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Mechanical Properties Of Fluids

Question
CBSEENPH11018301

Given, n identical drops (each of radius r) of liquid of surface tension T and density ρ coalesce to form single drop. The energy released in the processes is converted into heat. Find the increase in the temperature of drop.

Solution

Given that 'n' drops coalesce to form a single drop.
During the process, energy is released which is converted into heat. 
Energy released is given by,
straight E space equals space 4 πTr squared space left parenthesis straight n space minus space straight n to the power of begin inline style bevelled 2 over 3 end style end exponent right parenthesis space
This energy is converted to heat.

Therefore, 
space space space space space space space msJθ space equals space 4 πTr squared space left parenthesis straight n space minus space straight n to the power of 2 divided by 3 end exponent right parenthesis space

rightwards double arrow space straight rho space left parenthesis straight n 4 over 3 πr cubed right parenthesis sJθ space equals space 4 space straight pi space Tr squared space left parenthesis straight n space minus space straight n to the power of 2 divided by 3 end exponent right parenthesis space

straight i. straight e. comma space space space space straight theta space equals space fraction numerator 3 straight T over denominator rsρJ end fraction open parentheses fraction numerator straight n space minus space straight n to the power of 2 divided by 3 end exponent over denominator straight n end fraction close parentheses
where,  θ is the increase in temperature drop.