1000 small identical liquid droplets coalesce to form a single large drop of radius 1 mm . The surface tension of the liquid is 0.2 N m ⁻¹ , its density is 1000 kg m ⁻³ , and its specific heat capacity is 300 J kg ⁻¹ K ⁻¹ . Assuming all the released surface energy is completely converted into heat and retained by the drop, the rise in temperature of the large drop is x 10⁻³ K . The value of x is:

Step-by-step solution

Let r be the radius of the small droplets and R be the radius of the large drop. Number of droplets N = 1000 . By conservation of volume: 4 3 R^3 = N ( 4 3 r^3 ) R = N^ 1/3 r = 10r Thus, r = R 10 . The initial total surface area of the 1000 droplets is: A_i = 1000 4 r^2 = 1000 4 ( R 10 )^2 = 10 4 R^2 The final surface area of the large drop is: A_f = 4 R^2 The decrease in surface area is: A = A_i - A_f = 10(4 R^2) - 4 R^2 = 9(4 R^2) = 36 R^2 The released surface energy is: U = S A = 0.2 36 (10⁻³)^2 = 7.2 10⁻⁶ J The