If a number of little droplets of water, each of radius `r`, coalesce to form a single drop of radius `R`, show that the rise in temperature will be given by `(3T)/J(1/r-1/R)` where `T` is the surface tension of water and `J` is the mechanical equivalent of heat.

Answer

A

Explanation

Let n be the number of little droplets.
Since, volume will remain constant, hence volume
of n little droplets = volume of single drop
`:. n xx 4/3 pi r^(3) = 4/3 pi R^(3)`
or, `nr^(3)=r^(3)`
Decrease in surface area `=nxx4r^(2)-4piR^(2)`
or `Delta A =4pi[nr^(2)-R^(2)]`
`=4pi [(nr^(3))/(r)-R^(2)]=4pi[(R^(3))/(r)-R^(2)]`
`=4piR^(3)[(1)/(r)-(1)/(R)]` ,
Energy evolved `W =T xx` decrease in surface area
`=Txx4piR^(3)[(1)/(r)-(1)/(R)]`
Heat produced, `Q =W/J=(4pi T R^(3))/(J) [(1)/(r)-(1)/(R)]`
But `Q=msd theta`
Where, m is the mass of bid drop, s is the specific heat of water and `d thata` is the rise ain temperature.
`:. (4pi T R^(3))/(J) [(1)/(r)-(1)/(R)]`=Volume of big drop
`xx`densitty of water xx sp. heat of water `xxd theta`
or, `(4)/(3)pi R^(3)xx1xx1d theta=(4pi T R^(3))/(J)((1)/(r)-(1)/(R))`
or, `d theta=(3T)/(J)[(1)/(r)-(1)/(R)]`