Question

The heat flux (from outside to inside) across an insulating wall with thermal conductivity k = 0.04 W/m K and thickness 0.16m is 10 W/ m$^2$. The temperature of the inside wall is -5$^\circ$C. The outside wall temperature is______ $^\circ$C.

• A. 35
• B. 43.75
• C. 81.25
• D. 48.25

Hint:

Always keep track of negative signs when working with below-freezing temperatures.
Double check that \( T_{\text{out}} - T_{\text{in}} = 35 - (-5) = 40^\circ\text{C} \), which maintains the correct heat flux direction from outside (hot) to inside (cold).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks to calculate the outside temperature of a flat insulating wall under steady-state heat conduction.
We are given the thermal conductivity (\( k \)), the thickness of the wall (\( L \)), the heat flux (\( q \)), and the temperature of the inside surface (\( T_{\text{in}} \)).

Step 2: Key Formula or Approach:
For steady-state, one-dimensional heat conduction through a flat wall, Fourier's Law is expressed as:
\[ q = -k \cdot \frac{dT}{dx} \quad \implies \quad q = \frac{k \cdot (T_{\text{out}} - T_{\text{in}})}{L} \]
where \( q \) is the heat flux, \( k \) is the thermal conductivity, \( L \) is the thickness of the wall, \( T_{\text{out}} \) is the outside wall temperature, and \( T_{\text{in}} \) is the inside wall temperature.

Step 3: Detailed Explanation:

• Identify the given variables:
Heat flux, \( q = 10 \text{ W/m}^2 \)
Thermal conductivity, \( k = 0.04 \text{ W/mK} \)
Wall thickness, \( L = 0.16 \text{ m} \)
Inside wall temperature, \( T_{\text{in}} = -5^\circ\text{C} \)

• Rearrange the conduction equation to solve for \( T_{\text{out}} \):
\[ T_{\text{out}} - T_{\text{in}} = \frac{q \cdot L}{k} \]

• Substitute the known values:
\[ T_{\text{out}} - (-5) = \frac{10 \cdot 0.16}{0.04} \]

• Perform the arithmetic:
\[ T_{\text{out}} + 5 = \frac{1.6}{0.04} \]
\[ T_{\text{out}} + 5 = 40 \]
\[ T_{\text{out}} = 40 - 5 = 35^\circ\text{C} \]


Step 4: Final Answer:
The temperature of the outside wall is 35\(^\circ\)C.