Ice has formed on a shallow pond, and a steady state has...

Ice has formed on a shallow pond, and a steady state has been reached, with the air above the ice at $-5.0^{\circ} \mathrm{C}$ and the bottom of the pond at $4.0^{\circ} \mathrm{C}$. If the total depth of $i c e+$ water is $1.4 \mathrm{~m}$, how thick is the ice? (Assume that the thermal conductivities of ice and water are 0.40 and $0.12 \mathrm{cal} / \mathrm{m} \cdot \mathrm{C}^{\circ} \cdot \mathrm{s},$ respectively. $)$

Key Concepts

Composite Thermal Resistance

When heat transfers through layered materials, each layer contributes a thermal resistance which opposes the flow of heat. The total thermal resistance of a composite system is the sum of the individual resistances of the layers, each determined by its thickness and thermal conductivity. Understanding composite thermal resistance is essential for solving problems where heat transfer occurs through multiple materials with differing properties.

Fourier’s Law of Heat Conduction

Fourier’s Law provides the fundamental relationship between heat flux, thermal conductivity, and the temperature gradient. It states that the heat flux is proportional to the negative gradient of temperature and the thermal conductivity of the material. This law is key to establishing the equations needed to quantify heat flow in various scenarios involving conduction.

Thermal Conductivity

Thermal conductivity is an intrinsic property of materials that measures the rate at which heat is transferred through them when a temperature gradient is present. Materials with high thermal conductivity can transfer heat quickly, whereas materials with low conductivity act as insulators. This concept is crucial in analyzing how heat flows through different media in a layered structure.

Steady State Heat Conduction

This concept refers to a condition in which the temperatures within a system do not change over time, resulting in a constant heat flux through all parts of the system. In steady state, the heat entering one face of a medium is equal to the heat leaving the opposite face, allowing for the balance of energy and the use of conduction equations to solve for unknown thermal parameters in composite systems.

Transcript

00:01 In this problem, we have a pond.

00:02 We have a layer of ice, and we have them beneath that, the water.

00:07 We are told the depth of the ice and the water together.

00:10 We do not know the depth of the ice or the water individually.

00:15 We are going to find, our goal is to find the depth of the ice.

00:18 We are told the interphase temperatures, the bottom of the pond, between the water and the ice and the water in the air.

00:26 And we also told a very important thing.

00:29 Steady, stay condition.

00:30 That means these temperatures do not take.

00:32 Change.

00:33 So you could take a picture now of, say, three thermometers, and 10 minutes later, they're going to be the same.

00:39 That means that whatever leaves, whatever rate of heat leaving one, water has to equal whatever is leaving the ice.

00:51 Otherwise, there's going to be a buildup somewhere.

00:54 The temperatures will not stay the same, will not be a steady state condition.

00:58 So let's write the connectivity formula and equate what leaves the water and what leaves the ice.

01:10 Anything else is not steady state.

01:15 So the connectivity of water times the area.

01:21 And that would be the, you know, if you were looking down on this, what would be the, you know, if you're in a helicopter looking down on the pond, what the area would be.

01:30 That surface area.

01:33 It's not going to play any role here because we're assuming everything's nice and even.

01:39 Nothing funny in terms of the nature of the shape.

01:43 T1 minus t2, hotter, minus the colder, over dw, the thickness of the layer.

01:55 K -ice, t2, minus t3, over d, now we can make a, a very immediate update to this.

02:14 T2 is zero.

02:16 So these are gone.

02:19 And what else do we can we do before we write the next thing? but what is d? d is equal to d ice plus d water.

02:29 That tells me that d water, d minus d ice.

02:36 We're looking for d ice, remember.

02:38 This is our goal.

02:39 It's our goal.

02:41 So we have to get this with one variable in there.

02:44 So let's now rewrite it.

02:49 Kw and the a's, oh, i forgot the a here.

02:52 The a's will cancel out.

02:53 Like i said, nothing funny with the shape.

02:56 I just think of it as a perfect, perfect all the way around, from top to bottom.

03:03 K of the water, t1 over d minus d ice, is equal to k ice...

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