Problem 54 - College Physics

Question

\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?

Step-by-Step Solution

Answer

The temperature at the interface is approximately \(-5.78^{\circ} \text{C}, and the heat flow rate is 11.26 \text{W/m}^2.

Step 1: Understanding Heat Flow through Layers

To find the temperature at the interface and the heat flow through the wall, we must first understand that heat flows through each material sequentially. We have two materials: wood and Styrofoam, so we will use the concept of thermal resistance.

Step 2: Calculate Thermal Resistance

The thermal resistance (R) for each layer is given by the formula:\[ R = \frac{d}{kA} \]where \(d\) is the thickness, \(k\) is the thermal conductivity, and \(A\) is the area (considering per square meter here, thus \(A = 1 \text{ m}^2\)). For wood: \[ R_{\text{wood}} = \frac{0.030}{0.080 \times 1} = 0.375 \text{ Km}^2/W \]For Styrofoam:\[ R_{\text{styrofoam}} = \frac{0.022}{0.010 \times 1} = 2.2 \text{ Km}^2/W \]

Step 3: Calculate Total Resistance

The total thermal resistance is the sum of the resistances of each layer:\[ R_{\text{total}} = R_{\text{wood}} + R_{\text{styrofoam}} = 0.375 + 2.2 = 2.575 \text{ Km}^2/W \]

Step 4: Calculate Heat Flow per Square Meter

Using the temperature difference and the total resistance, we calculate the heat flow per square meter through the wall:\[ Q = \frac{T_{\text{inside}} - T_{\text{outside}}}{R_{\text{total}}} = \frac{19.0 - (-10.0)}{2.575} = \frac{29.0}{2.575} \approx 11.26 \text{ W/m}^2 \]

Step 5: Determine Temperature at Interface

The voltage drop due to heat flow across each layer can help determine the interface temperature. First, calculate heat flow per resistance:\[ Q = Q_{\text{wood}} = Q_{\text{styrofoam}} = 11.26 \text{ W/m}^2 \]Find temperature drop across wood:\[ \Delta T_{\text{wood}} = Q \times R_{\text{wood}} = 11.26 \times 0.375 \approx 4.223 \text{ K} \]With exterior temperature at -10°C, the temperature at the wood-styrofoam interface is:\[ T_{\text{interface}} = -10 + 4.223 \approx -5.777^{\circ} \text{C} \]

Key Concepts

Thermal Conductivity

Thermal conductivity is a measure of a material's ability to conduct heat. It is represented by the symbol \( k \) and is expressed in watts per meter-kelvin \( \text{W}/(\text{m} \cdot \text{K}) \). This property varies between different materials. Materials with high thermal conductivity transfer heat efficiently, while those with low thermal conductivity do the opposite, serving as better insulators.
When constructing a wall, the thermal conductivity of each material used, such as wood and Styrofoam, plays a crucial role in determining how heat will flow throughout the structure. For instance:

• Wood in the problem has a thermal conductivity of \(0.080 \text{ W}/(\text{m} \cdot \text{K})\), which means it's not as effective at transferring heat compared to metals like aluminum or copper.
• Styrofoam, on the other hand, has a very low thermal conductivity of \(0.010 \text{ W}/(\text{m} \cdot \text{K})\). This makes it an excellent insulator, which is why it's often used in thermal insulation.

Understanding these values helps predict how effective a material will be in maintaining different temperature zones within a building.

Temperature Interface

The temperature interface is the boundary where two materials meet. At this point, the temperature must be determined because it affects the overall heat flow through the layered wall. Knowing the interface temperature is crucial for understanding how energy moves through building materials, which is important for energy efficiency and cost-effectiveness in construction.
The temperature at this interface is affected by the thermal resistance of each material. In this problem, we find that:

• The total heat flowing through the layers affects the temperature drop across each material.
• The temperature drop across wood depends on its thermal resistance and the rate of heat flow.

To get the temperature at the interface, we first calculate the heat flow across the wood and multiply it by the wood's thermal resistance. Adding this temperature change to the exterior surface temperature gives the interface temperature. This step ensures we are correctly predicting the thermal behavior at the material junction, which is essential for preventing inefficiencies and structural damages.

Heat Flow Calculation

Calculating heat flow in thermal systems involves analyzing how heat moves across materials, specifically focusing on the rate per unit area. In the problem, this allows us to see how well the wall insulates the interior from the exterior.
The heat flow \( Q \) can be calculated using the formula:\[ Q = \frac{T_{\text{inside}} - T_{\text{outside}}}{R_{\text{total}}} \]Where:

• \( T_{\text{inside}} \) and \( T_{\text{outside}} \) are the temperatures inside and outside, respectively.
• \( R_{\text{total}} \) is the total thermal resistance of all layers in the wall combined.

Here’s why these steps matter:

• The formula gives the rate of heat transfer per square meter, showing efficiency in thermal insulation.
• The resistance of each material layer is accumulated to find how difficult it is for heat to pass through the entire wall structure.
• Understanding and calculating heat flow helps design effective insulation systems, crucial for maintaining comfortable and consistent indoor climates, and reducing energy consumption.

This thorough approach to heat flow calculation not only answers homework problems but also contributes to broader applications in building engineering and design.