Starting Data:
Initial Temperature = 150 K (see Footnote 1)
Asteroid Composition: One or more of the following minerals:
 
Laihunite: Fe+++Fe+++2(SiO4)^2
357.1 g/mole, 3.92 g/cm^3
47.64% Fe, 15.97% Si, 36.39% O

Magnesiocummintonite: (Mg,Fe++)7Si8O22(OH)2
836.02g/mole, 3.13g/cm^3
15.26% Mg, 11.69% Fe, 26.88% Si, .24% H, 45.93% O

Neotocite: (Mn,Fe++)SiO3·(H2O) (?)
149.26 g/mole, 2.8g/cm^3
27.6% Mn, 9.35% Fe, 18.82% Si, 1.35% H, 42.86% O

Now the above classes also incorporate pure Ni-Fe which is found in Taenite
and Kamacite which are:
7.9 g/cm^3
10.46 % Ni, 89.54 % Fe

Elements: Thermodynamic data on the elements present:

Manganese: Mn
Weight: 54.94 g/mol
Melting Point: 1519 K
Specific Heat: 32.0 J/K*mol
Latent Heat of Fusion: 13200 J/mol
Latent Heat of Vaporization: 220000 J/mol

Silicon: Si
Weight: 28.08 g/mol
Melting Point: 1687 K
Specific Heat: 18.81 J/K*mol
Latent Heat of Fusion: 50200 J/mol
Latenet Heat of Evaporation: 359000 J/mol

Magnesium: Mg
Weight: 24.3 g/mol
Melting Point: 923 K
Specific Heat: 32.67 J/K*mol
Latent Heat of Fusion: 8700 J/mol
Latent Heat of Evaporation: 128000 J/mol

Nickel: Ni
Weight: 58.69 g/mol
Melting Point: 1728 K
Specific Heat : 26.1 J/K*mol
Latent Heat of Fusion : 17200 J/mol
Latent Heat of Evaporation: 378000 J/mol

Iron: Fe
Weight: 55.84 g/mol
Melting Point: 1811 K
Specific Heat: 25.1 J/K*mol
Latent Heat of Fusion : 13800 J/mol
Latent Heat of Evaporation : 347000 J/mol

Hydrogen: H
Weight: 1 g/mol

Oxygen: O
Weight: 16 g/mol

 *note all elemental data taken from webelements.com*
*note all mineral data taken from  webmineral.com*

The Math Itself

Step One: The Asteroid

I have chosen, and continue to use the second vaporized asteroid. You will
note it because of its elongated ellipsoidal shape. The first step in
measuring this item's destruction was simply to measure the object. By using
the TL bolt comparison method (compare size of TL bolt with height of brim
trench then compare bolt with asteroid at impact) I obtained the following
measurements for the three axes of the asteroid:
x: 24.2 m
y: 31.9 m
z: 15.5 m

The equation for the volume of an ellipsoid is:
V=4/3*pi*(x*y*z)
This yields: 50121.7 m^3 as the volume of this asteroid.

Step Two Energy Requirements
Anyway here goes mineral by mineral:

---
LAIHUNITE
---

M=V*D
M=50121.7 m^3 * 3920 kg/ m^3
   = 196477064 kg

IRON
Fe=47.64%(M) =  93601673.3 kg
93601673.3 kg * 1 mol/ .05584 kg = 1676247731 mol

Melting Point Temp Change
[1676247731 mol] * [25.1 J/K*mol] * [(1811-150) K] = 6.99e13J

Latent Heat of Fusion
[1676247731 mol] * [13800 J/mol] = 2.313e13J

Latent Heat of Evaporation
[1676247731 mol] * [347000 J/mol] =5.81e14J

Total for Iron = 6.74e14 J

SILICON
Si = 15.97%(M) = 3137787.12 kg
3137787.12 kg * 1 mol / .02808 kg = 1117428316 mol

Melting Point Temp Change:
[1117428316 mol] * [18.81 J / K*mol] * [(1687-150) K] = 3.231e13 J

Latent Heat of Fusion
[1117428316 mol] * [50200 J/mol] = 5.609e13 J

Latent Heat of Evaporation
[1117428316 mol] * [359000 J/mol] = 4.011e14 J

Total For Silicon: 4.895e14 J

---
MAGNESIOCUMMINTONITE
---

M=V*D
M= 50121.7 m^3 * 3130 kg/ m^3
   = 156880921 kg

MAGNESIUM
Mg=15.26%(M)= 23940028.54 kg
23940028.54 kg * 1 mol/ .0243 kg = 985186359.86 mol

Melting Point Temp Change:
[985186359.86 mol] * [32.67 J / K*mol] * [(923-150) K] = 2.488e13J

