Leidenfrost Effect. A water drop that is slung onto a skillet with a temperature between $\mathbf{100}\mathbf{\text{°C}}$ and about $\mathbf{200}\mathbf{\text{°C}}$will last about . However, if the skillet is much hotter, the drop can last several minutes, an effect named after an early investigator. The longer lifetime is due to the support of a thin layer of air and water vapor that separates the drop from the metal (by distance L in Figure). Let $\mathit{L}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.100}\mathbf{}\mathbf{\text{mm}}$ , and assume that the drop is flat withheight $\mathit{h}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.50}\mathbf{}\mathbf{\text{mm}}$and bottom face area $\mathit{A}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.00}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}{\mathbf{\text{m}}}^{\mathbf{\text{2}}}$ . Also assume that the skillet has a constant temperature ${\mathit{T}}_{\mathbf{s}}\mathbf{=}\mathbf{300}\mathbf{\text{°C}}$ and the drop has a temperature of $\mathbf{100}\mathbf{\text{°C}}$ .Water has density $\mathbf{\text{ρ}}\mathbf{=}\mathbf{}\mathbf{1000}{\mathbf{\text{kg/m}}}^{\mathbf{\text{3}}}$, and the supporting layer has thermal conductivity $\mathit{k}\mathbf{=}\mathbf{0}\mathbf{.026}\mathbf{}\mathbf{\text{\hspace{0.17em}W/m.K}}$ . (a) At what rate is energy conducted from the skillet to the drop through the drop’s bottom surface? (b) If conduction is the primary way energy moves from the skillet to the drop, how long will the drop last?

1. The separation distance of the drop from the metal, $L=0.100\text{mmor}0.1\times {10}^{-3}\text{m}$
2. The height of the drop,$h=1.50\text{mmor}1.50\times {10}^{-3}\text{m}$
3. The bottom surface area of the drop,$A=4.00\times {10}^{-6}{\text{m}}^{\text{2}}$
4. The constant temperature of skillet is$T_s=300\text{°C}$
5. The temperature of the water drop is$T_w=100\text{°C}$
6. The density of the water is$\rho =1000{\text{kg/m}}^{\text{3}}$
7. The thermal conductivity of the supporting layer$k=0.026\frac{\text{W}}{\text{m.K}}$

According to Fourier's Law, the amount of time it takes for heat to go through a material is proportional to the negative gradient of the temperature as well as the cross-sectional area that is perpendicular to the gradient. We can use the concept of the rate at which energy has been conducted through the surface of the skillet. The heat transformation involved in the phase change from liquid to gas is called heat vaporization.

Formulae:

The conduction rate at which heat energy is transferred by a body,

$P_{cond}=\frac{kA(T_H-T_C)}{L}$ …(i)

The heat energy released by the body,$Q=L_{V}m$ …(ii)

Where,$L_V$ is latent heat of vaporization

$T_H$ is the temperature of skillet

$\begin{array}{c}{T}_H=T_s\\ =300\text{°C}\end{array}$

$T_C$ is the temperature of the water drop,

$\begin{array}{c}{T}_C=T_w\\ =100\text{°C}\end{array}$

So, the rate of conduction of the energy from the skillet to the drop is given using equation (i) as:

$\begin{array}{c}{P}_{\text{cond}}=\frac{\left(0.026\frac{\text{W}}{\text{m.K}}\right)\times (4.00\times {10}^{-6}{\text{m}}^{\text{2}})(300\text{°C}-100\text{°C})}{0.1\times {10}^{-3}\text{m}}\\ =0.208\text{W}\\ \approx 0.21\text{W}\end{array}$

Hence, the required rate of the conduction of the energy is $0.21\text{W}$

The water drop converts from liquid to vapor. The phase change is from liquid to vapor, hence heat of vaporization is given using equation (ii), where

$L_V$is the heat of vaporization for water,$L_V=2.256\times {10}^6\frac{\text{J}}{\text{kg}}$

Differentiating equation (ii) with respect to time, we can get the rate of conduction hence, the equation becomes

$P_{cond}t=L_{V}m\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \mathrm{..}\left(\text{iii}\right)\text{(}\because \frac{Q}{t}=P_{cond}\text{)}$

Again, we know that the mass of a body in terms of density and volume is given as:

$\begin{array}{c}m=\rho V\\ =\rho Ah\text{(}\because \text{Volume=area}\times height\text{)}\end{array}$

Substituting the above value of mass in equation (iii), we can get the time required for the drop to last is given as:

$\begin{array}{c}t=\frac{L_V\rho Ah}{P_{cond}}\\ =\frac{2.256\times {10}^6\frac{\text{J}}{\text{kg}}\times 1000{\text{kg/m}}^{\text{3}}\times 4.00\times {10}^{-6}{\text{m}}^{\text{2}}\times 1.50\times {10}^{-3}\text{m}}{0.208\text{W}}\\ =65\text{s}\end{array}$

Hence, the value of the required time is $65\text{s}$