Question

Leidenfrost effect. A water drop will last about $1 s$ on a hot skillet with a temperature between $100 ° C$ and about $200 ° C$ . 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 vapour that separates the drop from the metal (by distance $L$ in Fig.). Let $L = 0.0800 mm,$ and assume that the drop is flat with height $h = 1.50 mm$ and bottom face area $A = 5.00 \times 10^{- 6} m^{2} .$ Also assume that the skillet has a constant temperature $T_{s} = 300 ° C$ and the drop has a temperature of $100 ° C .$ Water has density $ρ = 1000 kg m^{- 3},$ and the supporting layer has a thermal conductivity $k = 0.026 W (m \cdot K)^{- 1} .$

$(a)$ 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?

Accepted Answer

According to Leindenfrost effect when a liquid, close to the surface that is significantly hotter than the boiling point of liquid than an insulating vapour layer that keeps the liquid from boiling quickly.

This happened because a repulsive force developed between droplets and surface to avoid physical contact with the hot surface.

$(a)$ Now we know that if, $H$ is the rate of heat flow then,

$H = \frac{k A (T_{1} - T_{2})}{L}$ , where thermal conductivity $k = 0.026 W m^{- 1} K^{- 1}$ , bottom face area $A = 5 \times 10^{- 6} m^{2}$ , temperature difference $(T_{1} - T_{2}) = 300 - 100 = 200 K$ , length $L = 8 \times 10^{- 5} m$ ,

Now put all values in the equation,

$H = \frac{k A (T_{1} - T_{2})}{L} = \frac{0.026 \times 200 \times 5 \times 10^{- 6}}{8 \times 10^{- 5}} = 32.5 \times 10^{- 2} = 0.325 W$

$(b)$ We know that power is energy per unit time,

$H = \frac{Q}{t} = \frac{m L_{vap}}{t} = \frac{L_{vap} ρ A h}{t}$

$\Rightarrow t =$ $\frac{L_{vap} ρ A h}{H} = \frac{2.256 \times 10^{6} \times 1000 \times 5 \times 10^{- 6} \times 1.50 \times 10^{- 3}}{0.325}$

$\Rightarrow t = 52.06 second$ .

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