Problem 115 - University Physics with Modern Physics

Question

A rustic cabin has a floor area of \(3.50 \mathrm{m} \times 3.00 \mathrm{m} .\) Its walls, which are 2.50 \(\mathrm{m}\) tall, are made of wood (thermal conductivity 0.0600 \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K} ) 1.80 \mathrm{cm}\) thick and are further insulated with 1.50 \(\mathrm{cm}\) of a synthetic material. When the outside temperature is \(2.00^{\circ} \mathrm{C},\) it is found necessary to heat the room at a rate of 1.25 \(\mathrm{kW}\) to maintain its temperature at \(19.0^{\circ} \mathrm{C}\) . Calculate the thermal conductivity of the insulating material. Neglect the heat lost through the ceiling and floor. Assume the inner and outer surfaces of the wall have the same temperature as the air inside and outside the cabin.

Step-by-Step Solution

Answer

The thermal conductivity of the insulating material is approximately 0.106 W/m·K.

Step 1: Identify Relevant Variables

We need to find the thermal conductivity of the insulating material. Given data includes: the floor area of the cabin, heights of the walls, thermal conductivity of wood, thickness of the wood and insulating material, heating rate, and temperature difference across walls.

Step 2: Calculate Heat Transfer Area

The wall's height is 2.50 meters and we have the floor area which is 3.50 m by 3.00 m, so the total heat transfer area for all the walls is 4 walls each with area \(\text{Height} \times (\text{Length} + \text{Width})\), which gives a total heat transfer area of \((3.50 + 3.00) \times 2 \times 2.50 = 32.5 \text{ m}^2\).

Step 3: Define Heat Transfer Rate Equation

Use the overall heat transfer equation for series resistances: \(Q = \frac{A(T_{in} - T_{out})}{R_{total}}\), where \(Q = 1.25 \text{ kW} = 1250 \text{ W}\), \(A = 32.5 \text{ m}^2\), and \(T_{in} - T_{out} = 19 - 2 = 17\text{ K}\).

Step 4: Calculate Total Thermal Resistance

The total thermal resistance is the sum of the resistance of the wood and the resistance of the insulating material: \(R_{total} = R_{wood} + R_{insulation}\).

Step 5: Calculate Thermal Resistance of Each Layer

The resistance for wood is \(R_{wood} = \frac{d_{wood}}{k_{wood} \times A}\), where \(d_{wood} = 0.018 \text{ m}\) and \(k_{wood} = 0.060 \text{ W/m} \cdot \text{K}\). Calculating gives: \(R_{wood} = \frac{0.018}{0.060 \times 32.5} = 0.00923 \text{ K/W}\).

Step 6: Set Up Equation for Insulation Layer

Let \(k_{insulation}\) be the thermal conductivity of the insulating material. Then, \(R_{insulation} = \frac{d_{insulation}}{k_{insulation} \times A}\) where \(d_{insulation} = 0.015 \text{ m}\). Use this in the equation: \(R_{total} = R_{wood} + \frac{0.015}{k_{insulation} \times 32.5}\).

Step 7: Solve for Thermal Conductivity of Insulation

Rearrange the heat transfer equation: \(\frac{A \cdot 17}{1250} = 0.00923 + \frac{0.015}{k_{insulation} \times 32.5}\). Solve for \(k_{insulation}\):\[ k_{insulation} = \frac{0.015}{32.5(\frac{17}{1250} - 0.00923)} \approx 0.106 \text{ W/m} \cdot \text{K} \]

Step 8: Conclusion

The thermal conductivity of the insulating material is approximately 0.106 W/m·K.

Key Concepts

Heat Transfer

Heat transfer refers to the movement of thermal energy from one object or material to another. In the context of the cabin described in the exercise, heat transfer occurs through its walls. The inside of the cabin is warm, while the outside is cold. As a result, heat flows from the warmer interior to the cooler exterior. This flow of heat is driven by a temperature difference, which is a fundamental principle in thermodynamics.
To calculate how heat moves through a material, we use the formula for heat conduction:
\[ Q = \frac{A(T_{in} - T_{out})}{R_{total}} \]
Where:

• \( Q \) is the heat transfer rate (Watts)
• \( A \) is the area through which heat is transferred (square meters)
• \( T_{in} - T_{out} \) is the temperature difference (Kelvin)
• \( R_{total} \) is the total thermal resistance (Kelvin/Watt)

In our exercise, we need to maintain the cabin at a stable internal temperature, despite the external cold. Thus, understanding and managing heat transfer is crucial for energy efficiency and comfort.

Insulation Materials

Insulation materials play a critical role in controlling heat transfer. They act as a barrier, reducing the rate at which heat escapes from the warm interior to the cold exterior, or vice-versa. Different insulation materials have different levels of efficiency, defined by their thermal conductivity. The lower the thermal conductivity, the better the material is at insulating.
In our cabin scenario, the walls are composed of wood and a layer of synthetic insulating material. While wood itself provides some insulation value, the addition of synthetic material greatly enhances its effectiveness.
The thermal conductivity of wood was given, but to sustain the desired room temperature with minimal energy usage, it was essential to calculate the thermal conductivity of the synthetic layer. This calculation helps in selecting the appropriate material to ensure efficiency and cost-effectiveness over time. Proper understanding of these materials can lead to better energy conservation strategies in building and architecture.

Thermal Resistance

Thermal resistance is a measure of a material's ability to resist the flow of heat. It is an essential concept when analyzing and designing buildings for energy efficiency. The higher the thermal resistance, the better the material is at preventing heat transfer.
In practical terms, thermal resistance is influenced by three factors:

• The thickness of the material (\(d\))
• The material's thermal conductivity (\(k\))
• The area through which heat is being transferred (\(A\))

The formula to calculate thermal resistance \(R\) is given by:
\[ R = \frac{d}{k \cdot A} \]
In the cabin problem, thermal resistance plays a crucial role in maintaining the desired temperature inside the cabin. By calculating the resistance of both the wood and the insulating layer, we can determine how effective the wall is at minimizing heat loss. The total thermal resistance is the sum of the resistances of all layers. This step ensures that the heating system isn't working harder than necessary, saving energy and costs.