The surface tension of water at $0^{0} C$ is $75.5 d y n e / c m$ . Find surface tension of water at $25^{0} C$
( $α$ for water $= 0.0021 / K$ )

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Solution

Given, $T_{0} = 75.5 d y n e / c m$ , $α = 0.0021$ , $θ = 25$
Thus, $T = T_{0} (1 - α θ)$
$= 75.5 [1 - (0.0021) (25)]$
$= 75.5 [1 - 0.0525]$
$= 75.5 (0.9475)$
$= 71.54 d y n e / c m$
Thus, the surface tension of water at $25$ is $71.54 d y n e / c m$ .
This is the surface tension of water at $25$ .

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