Answer:
B. 9025 J/s
Explanation:
thickness (d) = 0.367 cm = 0.00367 m
area (A) = 1.247 m^{2}
temperature difference (ΔT) =33.2 °C
thermal conductivity (K) = 0.8 J/s/m/°C
find the rate of heat flow
rate of heat flow = [tex]\frac{K.A.ΔT}{d}[/tex]
rate of heat flow = [tex]\frac{0.8x1.247x33.2}{0.00367}[/tex]
rate of heat flow = 9024.6 ≈ 9025 J/s