A plane slab of thickness 60 cm is made of a material of thermal conductivity k = 17.45 W/m K. Let us assume that one side of the slab absorbs a net amount of radiant energy at the rate q = 530.5 watt/m2. If the other face of the slab is at a constant temperature t2 = 38 degree Celsius. Comment on the temperature with respect to the slab?

Correct Answer: 47.08 degree Celsius
Heat flux, q = k (t s – t f) / Thickness. So, t s = 56.17 degree Celsius. Now, t = t s + (t f – t s) x/Thickness.