Question

If a number of little droplets of water, each of radius r, coalesce to form a single drop of radius R, show that the rise in temperature will be given by **3T/J (1/r - 1/R)** where T is the surface tension of water and J is the mechanical equivalent of heat.

Answer 1

$\Large\underline{\underline{\sf{\color{red}{GIVEN:-}}}}$

A number of little droplets of water, each of radius r, coalesce to form a single drop of radius R.

$\Large\underline{\underline{\sf{\color{red}{TO\ PROOF:-}}}}$

The rise in temperature will be given by :

$\sf{ \dfrac{3T}{J} \bigg(\dfrac{1}{r}-\dfrac{1}{R} \bigg)}$

$\Large\underline{\underline{\sf{\color{red}{PROOF:-}}}}$

Let n be the number of little droplets.

Since volume will remain constant, hence

• volume of n little droplets = volume of single drop

$\sf{\therefore n \times \dfrac{4}{3}\pi r^3= \dfrac{4}{3}\pi R^3}$

$\sf{\implies nr^3=R^3}$

Decrease in surface area = $\sf{n\times 4\pi r^2- \pi R^2}$

$\sf{\implies \Delta A=4 \pi(nr^2-R^2)}$

$\sf{\implies \Delta A= 4 \pi \bigg[\dfrac{nr^3}{r}-R^2 \bigg]}$

$\sf{\implies \Delta A= 4 \pi \bigg[ \dfrac{R^3}{r}-R^2 \bigg]}$

$\sf{\implies \Delta A= 4 \pi R ^3 \bigg[ \dfrac{1}{r}-\dfrac{1}{R} \bigg]}$

• Energy evolved, W = T × decrease in surface area

$\sf{\implies W=T \times 4\pi R^3 \bigg[\dfrac{1}{r}-\dfrac{1}{R} \bigg]}$

• Heat produced, Q = W/J

$\sf{\implies Q=\dfrac{4 \pi TR^3}{J}\bigg[\dfrac{1}{r}-\dfrac{1}{R} \bigg]}$

But Q = ms dθ,

• where m is the mass of big drop.
• s is the specific heat of water.
• dθ is the rise in temperature.

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$\sf{\therefore \dfrac{4\pi TR^3}{J}\bigg[\dfrac{1}{r}-\dfrac{1}{R} \bigg]=}$

volume of big drop × density of water × sp. heat of water × dθ

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OR

$\displaystyle \sf{\frac{4}{3} \pi R^3\times 1\times1\times d\theta=\frac{4\pi TR^3}{J} \bigg[ \frac{1}{r}-\frac{1}{R} \bigg]}$

OR

$\boxed{\displaystyle \sf{ d\theta=\frac{3T}{J}\bigg[ \frac{1}{r}-\frac{1}{R} \bigg] }}$

Hence, proved!