In the last problem set, the solution to the hurricane problem that some of you

Question

In the last problem set, the solution to the hurricane problem that some of you liked, some of you hated and some of you were blown away by (sorry, just couldn’t resist J), the solution was to simply determine the amount of total rainfall which is the amount of water vapor condensed and then determining the amount of heat released using DH­Vaporization. In actuality, much of the water vapor condenses at temperatures much lower than 0 C. Indeed, the high clouds of hurricanes can be -80 C or colder. Thus, a significant amount of water vapor goes directly to ice particles via vapor deposition. How and why will this affect the total amount of heat released by a hurricane?

Explanation / Answer

On average hurricane produces 1.5 cm/day of rain covering a radius of 665 km. Converting this in terms of volume of rainfall it gives 2.1 x1016 cm3/day. We know that 1 cm3 = 1 gm
so in terms of gm it gets converted to 2.1 x106 g/day. = 2.1 x1013 kg/day.
And we know that the energy released in the condensation of 1 kg of steam into 1 kg of liquid water is 2,260 kJ.

So amount of energy released =2.1x10^13 kg x2260 kJ = 4.7x10^16 kJ
Which is rough equals to 200 times world-wide energy generating capacity, a huge amount of energy is produced
the water vapor associated with latent heat have their roots from the ocean surface, as the ocean temperature is the warmer greater amount of water vapors are transferred in the atmosphere. In order to support a hurricane a continuous supply is required, with a greater depth of hot water, this ensures the mixing of ocean water by strong wind and it will not bring the cold water to hinder the supply of water vapors.

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