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Leidenfrost effect. A water drop that is slung onto a skillet with a temperature between \(100^{\circ} \mathrm{C}\) and about \(200^{\circ} \mathrm{C}\) will last Fig. \(18-46\) Problem \(61 .\) about \(1 \mathrm{~s}\). 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 sep- \(\quad\) Fig. \(18-47 \quad\) Problem 62 . arates the drop from the metal (by distance \(L\) in Fig. 18-47). Let \(L=\) \(0.100 \mathrm{~mm}\), and assume that the drop is flat with height \(h=1.50 \mathrm{~mm}\) and bottom face area \(A=4.00 \times 10^{-6} \mathrm{~m}^{2}\). Also assume that the skillet has a constant temperature \(T_{s}=300^{\circ} \mathrm{C}\) and the drop has a temperature of \(100^{\circ} \mathrm{C}\). Water has density \(\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}\), and the supporting layer has thermal conductivity \(k=0.026 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (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?

Step by step solution

01

Calculate Temperature Difference

The temperature difference between the skillet and the drop is given by \[ \Delta T = T_s - T_d \]where \( T_s = 300^{\circ} \mathrm{C} \) (temperature of the skillet) and \( T_d = 100^{\circ} \mathrm{C} \) (temperature of the drop). Thus, \[ \Delta T = 300^{\circ} \mathrm{C} - 100^{\circ} \mathrm{C} = 200^{\circ} \mathrm{C}. \]

02

Calculate Heat Conduction Rate

The rate of heat conduction can be calculated using the formula \[ \dot{Q} = \frac{k \cdot A \cdot \Delta T}{L} \]where\( k = 0.026 \, \text{W/m} \cdot \text{K} \) (thermal conductivity), \( A = 4.00 \times 10^{-6} \, \mathrm{m}^2 \) (area of the drop), \( \Delta T = 200 \, ^{\circ} \mathrm{C} \), and \( L = 0.1 \times 10^{-3} \, \mathrm{m} \) (thickness of the air layer). Substituting the values, we have\[ \dot{Q} = \frac{0.026 \times 4.00 \times 10^{-6} \times 200}{0.1 \times 10^{-3}} = 0.0208 \, \mathrm{W}. \]

03

Calculate Energy Required for Phase Change

The energy required to completely vaporize the water drop can be found using \[ Q = mL_v \] where \( m = \rho \cdot A \cdot h = 1000 \, \text{kg/m}^3 \times 4.00 \times 10^{-6} \, \text{m}^2 \times 1.50 \times 10^{-3} \, \text{m} = 6.00 \times 10^{-6} \, \text{kg} \) and \( L_v = 2.26 \times 10^6 \, \text{J/kg} \) (latent heat of vaporization for water). The energy needed is \[ Q = 6.00 \times 10^{-6} \times 2.26 \times 10^6 = 13.56 \, \text{J}. \]

04

Calculate Time for Drop to Last

The time \( t \) for which the drop will last can be found using \[ t = \frac{Q}{\dot{Q}} \]where \( Q = 13.56 \, \text{J} \) and \( \dot{Q} = 0.0208 \, \mathrm{W} \). Thus,\[ t = \frac{13.56}{0.0208} \approx 652.88 \, \text{seconds}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Conduction

Heat conduction is the process where heat energy is transferred through a substance without the movement of the substance itself. In simpler terms, it's like passing a secret note by hand through a row of people; the note gets to the end without the people moving. For the Leidenfrost effect, heat conduction plays a crucial role in determining how heat gets from the hot skillet up through the thin layer of vapor and into the water drop above.
In our problem, we're dealing with the transfer of heat energy from a hot skillet at 300掳C to a water drop at 100掳C. This occurs across a thin 0.1 mm layer of air and vapor. This setup creates a temperature gradient, which is necessary to drive heat conduction. The greater the difference in temperature, the faster the heat energy can be transferred. Understanding this concept gives us the big picture about why the water drop behaves as it does on a hot skillet.

Thermal Conductivity

Thermal conductivity is a property of a material that dictates how well it can conduct heat. Picture it as a material鈥檚 ability to let heat sneak through it quietly and quickly. Imagine a material like metal having high thermal conductivity compared to a material like rubber.
In our problem involving the Leidenfrost effect, the air-vapor mixture under the water drop has a thermal conductivity of 0.026 W/m路K. This value indicates that the layer isn't great at conducting heat. It's more like a thick winter blanket than a windowpane. The lower the thermal conductivity, the slower the heat moves through it. Consequently, the heat conduction rate in our problem is relatively low, contributing to the longevity of the water drop as the energy takes its time to reach through to vaporize it.

Latent Heat of Vaporization

The latent heat of vaporization is the amount of energy needed to turn a given mass of liquid into vapor without changing its temperature. It's like the secret key needed to change the state from water to gas. For water, this heat amounts to a hefty 2.26 million joules per kilogram. In our exercise, the entire water drop is converted into vapor, and we have to calculate the energy needed for this transformation. Since the mass of the water drop is 6.00 x 10鈦烩伓 kg, the latent heat needed is 13.56 joules. This energy must all be absorbed by the drop before it fully vaporizes, explaining why the drop remains on the skillet longer when the temperature is much higher.

Phase Change

A phase change occurs when a substance transitions between different states of matter: solid, liquid, and gas. It's like water turning into steam, our superhero transformation in the context of the Leidenfrost effect. In our scenario, the water drop undergoes a phase change from liquid to gas while sitting on the hot skillet. This happens only when sufficient energy, in this case, 13.56 joules, is absorbed by the drop. The energy needed for this phase change directly relates to the drop's longevity. Because the skillet is much hotter than the water drop, the energy transfer happens, leading eventually to this complete transformation from liquid water to vapor. This phase change ensures the water drop lasts those several minutes, fascinating observers with its defiance of heat.