Question

The surface tension of water at $0^{\circ }C$ is $75.5dyne/cm$. Calculate surface tension of water at ${25}^{\circ }C$. ($\alpha$ for water $=2.7\times {10}^{-3}/K$)

Answer 1

Given, $T_0=75.5dyne/cm$, $\alpha =0.0027$, $\theta ={25}^{\circ }C$

Thus, $T=T_0(1-\alpha \theta )$

$=75.5[1-\left(0.0027\right)\left(25\right)]$

$=75.5[1-0.0675]$

$=75.5\left(0.9325\right)$

$=70.40dyne/cm$

Thus, the surface tension of water at ${25}^{\circ }C$ is $70.40dyne/cm$.