Question

The lower end of a capillary tube of radius $r$ is "placed vertically in water. Then with the rise of water in the capillary, heat evolved is $+\frac{\pi r^{2} h^{2} d g}{n J}$ calorie, where $h$ is the height of capillary rise, $J$ is the joules mechanical equivalent of heat and $d$ is the density of water. Find $n$.

Answer 1

When the tube is placed vertically in water, water rises through height $h$ given by
$h=\frac{2 T \cos \theta}{r d g}$
Upward force $=2 \pi r \times T \cos \theta$
Work done by this force in raising water column through height $h$ is given by
$\Delta W=(2 \pi r T \cos \theta) h=(2 \pi r h \cos \theta) T$
$=(2 \pi r h \cos \theta)\left(\frac{r h d g}{2 \cos \theta}\right)=\pi r^{2} h^{2} d g$
However, the increase in potential energy $\Delta E_{P}$ of the raised water column $=m g \frac{h}{2}$
where $m$ is the mass of the raised column of water.
$\because m=\pi r^{2} h d $
So,$ \Delta E_{P}=\left(\pi r^{2} h d\right)\left(\frac{h g}{2}\right)=\frac{\pi r^{2} h^{2} d g}{2}$
Further, $\Delta W-\Delta E_{p}=\frac{\pi r^{2} h^{2} d g}{2}$
The part $\left(\Delta W-\Delta E_{P}\right)$ is used in doing work against viscous forces and frictional forces between water and glass surface and appears as heat. So heat released
$=\frac{\Delta W-\Delta E_{P}}{J}=\frac{\pi r^{2} h^{2} d g}{2 J}$