Question

Temperatures at two sides of a 0.4 m thick copper plate are 1000°C and 500°C. Assuming steady state, one-dimensional conductive heat transfer through the wall and ignoring end-effects, the magnitude of the heat flux through the wall is _________ \( \times 10^5 \, {W m}^{-2} \) (in integer). Given: Thermal conductivity of copper \( k = 400 \, {W m}^{-1} {K}^{-1} \).

Hint:

To calculate the heat flux through a material, use Fourier’s law of heat conduction, which relates the heat flux to the temperature gradient and material properties like thermal conductivity. Remember to use consistent units for temperature, thickness, and thermal conductivity.

Answer 1

The heat flux \( q \) through a thick wall can be calculated using the Fourier law for heat conduction: \[ q = \frac{k \Delta T}{L} \] where:
- \( k = 400 \, {W m}^{-1} {K}^{-1} \) is the thermal conductivity of copper,
- \( \Delta T = 1000°C - 500°C = 500°C = 500 \, {K} \) is the temperature difference,
- \( L = 0.4 \, {m} \) is the thickness of the copper plate.
Substituting the values into the equation: \[ q = \frac{400 \times 500}{0.4} = 500000 \, {W/m}^2 = 5 \times 10^5 \, {W/m}^2 \] Thus, the magnitude of the heat flux through the wall is \( 5 \times 10^5 \, {W/m}^2 \). Answer: 5