Question

The surface tension of water at 0°C is 70 dyne/cm. Find surface tension of water at 25°C (α for water = 0.00271°C)

Answer 1

Given :

t 1 = 0⁰C

t 2 = 25⁰C  

α = 0.0027/⁰C  

T = T0 (1 – α∆t)  

= T0 [1 – α (t2 – t1 )]

T = T0 [1 – α (t2 – t1 )]  

T = 70 [1 – 0.0027 (25 – 0)]

= 70 [1 – 0.0027 × 25]  

= 70 × 0.9325

∴ T = 65.275 dyne/cm

Answer 2

Answer:

65.279

Explanation:

t 1 = 0⁰C

t 2 = 25⁰C  

α = 0.0027/⁰C  

T = T0 (1 – α∆t)  

= T0 [1 – α (t2 – t1 )]

T = T0 [1 – α (t2 – t1 )]  

T = 70 [1 – 0.0027 (25 – 0)]

= 70 [1 – 0.0027 × 25]  

= 70 × 0.9325

∴ T = 65.275 dyne/cm