#### Question

The surface tension of water at 0°C is 75.5 dyne/cm. Calculate surface tension of water at 25°C.

(α for water = 2.7×10^{-3}/°C)

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#### Solution

**Given:**

T_{0} = 75.5 dyne/cm

α_{water} = 2.7 × 10^{-3}/°C

**To find:**

Surface tension of water at 25°C

**Formula:**

T_{1} = T_{0}(1 - αΔt)

**Solution:**

T_{25} = T_{0}(1 - αΔt)

T_{25} = T_{0}(1 - α(25 - 0))

T_{25} = 75.5(1 - 2.7 × 10^{-3} × 25)

T_{25} = 75.5(1 - 0.0675)

T_{25} = 70.4 dyne/cm

**The surface tension of water at 25°C is 70.4 dyne/cm.**

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#### Reference Material

Solution for question: Calculate Surface Tension of Water at 25°C concept: Surface Tension. For the courses HSC Science (Electronics), HSC Science (Computer Science), HSC Science (General)