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

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

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 $k=0.080 \mathrm{W} / \mathrm{m} \cdot \mathrm{K},$ and the Styrofoam has $k=0.010 \mathrm{W} / \mathrm{m} \cdot \mathrm{K}$. 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 Styrofoam? (b) What is the rate of heat flow per square meter through this wall?

Key Concepts

Fourier's Law of Heat Conduction

Fourier’s Law provides the mathematical framework for heat conduction, stating that the heat flux is proportional to the negative of the temperature gradient and the material's thermal conductivity. This law is the basis for calculating the rate of heat transfer in materials and is essential in understanding how temperature differences drive energy transfer.

Thermal Resistance in Composite Materials

When dealing with composite walls or systems, each layer offers a thermal resistance that opposes the flow of heat. The thermal resistance of a layer is calculated as the thickness divided by the thermal conductivity. In a series combination, the total resistance is the sum of the individual resistances, which helps determine how temperature differences are distributed across each layer.

Heat Conduction

Heat conduction is the process by which thermal energy is transferred within a material or between materials in direct contact due to a temperature gradient. This fundamental concept explains how heat moves from a region of higher temperature to one of lower temperature within solid objects.

Steady-State Heat Transfer

In steady-state conditions, the temperature distribution and heat flow remain constant over time, meaning that the rate of heat entering and leaving any section of the system is the same. This condition allows for the use of simplified equations, such as Fourier’s Law and series thermal resistances, to calculate key parameters like temperature at interfaces and the overall rate of heat transfer.

Transcript

00:01 In this question, we have a house with a layer of wood, three centimeters thick on the outside, and then a layer of styrofoam insulation, 2 .2 centimeters thick on the inside wall surface.

00:15 We're given the k values for the wood and the styrofoam, and we're told that the interior temperature is 19 degrees celsius, while the exterior temperature is negative 10 degrees celsius.

00:29 In part a, we're asked to calculate the temperature at the plane where the wood meets the styrofoam, and then in part b, we're asked to calculate the rate of heat flow per square meter through the wall.

00:41 So let's start off with just a little diagram to orient ourselves.

00:46 So let's say that this is our layer of wood, and then we'll have the layer of styrofoam.

00:53 So the layer of wood is three centimeters thick.

00:59 And the layer of styrofoam is 2 .2 centimeters.

01:08 And then the temperature on the outside, i'm going to call that th.

01:14 That was given as negative 10.

01:19 And then the temperature on the inside, we'll call that, or sorry, let's call that that one, th because that's hot, right? that was given as 19.

01:32 And then we'll call this one tc for tea cold.

01:41 Okay, and then in part a, we're asked to calculate what's the temperature at the point where the wood and the styrofoam meet, so right in between here.

01:52 So the idea behind this is that the flow rate of heat, the heat per unit time, that is going from the cyrofoam, or sorry, from the interior of the wall to the exterior of the wall, wall has to be constant all through the wall.

02:10 So the heat, the rate of heat flow through the styrofoam has to match the rate of heat flow through the wood.

02:20 So a way we can write that in terms of an equation is h.

02:25 Wood or hw is equal to h.

02:29 Styrofoam or hs.

02:31 And so that's basically because we can't just magically lose heat anywhere, so everything kind of has to be continuous.

02:42 So using this equivalency, we can go ahead and calculate what the temperature at the inner wall is going to be because both of these things depend on that temperature of the inner wall.

02:55 Now, before we do that, i'm just going to list out some givens.

02:59 So we have kw 0 .08.

03:05 I'm not going to rate in the units because everything is in standard units here.

03:14 The length of the wood is the 3 centimeters, so 0 .03 meters, and then the length of the styrofoam that is 0 .02 meters.

03:32 Okay.

03:33 So the flow rate of the heat, we can write that for each.

03:44 Section of the wall.

03:48 And that's going to depend on t h minus tc for each chunk, right? so the hot side of the wood part is going to be at temperature t, that intermediate temperature.

04:04 The cold side of the wood is going to be at tc.

04:08 And then we have the same equation, different k, different l, for the styrofoam, ls.

04:19 And this time, the t in the middle is going to be the cold temperature so it's going to be the minus one and then we'll have th there okay so this is the cross -sectional area through which the heat is flowing that should be the same through both sections of the wall and i'm just going to rearrange this a little bit to make this easier to solve so on the left -hand side i'm gonna multiply everything by ls so on the left -hand side so on that'll get kw ls and then i'm going to divide by ks.

04:58 So i'll get lwks.

05:01 And so that'll just leave me with th minus t.

05:04 So i can go ahead and plug everything in based on the values that i've already written out...

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