blog entry

[bwanaaa]

forgive me for listing this as an issue, but I am burning up with curiosity as to the final equilibrium temp of the earth after the white sky/hard rain peaks. Although you calculate the mass impacts, I dont really know how to accurately estimate the energy those meteorites will impart upon reentry. Some will entirely burn up and others not. I suppose one could grossly assume orbital reentry from an orbital velocity (3200 m/s) and calculate the total energy using the mass of the moon that re-enters (assume 1/3 of moon mass 7.34767309 × 10^22 kilograms = ~ 2.5 x 10^22 kg and calculate the total kinetic energy (1/2 * mv^2)
K= 1/2 mv^2
K= 1/2 * 2.5x10^22 * 3200^2 (units are kg * meters^2/s^2 = Joules)
K=1.28 x 10^30 joules
The thermal energy Q=mcDT where m is mass, c is specific heat, and DT is temp rise
So if all the kinetic energy goes into thermal energy
1/2mv^2=mcΔT
C for water is 4.179 Joules/(gram degrees C), m=2.5*10^25 grams
mcDT=1.04 x 10^26 Joules * DT
1.28 * 10^30 joules / 1.04 x 10^26 joules = 1.2 x 10^4 degrees C

12000 degrees?! but this does not take into account that much of the thermal energy will be used to change water from its liquid to vapor phase.

For this problem, there are only two heats to consider:
q1= heat required to warm the water from 20.0 °C to 100.0 °C.
q2= heat required to vaporize the water to steam at 100 °C.

q_1 = mcΔT = 1000.0 g × 4.184 J / (°C x g) × 80.0°C = 334,720 J
q_2 = mΔH_"vap" = 1000.0 g × 2260 J / g = 2,260,000 J
To convert 1 k g of water at 20.0 °C to steam at 100.0 °C requires 2.595 x 10^6 Joules of energy.

the mass of the oceans is 1.4 x 10^21 kg
To convert the oceans from 20deg C to steam requires 3.633 x 10^27 Joules
But the thermal energy from the white rain is 350x greater!
So not only will the oceans be converted to steam but the earth will become molten.

Please tell me I'm waaay off.