Problem 59 - University Physics with Modern Physics

Question

A carpenter builds an exterior house wall with a layer of wood 3.0 cm thick on the outside and a layer of Styrofoam insulation 2.2 cm thick on the inside wall surface. The wood has \(k\) = 0.080 W/m \(\cdot\) K, and the Styrofoam has \(k\) = 0.027 W /m \(\cdot\) K. The interior surface temperature is 19.0\(^\circ\)C, and the exterior surface temperature is -10.0\(^\circ\)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

The interface temperature is -0.88°C; heat flow rate is 24.37 W/m².

Step 1: Understanding the Heat Flow Equation

The temperature at the interface and the rate of heat flow can be determined using thermal conductivity and the concept of steady-state heat conduction. The formula \[ q = \frac{T_1 - T_2}{R} \] is used, where \( q \) is the heat flow, \( T_1 \) and \( T_2 \) are temperatures, and \( R \) is the thermal resistance sum.

Step 2: Calculate Thermal Resistance for Wood and Styrofoam

Thermal resistance \( R \) is calculated using the formula \( R = \frac{d}{k} \), where \( d \) is thickness and \( k \) is thermal conductivity. For wood, \( R_{\text{wood}} = \frac{0.030}{0.080} = 0.375 \) m²K/W. For Styrofoam, \( R_{\text{styrofoam}} = \frac{0.022}{0.027} = 0.815 \) m²K/W.

Step 3: Find Total Thermal Resistance

The total thermal resistance \( R \) is the sum of the resistances of the wood and Styrofoam: \[ R_{\text{total}} = R_{\text{wood}} + R_{\text{styrofoam}} = 0.375 + 0.815 = 1.190 \text{ m}^2\cdot\text{K/W}. \]

Step 4: Determine Temperature at Interface Using Temperature Drop Proportion

The temperature drop across each layer is proportional to its resistance. Given temperatures: \( T_{\text{inside}} = 19.0^\circ\text{C} \) and \( T_{\text{outside}} = -10.0^\circ\text{C} \). The total temperature drop is 29.0°C. Temperature drop across Styrofoam: \[ \Delta T_{\text{styrofoam}} = \frac{R_{\text{styrofoam}}}{R_{\text{total}}} \times 29 = \frac{0.815}{1.190} \times 29 = 19.88^\circ\text{C}. \] Thus, the interface temperature is \[ T_{\text{interface}} = T_{\text{inside}} - \Delta T_{\text{styrofoam}} = 19.0 - 19.88 = -0.88^\circ\text{C}. \]

Step 5: Calculate Rate of Heat Flow per Square Meter

Using the total thermal resistance, the heat flow rate per square meter can be calculated as:\[ q = \frac{T_{\text{inside}} - T_{\text{outside}}}{R_{\text{total}}} = \frac{19.0 - (-10.0)}{1.190} = \frac{29.0}{1.190} = 24.37 \text{ W/m}^2. \]

Key Concepts

Thermal Resistance

Thermal resistance is a way to measure how well a material can resist the flow of heat. Think of it as the thermal version of electrical resistance. Just like wires can resist the flow of electricity, materials can resist the flow of heat.
To calculate the thermal resistance (\( R \)) of a material, we use the formula: \[ R = \frac{d}{k} \]where:

• \( d \) is the thickness of the material
• \( k \) is the thermal conductivity, which tells us how well a material conducts heat

In our exercise with the wall, we have two layers: wood and Styrofoam. Each has its resistance based on its thickness and thermal conductivity:

• Wood: Thickness is 0.030 m, \( k = 0.080 \) W/m·K, so \( R_{\text{wood}} = 0.375 \) m²K/W
• Styrofoam: Thickness is 0.022 m, \( k = 0.027 \) W/m·K, so \( R_{\text{styrofoam}} = 0.815 \) m²K/W

By adding these resistances together, we get the total resistance of the wall, which is \( R_{\text{total}} = 1.190 \) m²K/W. This value shows how much the wall as a whole resists heat flow.

Heat Flow Equation

The heat flow equation helps us find out how much heat is moving through a material. It's like a formula that connects temperatures and thermal resistance.
The equation is: \[ q = \frac{T_1 - T_2}{R} \]where:

• \( q \) is the rate of heat flow (measured in watts per square meter, W/m²)
• \( T_1 \) and \( T_2 \) are the temperatures on either side of the material, measured in degrees Celsius
• \( R \) is the total thermal resistance of the materials

In the wall example, the inside temperature (\( T_{\text{inside}} \)) is 19.0°C, and the outside temperature (\( T_{\text{outside}} \)) is -10.0°C. We already calculated \( R_{\text{total}} = 1.190 \) m²K/W for the entire wall.
Using these numbers, we apply the formula to find the heat flow:\[ q = \frac{19.0 - (-10.0)}{1.190} = \frac{29.0}{1.190} = 24.37 \text{ W/m}^2 \]The answer, 24.37 W/m², tells us how much heat escapes through each square meter of the wall.

Steady-State Conduction

Steady-state conduction refers to a condition where the temperatures are constant over time. It is like when a system reaches equilibrium, and heat flows at a constant rate.
Imagine a situation where heat constantly flows through a wall, but the temperatures inside and outside remain stable. This steady condition means we can predict how heat behaves over time.
In our example problem, we used steady-state conduction assumptions. It helped us find the temperature at the interface between the wood and Styrofoam. By finding out how much the temperature drops across each material layer, based on its resistance, we can understand how heat spreads through the wall.
Let's break it down:

• The total temperature drop between the inside and outside of the wall is 29.0°C.
• The temperature drop across the Styrofoam alone is calculated using its resistance proportion:
\[ \Delta T_{\text{styrofoam}} = \frac{R_{\text{styrofoam}}}{R_{\text{total}}} \times 29 = \frac{0.815}{1.190} \times 29 = 19.88°C \]
• Thus, the temperature at the interface, going from inside to outside, is \( T_{\text{interface}} = 19.0 - 19.88 = -0.88°C \)

Understanding steady-state conduction helps predict real-world heat behavior in insulation and construction.