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 \rho and L is its latent heat of vaporization.

Engineering

Physics

SI Units

Calorimetry

Surface Tension and Surface Energy

Question

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 $ρ$ and L is its latent heat of vaporization.

$ρ$ L/T

T/ $ρ$ L

$\sqrt{T / ρL}$

2T/ $ρ$ L

$\frac{ΔQ}{Δt} = \frac{dm}{dt} L & \frac{dE}{dt} = 4 πT 2 R \frac{dr}{dt}$

$\frac{dm}{dt} L = 4 πT 2 R \frac{dr}{dt}$

$\frac{d}{dt} (ρ \frac{4}{3} {πr}^{3}) L = 8 πTR \frac{dr}{dt}$

$ρ \frac{4}{3} 3 r^{2} \frac{dr}{dt} L = 8 pT \frac{dr}{dt}$

$r = \frac{2 T}{ρL}$

This problem involves finding the minimum radius of a liquid drop that can evaporate solely by decreasing its surface energy, with temperature remaining constant. The surface tension is T, density is ρ, and latent heat of vaporization is L.

Key Concept: For evaporation to occur purely by surface energy decrease (without any heat exchange), the energy released due to reduction in surface area must be sufficient to provide the latent heat required for vaporization.

Step 1: Energy from Surface Area Reduction
Consider a spherical drop of radius r. Its surface area is $4 π r^{2}$ . If the radius decreases by a very small amount dr, the change in surface area dA is: $dA = d (4 π r^{2}) = 8 π r d r$ .
The surface energy released dE_surface is: $d E_{surface} = T \times dA = T \times 8 π r d r$ .

Step 2: Latent Heat Required for Evaporation
The decrease in radius dr corresponds to a loss of mass dm. The volume of liquid lost is $dV = 4 π r^{2} d r$ , so the mass lost is: $dm = ρ dV = ρ \times 4 π r^{2} d r$ .
The latent heat required to vaporize this mass is: $d E_{latent} = L \times dm = L \times ρ \times 4 π r^{2} d r$ .

Step 3: Equating the Energies
For evaporation to be possible solely by surface energy decrease: $d E_{surface} \geq d E_{latent}$ .
Substituting the expressions: $T \times 8 π r d r \geq L ρ 4 π r^{2} d r$ .
We can cancel common positive terms (π, dr, and 4): $2 T \geq L ρ r$ .
Rearranging for the radius r: $r \leq \frac{2 T}{L ρ}$ .

Step 4: Finding the Minimum Radius
The inequality $r \leq \frac{2 T}{L ρ}$ gives the maximum possible radius for which this process can happen. However, the question asks for the minimum radius for which it is possible. This might seem confusing.

Clarification: The derived condition is the upper limit. For any drop with a radius larger than $\frac{2 T}{L ρ}$ , the surface energy released is not sufficient to provide the latent heat. Therefore, for evaporation to be possible purely by surface energy decrease, the drop's radius must be less than or equal to $\frac{2 T}{L ρ}$ . The minimum radius for which this is possible is technically zero, but that's not practical. The question is interpreted as asking for the critical radius or the maximum possible radius that satisfies the condition, which is $\frac{2 T}{L ρ}$ . Among the options, this matches 2T/ρL.

Final Answer: The expression for the critical radius is $\frac{2 T}{ρ L}$ .

Detailed Solution

Related Topics & Formulae

Detailed Solution

Related Topics & Formulae

Student who searched for similar questions

Questions 1

Questions 2

Questions 3

Questions 4

Questions 5