depth: 14.1px
18.57ft/80px * 14.1px = 3.28 ft or 1m deep
wide: 30px
18.57ft/144px * 30px=3.87ft or 1.18m wide
18.57ft/3.28 = 5.66m
2(5.66m * 1m) = 13.36m^2
2(5.66m * 1.18m) = 9.11m^2
+
2(1m * 1.18m) = 2.36m^2
= 24.83m^2
K=80 J/msC (The conductivity of steel
A=18.4m
T= 9000?C (temperature of lava in Mt. St Helens) - 37?C (Megatron body temp)=8963?C
t= 21 seconds
L=.4125m (one half Megatron�s depth if you look at him straight on since the heat will move towards his center)
The temperature difference is tricky because it�s constantly changes as Megatron absorbs enegy. However, we can find what the final temperature will be and use that
q= (80J/smC * 18.4m^2 * 8963?C * 21s) � .4125m = 906.4 E6 joules
Megatron absorbed 36.5MJ for every square meter of his body.
Now the important thing to remember is that this is a HIGH end calculation. For every bit of energy that�s added to Megatron, his temperature rises, and the rate that he absorbs the energy decreases. The 906.4 MJ number is what he would absorb if the rate remained constant at its maximum.