Heat Conduction Through a Wall

Understanding how heat moves through materials is fundamental in thermodynamics and engineering applications. This problem focuses on determining the thermal conductivity of a wall, which is a key property that describes a material's ability to conduct heat. We are given the heat transfer rate (heat flux), the wall's thickness, and the temperature difference across it.

Key Concepts for Heat Transfer

• Heat Conduction: This is the transfer of thermal energy between substances in direct contact. Heat flows from an area of higher temperature to an area of lower temperature without any bulk movement of the material itself.
• Thermal Conductivity (\(k\)): It quantifies how efficiently a material conducts heat. Materials with high thermal conductivity (like metals) are good heat conductors, while those with low thermal conductivity (like insulation materials) are poor heat conductors. Its standard unit is Watts per meter Kelvin (W/m·K).
• Heat Flux (\(q\)): This term refers to the rate of heat energy transferred per unit area. It tells us how much heat is flowing through a specific surface area per unit time. The unit for heat flux is Watts per square meter (W/m²).
• Temperature Difference (\(\Delta T\)): The driving force for heat conduction. Heat always flows down a temperature gradient, from hotter to colder regions.

Formula for Heat Conduction Calculation

For steady-state heat conduction through a flat wall, Fourier's Law of Heat Conduction is used. The formula relates the heat flux to the thermal conductivity, temperature difference, and wall thickness:

\[ q = \frac{Q}{A} = k \frac{\Delta T}{L} \]

Where:

• \(q\) or \(\frac{Q}{A}\) represents the heat flux (rate of heat transfer per unit area) in W/m².
• \(k\) is the thermal conductivity of the wall material in W/(m·K).
• \(\Delta T\) is the temperature difference across the wall in °C or K (note that a temperature difference is the same in Celsius and Kelvin).
• \(L\) is the thickness of the wall in meters (m).

Given Information in the Problem

Let's list the known values from the question:

• Rate of heat conduction (Heat Flux, \(q = \frac{Q}{A}\)) = 30 W/m²
• Thickness of the wall (\(L\)) = 10 cm
• Temperature difference across the wall (\(\Delta T\)) = 10 °C

Step-by-Step Calculation of Wall's Thermal Conductivity

To accurately calculate the thermal conductivity, all units must be consistent with the SI system. The wall thickness is given in centimeters, so we need to convert it to meters:

• Wall thickness, \(L\) = 10 cm = \(10 \times 10^{-2}\) m = 0.1 m

Next, we need to rearrange Fourier's Law to solve for the thermal conductivity (\(k\)):

From \(q = k \frac{\Delta T}{L}\), we can isolate \(k\):

\[ k = \frac{q \cdot L}{\Delta T} \]

Now, substitute the numerical values into this rearranged formula:

\[ k = \frac{(30 \, \text{W/m}^2) \cdot (0.1 \, \text{m})}{10 \, \text{°C}} \]

Since a temperature difference of 10 °C is numerically equivalent to 10 K (for temperature differences, the value is the same whether in °C or K), the calculation proceeds as:

\[ k = \frac{(30 \, \text{W/m}^2) \cdot (0.1 \, \text{m})}{10 \, \text{K}} \]

Perform the multiplication in the numerator:

\[ k = \frac{3 \, \text{W/m}}{10 \, \text{K}} \]

Finally, divide to obtain the value of the thermal conductivity:

\[ k = 0.3 \, \text{W/(m·K)} \]

Summary of Results

The calculated thermal conductivity of the wall is 0.3 W/(m·K). This value helps characterize the material's heat transfer properties, indicating its efficiency in conducting heat.