1.

The surface tension of water at ` 0^(@) C` is 75.5 dyn/cm. Calculate the surface tension of water at ` 25^(@)C` .

Answer» Given : `T_(0)=75.5` dyne/cm, `alpha=0.0027//^@C, theta = 25^@C`
Thus, `T=T_(0)(1=-alphatheta)=75.5[1-(0.0027)(25)]=75.5[1-0.0675]`
`75.5(0.9325)`
`=70.40` dyne/cm
Thus, the surface tensiono of water at `25^@C` is 70.40 dyne/cm.


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