Question

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 ]

Answer 1

Given, T₀ = 75.5 dyne/cm, α = 0.0027, θ = 25°C
Thus, T = T₀ ( 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.