So we're talking 10^25 kJ of energy being released from a falling, condensing vapor canopy.
How would this affect the temperature of the atmosphere if it would not simply just re-vaporize any water? At about room temperature, air has a (constant-volume) heat capacity of about 0.718 kJ/kg/K. Using 5×10^18 kg for the mass of the atmosphere,
(10^25 kJ/5×10^18 kg)(1 kg·K/0.718 kJ) = 2.8×10^6 K
That is, Noah and family would be surrounded by water on the verge of boiling (100ºC) and air at 2.8 million degrees Celsius.
Regarding the above, I was recently asked,
Wait. Seriously. Does anything in this equation allow for the period of time over which the rain would have fallen? Doesn't it claim to have rained for forty days and forty nights? Would that have offset any of the heat caused by the release of energy?
I suppose it doesn't. Some of the heat would have radiated out into space. So let's take the first day's worth, 1/40th of the rainfall. Let's assume that a ridiculous 90% of the heat radiates away in the first 24 hours. So 2.8 million degrees divided by 40, times 10%, gives -- oh, damn -- only 7000 degrees. Maybe 99% radiated away? 700 degrees.
And this is ignoring the original temperature. Imagine it was 99.9% that radiated away. That's still a temperature 70 degrees above normal temperature, or about 95 degrees Celsius (200+ F).
Remember, that's after only ONE day throwing away 99.9% of the heat.