Leidenfrost effect. A water drop that is slung onto a skillet with a temperature

Question

Leidenfrost effect. A water drop that is slung onto a skillet with a temperature between 100 degree C and about 200 degree C will last about 1 s. 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 the figure). Let L = 0.219 mm, and assume that the drop is flat with height h = 2.68 mm and bottom face area A = 4.71 times 10^-6 m^2. Also assume that the skillet has a constant temperature T_s = 307 degree C and the drop has a temperature of 111 degree C. Water has density rho = 1000 kg/m^3, and the supporting layer has thermal conductivity k = 0.0364 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? The latent heat of vaporization of water is 2.256 times 10^6 J/kg. (a) Number Units (b) Number Units

Explanation / Answer

For a) The heat conduction equation is

dQ/dt = k*A*T/x

The skillet temperature is constant, and since the drop is at the boiling point of water, its temperature is also constant. Thus

dQ/dt = k*A*(Ts - Td)/L = 0.0364*4.71e-6*(307-111)/0.219e-3 = 0.171444e-6*469/0.219e-3 = 0.367 W

b) The drop will vanish when the amount of heat transferred equals the heat of evaporation of the drop.

vh = 2257 J/g for water
mass of drop= density * volume = 1000* 4.71e-6*0.219e-3 = 1.03 e-3 g

amount of heat transferred= 1.03 e-3*2257=2.33 J
Time taken by drop to vanish = 2.33/0.367= 6.34 seconds

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