Latent Heat of Fusion
[985186359.86  mol] * [8700 J/mol] = 8.57e12 J

Latent Heat of Evaporation
[985186359.86  mol] * [128000 J/mol] = 1.261e14 J

Total For Magnesium: 1.595e14 J

IRON
Fe=11.69%(M) = 18339379.66 kg
18339379.66 kg * 1 mol/ .05584 kg = 328427286 mol

Melting Point Temp Change
[328427286 mol] * [25.1 J/K*mol] * [(1811-150) K] = 1.369e13J

Latent Heat of Fusion
[328427286 mol] * [13800 J/mol] = 4.53e12J

Latent Heat of Evaporation
[328427286 mol] * [347000 J/mol] = 1.14e14J

Total for Iron = 1.32e14 J

SILICON
Si= 26.88%(M) = 42169591.56 kg
42169591.56 kg * 1 mol/ .02808 kg = 1501766081 mol

Melting Point Temp Change:
[1501766081 mol] * [18.81 J / K*mol] * [(1687-150) K] = 4.342e13 J

Latent Heat of Fusion
[1501766081 mol] * [50200 J/mol] = 7.539e13 J

Latent Heat of Evaporation
[1501766081 mol] * [359000 J/mol] = 5.391e14 J

Total For Silicon: 6.579e14J

---
NEOTOCITE
---

M=V*D
M= 50121.7 m^3 * 2800 kg / m^3
   = 140340760 kg

MANGANESE
Mn=27.6%(M)= 38734049.76 kg
38734049.76 kg * 1 mol/ .05494 kg = 705024567.9 mol

Melting Point Temp Change
[705024567.9 mol] * [32.0 J / K*mol] * [(1519-150) K] = 3.088e13J

Latent Heat of Fusion
[705024567.9 mol] * [13200 J/mol] = 9.306e12J

Latent Heat of Evaporation
[705024567.9 mol] * [220000 J/mol] = 1.551e14J

Total For Manganese: 1.953e14J

IRON
Fe=9.35%(M) = 13121861.06 kg
13121861.06 kg * 1 mol/ .05584 kg = 234990348.5 mol

Melting Point Temp Change
[234990348.5 mol] * [25.1 J/K*mol] * [(1811-150) K] = 9.797e12J

Latent Heat of Fusion
[234990348.5 mol] * [13800 J/mol] =  3.242e12J

Latent Heat of Evaporation
[234990348.5 mol] * [347000 J/mol] =  8.154e13J

Total for Iron = 9.458e13J

SILICON
Si= 18.82%(M) = 26412131 kg
26412131 kg * 1 mol/ .02808 kg = 940602955.8 mol

Melting Point Temp Change:
[940602955.8 mol] * [18.81 J / K*mol] * [(1687-150) K] = 2.719e13J

Latent Heat of Fusion
[940602955.8 mol] * [50200 J/mol] = 4.722e13 J

Latent Heat of Evaporation
[940602955.8 mol] * [359000 J/mol] = 3.377e14 J

Total For Silicon: 4.121e14J

---
TAENITE/KAMACITE
---

M=V*D
M= 50121.7 m^3 * 7900 kg / m^3
   = 395961430 kg

NICKEL
Ni = 10.46% (M) = 41417565.5 kg
41417565.578 kg * 1 mol / .05869 kg = 705700555 mol

Melting Point Temp Change:
[705700555 mol] * [26.1 J / K*mol] * [(1728-150) K] = 2.906e13 J

Latent Heat of Fusion
[705700555 mol] * [17200 J/mol] = 1.214e13 J

Latent Heat of Evaporation
[705700555 mol] * [378000 J/mol] = 2.667e14 J

Total For Nickel: 3.079e14 J

IRON
Fe = 89.54% (M) = 354543864.4 kg
354543864.4 kg * 1 mol/ .05584 kg = 6349281239 mol

Melting Point Temp Change
[6349281239 mol] * [25.1 J/K*mol] * [(1811-150) K] = 2.647e14 J

Latent Heat of Fusion
[6349281239 mol] * [13800 J/mol] = 8.762e13 J

Latent Heat of Evaporation
[6349281239 mol] * [347000 J/mol] =2.203e15 J

Total for Iron = 2.555e15 J

Step Three: Fourth Grade Addition

Now we have element by element energy requirements for each mineral, we
simply total each mineral's constituent elements to get a total for that
mineral IF the asteroid were purely of that mineral:

Laihunite: 6.74e14J + 4.895e14J = 1.16e15J
Magnesiocummintonite: 1.595e14J + 1.32e14 J + 6.579e14 J = 9.49e14J
Neotocite: 1.953e14J + 9.45e13J + 4.121e14J = 7.01e14J
Taenite/Kamacite: 3.079e14J + 2.555e15J = 2.863e15J

Now this leads to a range of solutions from 701 TJ to 2863 TJ based on the
asteroid being purely of one mineral or the other.