Problem 113 - 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 termperature 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: Calculate Wall Area

First, determine the total area of the walls of the cabin. Since the cabin is rectangular, there are two walls of 3.50 m width and two walls of 3.00 m width. The height of the walls is 2.50 m.The total area of the walls is calculated by:\[ A = 2\times (3.50\, \mathrm{m} \times 2.50\, \mathrm{m}) + 2\times (3.00\, \mathrm{m} \times 2.50\, \mathrm{m}) \]\[ A = 2\times (8.75\, \mathrm{m^2}) + 2\times (7.5\, \mathrm{m^2}) \]\[ A = 17.5\, \mathrm{m^2} + 15\, \mathrm{m^2} \]\[ A = 32.5\, \mathrm{m^2} \]

Step 2: Calculate Total Thermal Resistance

Use the formula for total thermal resistance in series for two materials:\[ R_{\text{total}} = R_{\text{wood}} + R_{\text{insulation}} \]Each resistance is calculated as:\[ R = \frac{d}{k \times A} \]where \( d \) is thickness, \( k \) is thermal conductivity, and \( A \) is area. The wood thickness is 0.0180 m and the insulation thickness is 0.0150 m.

Step 3: Calculate Thermal Resistance for Wood

Calculate the thermal resistance of the wood:\[ R_{\text{wood}} = \frac{d_{\text{wood}}}{k_{\text{wood}} \times A} = \frac{0.0180\, \mathrm{m}}{0.0600\, \mathrm{W/m \cdot K} \times 32.5\, \mathrm{m^2}} \]\[ R_{\text{wood}} = \frac{0.0180}{1.95} \approx 0.00923\, \mathrm{K/W} \]

Step 4: Use Heat Transfer Equation

The heat transfer must satisfy:\[ P = \frac{\Delta T}{R_{\text{total}}} \]where \( \Delta T \) is the temperature difference of \( 19.0^{\circ} \mathrm{C} - 2.00^{\circ} \mathrm{C} = 17.0^{\circ} \mathrm{C} \) and \( P \) is the power required to maintain temperature, given as 1.25 kW.

Step 5: Solve for Insulation Thermal Conductivity

Rearrange the heat transfer equation to solve for the thermal conductivity of the insulating material:\[ R_{\text{insulation}} = \frac{\Delta T}{P} - R_{\text{wood}} \]\[ R_{\text{insulation}} = \frac{17.0}{1250} - 0.00923 \approx 0.00437 \mathrm{K/W} \]Now solve for \( k_{\text{insulation}} \):\[ k_{\text{insulation}} = \frac{d_{\text{insulation}}}{R_{\text{insulation}} \times A} = \frac{0.0150}{0.00437 \times 32.5} \]\[ k_{\text{insulation}} \approx 0.106 \mathrm{W/m \cdot K} \]

Key Concepts

Heat Transfer

Heat transfer is the movement of thermal energy from one object or material to another due to a temperature difference. Understanding how heat moves is crucial, especially when dealing with temperature regulation, such as in a cabin. There are three main types of heat transfer: conduction, convection, and radiation.

• Conduction: This is the process where heat moves through materials. In the cabin, heat is transferred from the warmer inside air to the cooler outside through the walls.
• Convection: This type of heat transfer happens when a fluid (like air or water) carries heat from one place to another. While not directly involved in the wall's heat transfer, it's crucial for heating air inside the cabin.
• Radiation: This form of energy transfer occurs through electromagnetic waves and doesn't rely on contact. Although not calculated here, it can affect heating loads.

Conduction is the primary heat transfer considered in this scenario. Knowing the rate at which heat transfers through materials helps control the cabin temperature more efficiently.

Thermal Resistance

Thermal resistance is a measure of a material's ability to resist the flow of heat. It is crucial in evaluating how effective materials are in insulating a building or structure. Thermal resistance is usually calculated using the formula:\[R = \frac{d}{k \times A}\]where:

• \( d \) is the thickness of the material,
• \( k \) the thermal conductivity, and
• \( A \) the area through which heat is transferred.

In our example, the total thermal resistance is the sum of the resistances of both wood and synthetic insulation. Each material contributes to slowing down heat transfer. The thicker the material or the lower its thermal conductivity, the higher the thermal resistance.

Insulation Material

Insulation material is used to reduce heat transfer between the inside and outside of a structure, making it easier to maintain a specific temperature. It's particularly vital in environments with significant temperature differences like in our cabin example.

Materials chosen for insulation should have low thermal conductivity to ensure minimal heat loss. In the problem, synthetic material is added to bolster the wood's insulation capabilities. Lower rates of heat transfer mean less energy is required to maintain the inside temperature.

When selecting insulation material, consider:

• Thermal conductivity – lower is better for insulation.
• Durability and lifespan in relation to environmental conditions.
• Safety and environmental factors, like fire resistance and recyclability.

This way, the structure becomes more energy-efficient, and heating or cooling costs can be significantly reduced.

Temperature Gradient

A temperature gradient refers to how the temperature changes from one location to another. It's a crucial concept because it describes the driving force for heat transfer. In any heat transfer scenario, the greater the temperature difference between two points, the faster the heat will transfer from the warm area to the cooler area.

In the cabin example, there is a temperature gradient between the inside at \( 19.0^{\circ} \mathrm{C} \) and the outside at \( 2.0^{\circ} \mathrm{C} \). This gradient is what causes heat to flow out of the cabin through its walls.

• A larger temperature gradient results in increased heat flow, making it essential to have effective insulation to counteract this.
• This principle guides the management of heat loss or gain in buildings and environmental systems.

By calculating and managing the temperature gradient effectively, energy use can be minimized, resulting in more sustainable and economical building operations.