Question

if a drop of liquid breaks into smaller droplets is results in lowering of temperature of the droplets let a be the drop of radius R break into N small droplets each of radius r. estimate the drop in temperature

Answer 1

Drop in temperature is $\bf{\frac{3T}{\rho c}(\frac{1}r-\frac{1}{R})}$

Explanation:

Suppose, there are N droplets obtained form a bigger drop.

Thus volume of a big drop, $V_2$  is equivalent to  addition of the volumes of the smaller droplets, $V_1$.

Volume of a big drop of radius R = N X  volume of small drop of radius r

$V_2 = N \times V_1$

$\frac{4}{3}\pi R^3 = N \times \frac{4}{3}\pi r^3$

$R^3 = Nr^3$

Energy released i= difference in their surface energy

$E = T\triangle A = T4\pi( R^2-r^2)$

Temperature drops, m denotes mass and c denotes specific heat then,

$E = mc\triangle t$

$\triangle t=\frac{E}{mc} = \frac{T4\pi (R^2-Nr^2)}{\frac{4}{3}\pi R^3\rho c}$

=$\frac{3T}{\rho c}(\frac{1}{R} - \frac{Nr^2}{R^3}) = \frac{3T}{\rho c}({\frac{1}{R}-\frac{Nr^2}{Nr^3})$

$=\frac{3T}{\rho c}(\frac{1}R-\frac{1}{r})$

$\frac{3T}{\rho c}(\frac{1}R-\frac{1}{r})$ is negative because r<R

Thus, drop in temperature is $\bf{\frac{3T}{\rho c}(\frac{1}r-\frac{1}{R})}$