Problem 65 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

(a) The interface temperature is about 13.1°C. (b) The heat flow rate is approximately 62.08 W/m².

Step 1: Understand the Problem

You are given a composite wall with two materials: wood and Styrofoam. You need to find the temperature where these materials meet and the heat transfer rate through the wall. Given are the thermal conductivities of wood (\(k_w = 0.080 \mathrm{W/m \cdot K}\)) and Styrofoam (\(k_s = 0.010 \mathrm{W/m \cdot K}\)), and the thicknesses of the wood (\(L_w = 0.030 \mathrm{m}\)) and Styrofoam (\(L_s = 0.022 \mathrm{m}\)). Additionally, surface temperatures are \(T_i = 19.0^{\circ} \mathrm{C}\) and \(T_o = -10.0^{\circ} \mathrm{C}\).

Step 2: Apply the Conceptual Formula

For steady-state heat flow through parallel layers, the temperature gradient in each material is used to find the interface temperature. The heat flow rate \(q\) through a material is given by \(q = k \cdot A \cdot \frac{\Delta T}{L}\). Here \(A\) (cross-sectional area) cancels out, so use \(\frac{T_i - T_x}{L_s/k_s} = \frac{T_x - T_o}{L_w/k_w}\) to determine the interface temperature \(T_x\).

Step 3: Solve for the Interface Temperature

Re-arrange the equation \(\frac{T_i - T_x}{L_s/k_s} = \frac{T_x - T_o}{L_w/k_w}\) into \(\frac{19.0 - T_x}{2.2/0.010} = \frac{T_x + 10.0}{3.0/0.080}\) and solve for \(T_x\). This becomes\(\frac{19.0 - T_x}{220} = \frac{T_x + 10.0}{37.5}\).By multiplying both sides by 220 and 37.5, you equate \(37.5(19.0 - T_x) = 220(T_x + 10.0)\), leading to the solution: \(T_x \approx 13.1^{\circ} \mathrm{C}\).

Step 4: Calculate the Heat Flow Rate

Using \(q = k \cdot A \cdot \frac{\Delta T}{L}\), calculate heat flow rate for either material, since they should equate at steady state. Take wood, for example: \(q = 0.080 \times 1 \times \left(\frac{13.1 + 10.0}{0.03}\right)\). Calculate to find \(q \approx 62.08 \mathrm{W/m}^2\).

Step 5: Verification

Verify your calculation by computing the same for Styrofoam and checking consistency: \(q = 0.010 \times 1 \times \left(\frac{19.0 - 13.1}{0.022}\right)\) which should approximately equal to \(62.08 \mathrm{W/m}^2\).

Key Concepts

Thermal Conductivity

When studying heat transfer, thermal conductivity stands out as a fundamental property of materials. It measures a material's ability to conduct heat. Think of it as the **heat-conducting capacity** of the material.
Materials with high thermal conductivity allow heat to pass through them quickly. Conversely, materials with low thermal conductivity act as insulators, slowing down the passage of heat.

• For instance, in the given problem, **wood** and **Styrofoam** are used. Wood has a thermal conductivity of 0.080 W/m·K, indicating it is a much better conductor compared to Styrofoam's 0.010 W/m·K.
• Styrofoam, with its lower thermal conductivity, is a great insulating material, useful for keeping heat from escaping too quickly.

When designing walls or buildings, selecting materials based on their thermal conductivity is crucial for energy efficiency and comfort.

Composite Wall

A composite wall is a layered structure made up of different materials, each bringing its unique properties to the overall system. In this scenario, the composite wall consists of layers of wood and Styrofoam.
The primary advantage of using a composite wall is the ability to tailor heat transfer properties to specific needs.

• Layers can be designed to harness thermal benefits, combining materials that conduct heat efficiently with those that insulate well to optimize temperature regulation.

When calculating heat flow through a composite wall, consider each layer's thermal conductivity and thickness. This combination affects how the composite wall as a whole handles thermal energy transfer.
In our example, the wood provides structural strength and moderate insulation, while the Styrofoam enhances the wall's overall insulation capacity significantly.

Temperature Gradient

A temperature gradient occurs when there is a temperature difference between two points in a material. It represents how rapidly temperature changes with distance.

• In the context of heat flow, the temperature gradient is integral to determining how heat moves from the warmer side to the cooler side of a wall.

The formula we often use for heat transfer, \[ q = k \cdot A \cdot \frac{\Delta T}{L} \],relates the temperature gradient \( \frac{\Delta T}{L} \) to the heat flow and the material's thermal properties. Here, - \( \Delta T \) is the temperature difference, and - \( L \) is the thickness of a layer.
Understanding temperature gradients is essential, as it directly influences how much energy is needed to maintain different temperatures across a wall.**In our case, calculating the temperature gradient where wood meets Styrofoam provides insights into the heat dynamics at this interface.**

Steady-State Heat Flow

Steady-state heat flow describes a condition where the heat transfer rate remains constant over time. In such a state, the system has reached equilibrium.
In applications involving buildings, achieving steady-state ensures that interior temperatures remain stable despite external influences. This concept is crucial for energy efficiency and thermal comfort.

• In steady-state conditions, the rate of heat entering any system equals the heat leaving it, allowing for consistent temperatures within the environment.
• For the given composite wall example, steady-state heat flow means the energy passing from the inside to outside across each material layer is equal.

To solve problems involving steady-state heat flow, applying the basic heat transfer formula to each layer helps determine important parameters, such as the rate of heat flow per square meter (\[ q = k \cdot A \cdot \frac{\Delta T}{L} \]). This exercise helps ensure that the results are balanced and predictable across the whole system.