∙A carpenter builds an exterior house wall with a layer of wood 3.0 cm thick on the outside and

step 1 Hey everyone, this is question number 54 from Chapter 14. In this problem, we're given that we h

∙A carpenter builds an exterior house wall with a layer of...

\bullet A carpenter builds an exterior house wall with a layer of wood 3.0 $\mathrm{cm}$ thick on the outside and a layer of Styrofoam"" insulation 2.2 $\mathrm{cm}$ thick on the inside wall surface. The wood has a thermal conductivity of $0.080 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K}),$ and the Sty- rofoam TM has a thermal conductivity of 0.010 $\mathrm{W} /(\mathrm{m} \cdot \mathrm{K})$ . The interior surface temperature is $19.0^{\circ} \mathrm{C},$ and the exterior surface temperature is $-10.0^{\circ} \mathrm{C}$ . (a) What is the temperature at the plane where the wood meets the Styrofoamm? (b) What is the rate of heat flow per square meter through this wall?

\bullet A carpenter builds an exterior house wall with a layer of wood 3.0 $\mathrm{cm}$ thick on the outside and a layer of Styrofoam"" insulation 2.2 $\mathrm{cm}$ thick on the inside wall surface. The wood has a thermal conductivity of $0.080 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K}),$ and the Sty-
rofoam TM has a thermal conductivity of 0.010 $\mathrm{W} /(\mathrm{m} \cdot \mathrm{K})$ . The interior surface temperature is $19.0^{\circ} \mathrm{C},$ and the exterior surface temperature is $-10.0^{\circ} \mathrm{C}$ . (a) What is the temperature at the plane where the wood meets the Styrofoamm? (b) What is the rate of heat flow per square meter through this wall?

Key Concepts

Heat Flow Rate (Heat Flux)

Heat flow rate per unit area, or heat flux, represents the amount of heat energy passing through a surface in a given time, typically measured in W/m². It is determined by the temperature difference across the material and the total thermal resistance of the system, according to Fourier’s law. Understanding this concept is crucial for assessing the energy efficiency and thermal performance of building envelopes and other insulated structures.

Composite Wall Heat Transfer

Composite wall heat transfer involves the analysis of heat conduction through structures consisting of multiple layers of different materials. In such systems, each layer adds its own thermal resistance to the overall heat transfer, and the temperature drop occurs across each material proportionally to its resistance. This allows for the determination of interface temperatures and overall heat flow through the composite wall.

Thermal Conductivity

Thermal conductivity is a material property that indicates how well a substance can conduct heat. It quantifies the rate at which heat is transferred through a material due to a temperature difference, with units typically expressed in watts per meter-kelvin (W/(m·K)). High thermal conductivity means the material allows heat to flow easily, whereas low thermal conductivity indicates that the material is a good thermal insulator.

Thermal Conduction

Thermal conduction is the process by which heat energy is transmitted through collisions between neighboring molecules or electrons within a body. It is governed by Fourier's law, which establishes that the heat flux is proportional to the negative gradient of temperatures and the material's thermal conductivity. This concept is central to understanding how temperature differences drive heat transfer in solid materials.

Thermal Resistance

Thermal resistance is a measure of a material's opposition to heat flow. It is calculated based on the thickness of the material and its thermal conductivity and is analogous to electrical resistance in circuits. The concept is especially useful for analyzing layered systems, as the total thermal resistance of a composite structure is the sum of the individual resistances of each layer.

Transcript

00:01 Hey everyone, this is question number 54 from chapter 14.

00:04 In this problem, we're given that we have two sections of a wall.

00:07 One is wood.

00:08 We're given a thickness and a conductivity value.

00:12 And then the other is styrofoam.

00:13 We're given a thickness and a conductivity value.

00:16 And then we're told the interior temp and the exterior temp.

00:19 And then part a, we're asked to find the temperature where the wood and styrofoam meet at that plane.

00:25 Okay, so we're dealing with conductivity.

00:28 So immediately we can have our q over t equation in mind, which is equal to k -a -t -h -minus t -c over l.

00:50 And we're trying to find the temperature at the plane, and the temperature at that plane is going to be the same for the cyrofoam and the wood, so we can set q over t of the wood equal to q over t of the wood equal to q -over -t.

01:05 T of the styrofoam.

01:08 And once we round out, those two equations are both going to involve the unknown t that we need, and we can solve for that t.

01:14 So let's go ahead and write out this equation for the wood and for the styrofoam.

01:23 Okay, so we have k of, let's do wood first, k of wood.

01:33 We're given in the question 0 .08 watts per meter.

01:39 K and then we're going to end up having area on both sides and that's the cross -sectional area of the wall so that's going to cancel so then we go to temperature and the wood is on the wood is on the outside and is exposed to the exterior temperature which we're given so you notice by the temperatures that we're given the exterior is colder than the interior so our high temperature for the wood is going to be what we're looking for t and then it's going to be minus the interior or the exterior excuse me minus negative 10 so minus 10 or plus 10 degrees celsius the two minuses will cancel and then over l we're given a thickness of three centimeters which is 0 .03 meters and that is equal to k of the styrofoam which is 0 .01 that's watts watts per meter k and then for the styrofoam it's on the inside so the high temperature is going to be the interior temperature, which is 19 degrees c, and the low temperature is going to be that plain temperature, which is d that we're looking for.

02:57 And then l for this is 0 .022 meters.

03:05 Okay, so we have this equation, and we can cross multiply the bottoms over and start to solve this...

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