Question

The energy of a thunderstorm results from the condensation of water vapor in humid air. Suppose a thunderstorm could condense all the water vapor in 10 km3 of air.How much heat does this release? (You may assume each cubic meter of air contains 0.017 kg of water vapor.)How does this compare to an atomic bomb which releases an energy of 2 x 1010 kcal?

Answer 1

The energy released from condensing all the water vapor in 10 km3 of air can be calculated by multiplying the mass of the water vapor by the latent heat of condensation. The total energy released is 2.5 x 1016 J, which is much smaller than the energy released by an atomic bomb.

Step-by-step explanation:

The energy released from condensing all the water vapor in 10 km3 of air can be calculated by multiplying the mass of the water vapor by the latent heat of condensation. Each cubic meter of air contains 0.017 kg of water vapor, so the total mass of water vapor in 10 km3 of air is 10,000,000,000 kg. The latent heat of condensation is approximately 2.5 x 106 J/kg. Therefore, the total energy released is:

Energy released = mass of water vapor * latent heat of condensation

= 10,000,000,000 kg * 2.5 x 106 J/kg

= 2.5 x 1016 J

This is equivalent to 2.38 x 106 kcal. Therefore, the energy released by a thunderstorm condensing all the water vapor in 10 km3 of air is much smaller than that released by an atomic bomb, which releases an energy of 2 x 1010 kcal.

Answer 2

To solve this problem we will apply the equations related to the loss of heat produced in the mass and quantified by the latent heat. From this value it will be possible to find the energy of an atomic bomb and the release of energy. Let's start with the given values, latent heat required for condensation is

`L_c = 2264.76KJ/Kg`

Amount of water in
`10km^3`of air

`m = 10*(1000)^3*0.017`

`m = 1.7*10^8kg`

Therefore the required amount heat release is

`Q=mL`

`Q = 1.7*10^8(2264.76)`

`Q = 3.85*10^(11)kJ`

Now of the values given for the release of an atomic bomb and making the conversion to KiloJules we would have to

One atomic bomb release
`2*10^(10)kCal`

`E = 2*10^(10)*4.184`

`E= 8.368*10^(10)kJ`

Therefore to condense water

`\eta = \frac{\text{Energy in thunerstrom}}{\text{Energy released by atomic bom}}`

`\eta = (3.85*10^11)/(8.368*10^(10))`

`\eta = 4.6\text{bomb energy}`