A copper slab is 2 mm thick. It is protected by a 2 mm layer of stainless steel on both sides. The temperature on one side of this composite slab is `400^(@)C` and `200^(@)C` on the other side. Value of thermal conductivities are-
`k_(cu)=400 Wm^(-1)K^(-1) ` and `k_(s)=16 Wm^(-1)K^(-1)`
(a) Just by knowing that thermal conductivity of steel is much less than copper, find (approximately) the temperature of the copper slab.
(b) Plot the variation of temperature across the thickness of the composite wall.

Answer

(a) Temperature drop will take palace almost entirely in steel. The copper plate will have almost uniform temperature pf `(400+200)/(2)=300^(@)`
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