Problem 54 - College 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 a thermal conductivity of \(0.080 \mathrm{~W} /(\mathrm{m} \cdot \mathrm{K}),\) and the Styrofoam 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 Styrofoam? (b) What is the rate of heat flow per square meter through this wall?

Step-by-Step Solution

Answer

(a) 14.8°C at the interface; (b) heat flow is 11.26 W/m².

Step 1: Understanding the Problem

The problem involves two materials in a composite wall, each with its own thickness and thermal conductivity. We need to find the temperature at the interface between wood and Styrofoam, and the heat flow through the wall.

Step 2: Set Up Thermal Circuit

Model the wall as a thermal circuit with resistances in series. The thermal resistance for a material is given by \( R = \frac{d}{k} \), where \( d \) is the thickness and \( k \) is the thermal conductivity.

Step 3: Calculate Resistance of Each Material

Calculate the resistance for wood: \[ R_{wood} = \frac{0.03 \text{ m}}{0.080 \text{ W/m} \cdot \text{K}} = 0.375 \text{ m}^2 \cdot \text{K/W}. \]Calculate the resistance for Styrofoam:\[ R_{styrofoam} = \frac{0.022 \text{ m}}{0.010 \text{ W/m} \cdot \text{K}} = 2.2 \text{ m}^2 \cdot \text{K/W}. \]

Step 4: Total Resistance and Heat Flow

The total resistance is the sum of the resistances:\[ R_{total} = R_{wood} + R_{styrofoam} = 0.375 + 2.2 = 2.575 \text{ m}^2 \cdot \text{K/W}. \]The rate of heat flow, \( Q \), is given by:\[ Q = \frac{T_{hot} - T_{cold}}{R_{total}} = \frac{19 - (-10)}{2.575} = \frac{29}{2.575} \approx 11.26 \text{ W/m}^2. \]

Step 5: Calculate Temperature at the Interface

Using the thermal resistance concept, the temperature drop across wood is:\[ \Delta T_{wood} = Q \cdot R_{wood} = 11.26 \times 0.375 = 4.223 \text{ K}. \]The temperature at the wood-Styrofoam interface is then:\[ T_{interface} = 19 - 4.223 = 14.777^{\circ}C. \]

Key Concepts

Composite Wall

A composite wall is like a multi-layered sandwich, where each layer is made of different materials. In our example, the wall consists of a wood layer and a Styrofoam insulation layer. Each material in the wall has its own specific properties, like thickness and thermal conductivity.

In a composite wall, heat passes through each layer sequentially from the inside to the outside or vice versa, depending on temperature differences. The concept of a composite wall is important because it helps us analyze how different materials interact thermally with each other, influencing the temperature distribution and heat flow within the wall. This interaction determines how effective the wall is at insulating, given the differing properties of each material involved.

Thermal Resistance

Thermal resistance is like the wall's way of resisting the flow of heat through it. Think of it as a barrier that slows down heat transfer. For any material, thermal resistance \( R \) is calculated using the formula: \[ R = \frac{d}{k} \]where \( d \) is the thickness of the material and \( k \) is the thermal conductivity.

In our exercise, each material, wood and Styrofoam, offers resistance to heat flow through the wall. The total thermal resistance of the composite wall is the sum of the resistances of both materials. This sum determines how much the wall can slow down the heat exchange. Calculating thermal resistance is crucial for evaluating insulation performance, which helps in maintaining the desired indoor temperature.

Heat Flow

Heat flow quantifies how much thermal energy moves from one side of the composite wall to the other. The formula for calculating the rate of heat flow \( Q \) through a wall is: \[ Q = \frac{T_{hot} - T_{cold}}{R_{total}} \]where \( T_{hot} \) and \( T_{cold} \) are the temperatures on either side of the wall, and \( R_{total} \) is the total thermal resistance.

Here, the calculated heat flow per square meter gives us the rate of energy transfer through the wall surface. Lower resistance or a bigger temperature difference will increase this flow, meaning more heat energy gets transferred quickly. Understanding heat flow helps in designing efficient insulation systems that optimize energy use and indoor comfort.

Temperature Gradient

The temperature gradient represents the change in temperature across the composite wall, from one side to the other. It's the slope of temperature change across the wall thickness. In essence, the temperature gradient tells us how temperature varies across different layers of the wall.

Using the concept of thermal resistance, we can calculate the temperature drop across each material layer. For instance, in our problem, we first calculate the drop across the wood. This drop helps us find the interface temperature between the wood and the Styrofoam. A steeper gradient indicates faster heat loss, while a shallower gradient suggests better insulation effectiveness. The temperature gradient is vital for understanding how temperature is distributed in the wall, allowing us to predict where heat loss might be minimized through effective material choices.