If a large number of water droplets of diameter 2r cm each coalesce to form a large water drop of diameter 2 R cm, then prove that the rise in temperature of water is (3T)/(J)((1)/(r )-(1)/(R )). Here, T is the surface tension of water and J is the mechanical equivalent of heat.

If a large number of water droplets of diameter 2r cm each coalesce to

1.	If a large number of water droplets of diameter 2r cm each coalesce to form a large water drop of diameter 2 R cm, then prove that the rise in temperature of water is (3T)/(J)((1)/(r )-(1)/(R )). Here, T is the surface tension of water and J is the mechanical equivalent of heat.
Answer» Solution :If the number of SMALL droplets is n , then . The dissipation of surface energy, W = `(4pir^2 n-4piR^2) T.` We KNOW thatW = JH and H = heat absorbed = `mstheta` [where m = MASS of the larger water drop, s = SPECIFIC heat of water, `theta` = temperature INCREASES] `therefore H=(W)/(J)=(T)/(J)xx4pi(nr^2-R^2)` or, `(3)/(4)piR^3.1.theta=(T)/(J)xx4pi(nr^2-R^2)` [`because m =(4)/(3)piR^3. 1 =(4)/(3)piR^3 and s=1 cal. g^-1. ^@ C^-1`] or, `theta=(3T)/(J)((nr^2-R^2))/(R^3)` `["here", n.(4)/(3)pir^3=(4)/(3)piR^3 or, R^3=nr^3 or, n=(R^3)/(r^3)]` `=(3T)/(J)[(R^3)/(r^3).(r^2)/(R^3)-(1)/(R)]` `=(3T)/(J)(1/r-1/R)`.