Question

A number of little droplets of water, each of radius 0.1 mm, coalesce to form a single drop of radius 1 mm. Assume the surface tension of water $T=0.07N/m$ and the mechanical equivalent of heat J$=4.2J/cal$. The change in temperature is (answer upto two decimal places)

Answer 1

Let n be the number of little droplets.
Since volume will remain constant,
volume of n little droplets = $n\times$volume of single drop

$n\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3\Rightarrow n{r}^3=R^3\text{Decrease in surface area}\Delta A=4\pi (n{r}^2-R^2)=4\pi (\frac{n{r}^3}{r}-R^2)=4\pi (\frac{R^3}{r}-R^2)=4\pi R^3(\frac{1}{r}-\frac{1}{R})\text{Energy evolved}W=T\times \Delta A=T\times 4\pi R^3(\frac{1}{r}-\frac{1}{R})\text{Heat produced}Q=\frac{W}{J}=\frac{4\pi R^{3}T}{J}(\frac{1}{r}-\frac{1}{R})\text{But}Q=mSd\theta \text{where}m=\frac{4}{3}\pi R^3{\rho }_w=\frac{4}{3}\pi R^3({\rho }_w=1kg{m}^{-3})S=\text{specific heat of water}=1000cal/kg°C\therefore \frac{4}{3}\pi R^3\times 1000\times d\theta =\frac{4\pi R^{3}T}{J}(\frac{1}{r}-\frac{1}{R})\Rightarrow d\theta =\frac{3T}{1000\times J}(\frac{1}{r}-\frac{1}{R})\text{Substituting}T=0.07N/mJ=4.2J/calR={10}^{-3}mr={10}^{-4}md\theta =\frac{3\times 0.07}{1000\times 4.2}(\frac{1}{{10}^{-4}}-\frac{1}{{10}^{-3}})=\frac{1}{2}=0.5°C$