Numerical

The surface tension of water at 0ºc is 75·5 dyne/cm. Find surface tension of water at 25°C. [ α for water = 0·0021/°C ]

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#### Solution

Given, T

Thus, T = T

= 75.5 [ 1 - (0.0027)(25) ]

= 75.5 [ 1 - 0.0675 ]

= 75.5( 0.9325 )

= 70.40 dyne/cm

Thus, the surface tension of water at 25°C is 70.40 dyne/cm.

_{0}= 75.5 dyne/cm, α = 0.0027, θ = 25°CThus, T = T

_{0 }( 1 - αθ )= 75.5 [ 1 - (0.0027)(25) ]

= 75.5 [ 1 - 0.0675 ]

= 75.5( 0.9325 )

= 70.40 dyne/cm

Thus, the surface tension of water at 25°C is 70.40 dyne/cm.

Concept: Surface Tension

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