Two water drops each of radius 'r' coalesce to from a bigger drop. If ' T ' is the surface tension, the surface energy released in this process is :

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​ 2 × 3 4 ​ π R 3 = 3 4 ​ π r 3 ⇒ r = 2 1/3 R U i ​ = 2 × 4 π R 2 T U f ​ = 4 π r 2 T = 4 π R 2 T 2 2/3 ∴ Heat lost = u i ​ − u f ​ = 4 π R 2 T [ 2 − 2 2/3 ] ​

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