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As a spherical ammonia vapor bubble rises in liquid ammonia, its diameter changes from 1 cm to 3 cm. Calculate the amount of work produced by this bubble, in kJ, if the surface tension of ammonia is 0.07 N/m.

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princessesther2011

Answer:

8.7976×10^-8 kJ

Explanation:

Change in diameter = 3 cm - 1 cm = 2 cm = 2/100 = 0.02 m

Area of spherical ammonia = πD^2 = 3.142×0.02^2 = 1.2568×10^-3 m^2

Work produced = surface tension × area = 0.07 × 1.2568×10^-3 = 8.7976×10^-5 Nm = 8.7976×10^-5 J = 8.7976×10^-5/1000 = 8.7976×10^-8 kJ

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