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 involved is:

• A. $+\frac{\pi r^2{h}^{2}dg}{2J}$
• B. $+\frac{\pi r^2{h}^{2}dg}{J}$
• C. $-\frac{\pi r^2{h}^{2}dg}{2J}$
• D. $-\frac{\pi r^2{h}^{2}dg}{J}$

Answer 1

Answer: (A)

$+\frac{\pi r^2{h}^{2}dg}{2J}$

When the tube is placed vertically in water, water rises through height h given by

$h=\frac{2T\cos \theta }{rdg}$

Upward force $=2\pi r\times T\cos \theta$

Work done by this force in raising water column through height $h$ is given by

$\Delta W=\left(2\pi rT\cos \theta \right)h=\left(2\pi rh\cos \theta \right)T=\left(2\pi rh\cos \theta \right)\left(\frac{rhdg}{2\cos \theta }\right)=\pi r^2{h}^{2}dg$

However, the increase in potential energy $\Delta E_p$ of the raised water column $=mg\frac{h}{2}$

where $m$ is the mass of the raised column of water

Thus $m=\pi r^{2}hd$

So $\Delta E_p=\left(\pi r^{2}hd\right)\left(\frac{hg}{2}\right)=\frac{\pi r^2{h}^{2}dg}{2}$

Further

$\Delta W-\Delta E_p=\frac{\pi r^2{h}^{2}dg}{2}$

The part $\Delta W-\Delta E_p$ is used in doing work against viscous forces and frictional forces between water and

glass surface and appears as heat. So heat released $=\frac{\Delta W-\Delta E_p}{J}=\frac{\pi r^2{h}^{2}dg}{2J}$