Question

A slab of 250 mm thick is made up of material ‘X’ and its thermal conductivity is 500 W/m·K. One of the surface is kept at 100°C and another surface is maintained at 25°C. The net heat flux across the surface is:

• A. 150 kW/m²
• B. 250 kW/m²
• C. 100 kW/m²
• D. 200 kW/m²

Hint:

For steady-state heat conduction, the heat flux is directly proportional to the temperature difference and inversely proportional to the material thickness.

Answer 1

The Correct Option is C

The heat flux \( q \) across a slab is given by Fourier’s law of heat conduction: \[ q = \frac{k \Delta T}{L} \] Where: - \(k = 500 \, \text{W/m.K}\) is the thermal conductivity, - \( \Delta T = 100°C - 25°C = 75°C \), - \( L = 0.25 \, \text{m} \) is the thickness of the slab. Substituting the values: \[ q = \frac{500 \times 75}{0.25} = 100,000 \, \text{W/m²} = 100 \, \text{kW/m²} \]