Q1: One face of a furnace wall is at 1030°C and the other face is exposed to room temperature (30°C). If the thermal conductivity of furnace wall is 3 W . m^-1 . k^-1 and the wall thickness is 0.3 m, the maximum heat loss (in W/m) is

ANS:D - 10000

To calculate the maximum heat loss through the furnace wall, we can use Fourier's law of heat conduction: Q=dk⋅A⋅ΔT​ Where:

• Q = Heat transfer rate (W)
• k = Thermal conductivity of the material (W/m·K)
• A = Surface area perpendicular to heat flow (m²)
• ΔT = Temperature difference across the wall (K)
• d = Thickness of the wall (m)

• Thermal conductivity (k) = 3 W/m·K
• Thickness of the wall (d) = 0.3 m
• Temperature difference (ΔT) = 1030°−30°=1000°=10001030°C−30°C=1000°C=1000K (since 1°C = 1K)
• Surface area (A) = Area of one face of the wall (since heat is transferred through one face to the other, we consider only one face) =1×1=1 m2=1×1=1m2