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

Ice has formed on a shallow pond and a steady state has been reached with the air above the ice at $-5.20^{\circ} \mathrm{C}$ and the bottom of the pond at $3.98^{\circ} \mathrm{C}$. If the total depth of ice $+$ water is $1.42 \mathrm{~m}$, how thick is the ice? (Assume that the thermal conductivities of ice and water are $1.67$ and $0.502 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$, respectively.)

Key Concepts

Thermal Conductivity

Thermal conductivity is a property of a material that indicates its ability to conduct heat. It appears in the formula for conductive heat transfer, where the rate of heat transfer is directly proportional to the thermal conductivity and the temperature difference across a material, and inversely proportional to the material’s thickness. This concept helps determine how fast a material can transfer heat, which is critical in problems involving different materials such as ice and water.

Steady-State Heat Transfer

Steady-state heat transfer refers to the condition when the temperature distribution in a system does not change over time because the heat entering any region equals the heat leaving that region. In this state, the heat flux (rate of heat transfer per unit area) remains constant across different layers or materials. This principle is essential for setting up equations to solve for unknown parameters in multilayer systems.

Thermal Resistance in Composite Systems

Thermal resistance represents the opposition to heat flow through a material and is analogous to electrical resistance in circuits. In problems involving composite systems (layers of different materials), the total thermal resistance is the sum of the individual resistances. This concept is particularly useful when determining the overall heat transfer through systems with multiple materials, as it allows one to relate temperature differences to the rate of heat conduction.

Heat Flux Continuity

Heat flux continuity is the principle that, under steady-state conditions, the rate of heat flow must be the same through all layers in a composite system. This implies that the heat flux across one material is equal to the heat flux in an adjacent material, even if their individual thermal conductivities and thicknesses differ. This concept enables the formulation of equations that link the temperature gradients and thicknesses of the layers, such as when determining the unknown thickness of ice in a layered system.

Transcript

0:00 Hi there.

00:01 So for this problem, we have ice that has formed on a shallow pond and a steady state has been reached with the earth above the ice.

00:11 And the temperature 1 is given and that value is minus 5 .20 celsius degrees.

00:20 And the button of the pond is a temperature 2, which is equal to 3 .98 degrees.

00:30 Degrees.

00:32 Now if the total depth of the ice plus the water is equal to the length l -t that is equal to 1 .42 meters, how thick is the ice? so we need to find the value for the thickness of the ice l2.

01:00 Now, to solve this problem, we use the result from the exercise sets.

01:07 Now, at the interface between the ice and the water, we will have that temperature is zero celsius degrees.

01:18 Then we will have that the rate, the constant r1 times the temperature 2 plus the constant r2 times the temperature 1 is equal to 0.

01:32 And what we know is that this constant r1, the constant r1 is the constant k over the length l1, that is the thickness, in this case, that is the thickness of the water, times the temperature 1, plus the constant k2 times the temperature 2 of the ice over l2, which is the thickness of the ice, the one that we want to found.

02:06 And we also know that the sum of these two thickness give us the total value l, so we can write l1 as l minus l2.

02:18 So substituting this into the previous equation, we will find the following.

02:24 That is that this, which is the thermal conductivity, the thermal conductivity one, the temperature of 1 times the length, plus the length l2 times the thermal conductivity 2 times the temperature 2 is equal to 0...

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