Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is T, density of liquid is ? and L is its latent heat of vaporization.

Question

Saurabh Koranglekar · Answer

Vikas TU · Answer

When radius is by ∆R, 4πR 2 ∆RpL = 4πT[R 2 – (R - ∆R] pR 2 ∆RL = T[R 2 – R 2 + 2R∆R - ∆R 2] pR 2 ∆RL =T2R∆R [ ∆R is very small ] R = 2T/pL

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Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is T, density of liquid is ? and L is its latent heat of vaporization.

Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is T, density of liquid is ? and L is its latent heat of vaporization.

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Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is T, density of liquid is ? and L is its latent heat of vaporization.

Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is T, density of liquid is ? and L is its latent heat of vaporization.

2 Answers

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