertices of the pentagon as its points. The star is shattered into 5 identical kites. The smallest distance from a "trough" of the star to the edge of the pentagon is 5 millimeters. The star is made of element Q, which is identical to gold but when in rectangular strips, varies its dimensions Width and Length as a function of temperature in the relation (Change in Width per second) = (Change in Length per second) = ln((Temperature in Kelvin)/298) + 1. Two kites are deformed into rectangular strips, of the same width as the shorter side of the kite, and dropped from an altitude X. Assuming that the atmosphere is infinite, what is the necessary altitude X to drop the kites from such that the two rectangular strips emit 500 nm light as a result of blackbody radiation when both in series with an ideal voltage source of 100 kilo-Volts?

First off, I am going to ignore the geometrical error. It turns out, the distance from the "trough" of a regular pentagram to the pentagon must be 7.2654mm, not 5mm. Other than that, it does seem quite solvable.
Regardless of that, each kite has two sides of length 12.360mm and two sides of length 8.981mm. It turns out the toal area is 89.81mm^2, which results in a strip of Q 8.981mm by 10mm.
Skipping forward some, we can calculate that the blackbody temperature needed is 5795K, which leads to a emissivity of 63.95W/mm^2. Since both resistors are the same, the voltage drop in each is 50kV, then the power dissipated across the resistor is 2500000000V^2/R. It turns out, the sheet resistance for gold is about .27 ohms per square, so solving, we can find out the actual size of the sheet needed. This is just solving the equation 63.95(8.981+x)(10+x)=2.5e9/(.27*(10+x)/(8.981+x)), which solves to x=12022.832mm.
From this, we know that the integral from 0 to t_f of ln(T(t)/298)+1 dt is equal to 12022.832. We know that gold has a specific heat of .126J/gK, and given the density, leads to a heat capacity of .2235J/K. Then, we know that .2235dT/dt=63.95W/mm^2. Since the expansion is independent of size, for any 1mm portion, .2235dT/dt=63.95W, which has solution dT/dt=286.15W/s, or T=e^286.15t. This turns to mean that the growth function is 1.9602dt, and thus, takes 6133.57 seconds. Assuming Earth gravity, this would require a fall distance of .5g(6133.57s)^2, or 184467km.