A large number of water droplets each of radius ' t ' combine to form a large drop of Radius ' R '. If the surface tension of water is ' T ' & mechanical equivalent of heat is ' J ' then the rise in temperature due to this is

Step-by-step solution

Radius of each droplet = r Radius of the drop = R As volume remains constant, aligned & n 4 3 r ^3= 4 3 R ^3 & n = R ^3 r ^3 aligned Decrease in surface area =4 r^2 n-4 R^2 n aligned A & =4 [ nr ^2- R ^2 ] & =4 [ R ^3 r ^3 r ^2- R ^2 ] & =4 R ^3 [ 1 r - 1 R ] aligned Enery released W = T T aligned & Heat produced Q = W J & H aligned Heat produced Q = W J Q = m s Put (i) and (iii) into (ii) aligned & m S = 4 R ^3 ~T ~J [ 1 r - 1 R ] & 4 3 R _ water S _ water = 4 R ^3 ~T ~J [ 1 r - 1 R ] & Q = 3 ~T ~J [ 1 r - 1 R ] a