Question

The end of a capillary tube (glass) with radius $r$ is immersed into water. What amount of heat in the joule will be evolved when the water rises in the tube? Take surface tension, $T = \sqrt {\pi \rho } $ where $\rho $ is the density of water.

Answer 1

Hint:In order to solve this question we need to understand surface tension and heat. Consider a water bubble inside water, it experiences force from all directions but if we consider water bubbles on surface of water then it experience force from only three directions, not from above as above the bubble there is no water, so a unbalanced downward force acts on bubble and due to this tension developed on the surface, this is known as surface tension. It is mathematically defined as the ratio of force per unit length. Heat is defined as energy which flows due to temperature difference.

Complete step by step answer:
Consider a water bubble of spherical shape having radius $r$.Given the surface tension is, $T$ and according to the problem, $T = \sqrt {\pi \rho } $.
So a force due to surface tension is given by, $F = Tl$
Here $l$ is length of bubble which is equal to, $l = 2\pi r$
So force is, $F = 2\pi rT$

Since the water level rises by an amount of height, “h”.So Volume of water, as it forms a cylinder of height “h” and radius “r” so, $V = \pi {r^2}h$.Since density of water is, $\rho $.So mass of water which rise is, $m = \rho V$
$m = \pi {r^2}h\rho $
So force of gravity is,
${F_g} = mg$
$\Rightarrow {F_g} = \pi {r^2}h\rho g$
Considering water bubble to at equilibrium we equate both forces, we get, $F = {F_g}$
$2\pi rT = \pi {r^2}h\rho g$

Putting values we get, $2\pi r\sqrt {\pi \rho } = \pi {r^2}h\rho g$
$h = \dfrac{{2\pi r\sqrt {\pi \rho } }}{{\pi {r^2}\rho g}}$
$\Rightarrow h = \dfrac{2}{{rg}}\sqrt {\dfrac{\pi }{\rho }} $
So work done by force due to surface tension in raising height “h” is, ${W_1} = Fh$
Putting values we get, ${W_1} = 2\pi rTh$
${W_1} = 2\pi r(\sqrt {\pi \rho } )(\dfrac{2}{{rg}}\sqrt {\dfrac{\pi }{\rho }} )$
$\Rightarrow {W_1} = \dfrac{{4{\pi ^2}}}{g}$

Also Potential energy stored by gravity is , ${U_g} = \dfrac{{mgh}}{2}$
${U_g} = \dfrac{{(\pi {r^2}h)\rho gh}}{2}$
$\Rightarrow {U_g} = \pi {r^2}\rho g{h^2}(\dfrac{1}{2})$
$\Rightarrow {U_g} = \pi {r^2}\rho g{(\dfrac{2}{{rg}}\sqrt {\dfrac{\pi }{\rho }} )^2}(\dfrac{1}{2})$
$\Rightarrow {U_g} = \dfrac{{2{\pi ^2}\rho g{r^2}}}{{{r^2}{g^2}\rho }}$
$\Rightarrow {U_g} = \dfrac{{2{\pi ^2}}}{g}$
So heat evolved is, given as, $H = {W_1} - {U_g}$
Putting values we get, $H = \dfrac{{4{\pi ^2}}}{g} - \dfrac{{2{\pi ^2}}}{g}$
$\therefore H = \dfrac{{2{\pi ^2}}}{g}$

So the amount of heat evolved in joules is, $H = \dfrac{{2{\pi ^2}}}{g}\,J$.

Note: It should be remembered that gravity will act only on the center of mass of water raised which is at height $\dfrac{h}{2}$ so , potential energy stored in water contains a half factor. Also energy is released in form of heat if the eater raises, but if water falls in the capillary tube then heat would be required to do work against gravity.