A large number of small identical liquid drops, each of radius 0.1 mm , coalesce to form a single large drop of radius 10 mm . The released surface energy is entirely absorbed by the liquid, raising its temperature. If the surface tension of the liquid is 0.1 N m ⁻¹ , its density is 1000 kg m ⁻³ , and its specific heat capacity is 3000 J kg ⁻¹ K ⁻¹ , the rise in temperature of the liquid is

Step-by-step solution

Let n be the number of small drops, r be the radius of each small drop, and R be the radius of the large drop. By conservation of volume: n 4 3 r^3 = 4 3 R^3 n = ( R r )^3 The decrease in surface area is: A = n(4 r^2) - 4 R^2 = 4 [ ( R r )^3 r^2 - R^2 ] = 4 R^3 ( 1 r - 1 R ) The released surface energy is: E = T A = T 4 R^3 ( 1 r - 1 R ) This energy is used to heat the liquid. The heat absorbed is: Q = m s = ( 4 3 R^3 ) s Equating E and Q : T 4 R^3 ( 1 r - 1 R ) = 4 3 R^3 s = 3T s ( 1 r - 1 R ) Given values: T = 0.