Question

The opposite faces of a metal slab of thickness \( 5 \, \text{cm} \) and thermal conductivity \( 400 \, \text{W} \, \text{m}^{-1} \, ^\circ\text{C}^{-1} \) are maintained at \( 500 \, ^\circ\text{C} \) and \( 200 \, ^\circ\text{C} \). The area of each face is \( 0.02 \, \text{m}^2 \). Assume that the heat transfer is steady and occurs only in the direction perpendicular to the faces. The magnitude of the heat transfer rate, in \( \text{kW} \), rounded off to the nearest integer, is \_\_\_\_\_\_\_.

Hint:

For steady-state heat conduction, ensure consistent units for area, thickness, and thermal conductivity when using Fourier's law.

Answer 1

Step 1: Fourier's law of heat conduction. The rate of heat transfer through a slab is given by: \[ Q = -k A \frac{\Delta T}{L}, \] where: - \( Q \) is the rate of heat transfer (\( \text{W} \)), - \( k \) is the thermal conductivity of the slab (\( \text{W} \, \text{m}^{-1} \, ^\circ\text{C}^{-1} \)), - \( A \) is the area of the slab (\( \text{m}^2 \)), - \( \Delta T \) is the temperature difference between the faces (\( ^\circ\text{C} \)), - \( L \) is the thickness of the slab (\( \text{m} \)). Step 2: Substitute the given values. Given: \[ k = 400 \, \text{W} \, \text{m}^{-1} \, ^\circ\text{C}^{-1}, \quad A = 0.02 \, \text{m}^2, \quad \Delta T = 500 - 200 = 300 \, ^\circ\text{C}, \quad L = 5 \, \text{cm} = 0.05 \, \text{m}. \] Substitute into the equation: \[ Q = -400 \cdot 0.02 \cdot \frac{300}{0.05}. \] Step 3: Simplify the expression. \[ Q = -400 \cdot 0.02 \cdot 6000 = 48,000 \, \text{W}. \] Step 4: Convert to kilowatts. \[ Q = 48,000 \, \text{W} = 48 \, \text{kW}. \] Step 5: Conclusion. The magnitude of the heat transfer rate is \( 48 \, \text{kW} \).