Change in Droplet Temperature

Question

A drop of rain falls from a height of 300 m in relation to the ground. The drop first falls uniformly accelerated with an acceleration of 6 m/s^2. At a height of 50 m from the ground the drop is dispersed into 6 small droplets which continue to fall at a constant speed towards the ground. The specific heat capacity of water is 4190 J/(kgK).

a) by how much is the temperature of the drops at the time of getting dispersed higher than the initial temperature of the drops at a height of 300 m?

b) by how much the temperature of each droplet formed increases during the fall?

Answer 1

Fall under constant acceleration of 6m/s2

Height fallen = 300- 50 = 250 m

\Using

vf² = Vi² +2aH

Vi = 0

Vf² = 2aH

Vf² = 2*6*250 = 3000

Velocity at 250m height = Vf = 54.77m/s

KE at this height = 1/2 (mV² )

This KE get converted into heat

1/2 (mV² ) = m*S* Change in temperature

Change in temperature = 1/2(v2)/S , Where S= specific heat capacity of water

Change in Temperature = 1/2 (3000/ 4190) = 0.358 ⁰C = 0.358 K

It gets split into six droplets from here ( We need to know loss of heat in splitting here, due to change in surface energy due to Surface tension )

If we ignore loss of temperature on splitting, Then there is no further change in temperature due to fall with constant velocity till the droplets touch the ground

So net increase in temperature of each droplet = 0.358 ⁰C = 0.358 K