Answer:
The temperature difference across the slab is 27.5 K.
Explanation:
Q/A = k∆T/t
Q/A is heat flux = 16,500 W/m^2
k is thermal conductivity = 18 W/m.K
∆T is temperature difference
t is thickness of the slab = 3 cm = 3/100 = 0.03 m
∆T = (heat flux × thickness)/thermal conductivity = 16,500×0.03/18 = 27.5 K
