Thunderstorms that form east of the Rocky Mountains obtain their moisture primarily from evaporated water from the Gulf of Mexico. That water vapor, when condensed within a thunderstorm, creates the thunderstorm’s clouds and vast amounts of energy are released into the atmosphere in the form of latent heat. Let’s calculate how much water and energy is in a large thunderstorm complex. Determine the total number of gallons of water in this storm and also the total energy released in the storm expressed in megatons and joules. You must show your work to get credit.

Assumptions:
Each cubic centimeter (cm3) contains 7.5*10-7 grams of water (7.5*10-7 g/cm3)
The storm’s dimensions are 60 km wide, 400 km long and 12 km tall.
1 gallon of water = 3785 grams of water
Each gram of water in the cloud releases 2260 Joules of energy as it condenses (2260 J/g)
A 1-megaton nuclear bomb releases 4.18 x 1015 Joules of energy

5.71×10¹¹ gallons

4.88×10¹⁷ J

1.17×10² megatons

Step-by-step explanation:

Volume of the storm:

(60 km) (400 km) (12 km) (1000 m / km)³ (100 cm / m)³ = 2.88×10²⁰ cm³

Mass of water:

(2.88×10²⁰ cm³) (7.5×10⁻⁷ g/cm³) = 2.16×10¹⁴ g

Volume of water:

(2.16×10¹⁴ g) (1 gallon / 3785 g) = 5.71×10¹¹ gallons

Total energy:

(2.16×10¹⁴ g) (2260 J/g) = 4.88×10¹⁷ J

(4.88×10¹⁷ J) (1 megaton / 4.18×10¹⁵ J) = 1.17×10² megatons

Step-by-step explanation:

a) The volume of the storm is ...

  (60 km)(400 km)(12 km) = 2.88×10^5 km^3

  = (2.88×10^5 km^3)×((10^5 cm)/(1 km))^3 = 2.88×10^20 cm^3

Then the volume of water is ...

  (2.88×10^20 cm^3)×(7.5×10^-7 g/cm^3) = 2.16×10^14 g

In gallons, this is ...

  (2.16×10^14 g)×(1 gal)/(3.785×10^3 g) ≈ 5.71×10^10 gal

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b) The energy released is ...

  (2.16×10^14 g)(2.260×10^3 J/g) = 4.8816×10^17 J

In megatons, this is ...

  (4.8816×10^17 J)/(4.18×10^15 J/megaton) ≈ 117 megatons

Comment on the storm

The average rainfall over the storm area would be about 0.9 cm, about 0.35 inches.

e) none of the above because the measure of angle cba is 101°