Question

A box with a total surface area of 0.55 m^2 and a wall thickness of .026 m is made of an insulating material. A 6.3 W electric heater inside the box maintains the inside temperature at 11.4 ◦C above the outside temperature. Find the thermal conductivity of the insulating material. Answer in units of W/m ·◦ C.

Answer 1

To find the thermal conductivity of the insulating material, we can use Fourier's law of heat conduction, which relates the heat transfer through a material to its thermal conductivity, the area through which heat is conducted, and the temperature difference across that material.

The formula is:
\[
Q = k \cdot A \cdot \frac{\Delta T}{d}
\]
where:
- \( Q \) is the heat transfer rate (in Watts),
- \( k \) is the thermal conductivity (in W/m·°C),
- \( A \) is the surface area (in m²),
- \( \Delta T \) is the temperature difference (in °C),
- \( d \) is the thickness of the insulating material (in meters).

From the problem:
- The total surface area \( A = 0.55 \, \text{m}^2 \).
- The wall thickness \( d = 0.026 \, \text{m} \).
- The heat transfer rate \( Q = 6.3 \, \text{W} \).
- The temperature difference \( \Delta T = 11.4 \, \text{°C} \).

We can rearrange the formula to solve for \( k \):

\[
k = \frac{Q \cdot d}{A \cdot \Delta T}
\]

Now, substituting the values into the equation:

\[
k = \frac{6.3 \, \text{W} \cdot 0.026 \, \text{m}}{0.55 \, \text{m}^2 \cdot 11.4 \, \text{°C}}
\]

Calculating the numerator:

\[
6.3 \, \text{W} \cdot 0.026 \, \text{m} = 0.1638 \, \text{W·m}
\]

Calculating the denominator:

\[
0.55 \, \text{m}^2 \cdot 11.4 \, \text{°C} = 6.297 \, \text{m}^2·°C
\]

Now, substituting these results back into the equation for \( k \):

\[
k = \frac{0.1638}{6.297}
\]
\[
k \approx 0.026 \, \text{W/m·°C}
\]

Thus, the thermal conductivity of the insulating material is approximately **0.026 W/m·°C**.