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)
Answers
Answered by
42
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
Answered by
8
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
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