A slab consists of two parallel layers of cop | vedclass

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(A) Let the thickness of each layer be $d$. Let $K_c$ be the thermal conductivity of copper and $K_b$ be the thermal conductivity of brass. Given $K_c

Vedclass · Accepted Answer

$80$

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(A) Let the thickness of each layer be $d$. Let $K_c$ be the thermal conductivity of copper and $K_b$ be the thermal conductivity of brass. Given $K_c : K_b = 1 : 4$,so $K_b = 4K_c$.In a steady state,the rate of heat flow through the copper layer must equal the rate of heat flow through the brass layer.Let $	heta$ be the temperature of the interface.Rate of heat flow $H = \frac{KA(\Delta T)}{d}$.For copper: $H_c = \frac{K_c A(	heta - 0)}{d}$.For brass: $H_b = \frac{K_b A(100 - 	heta)}{d}$.Since $H_c = H_b$,we have $\frac{K_c A 	heta}{d} = \frac{4K_c A(100 - 	heta)}{d}$.Canceling common terms $K_c, A, d$: $	heta = 4(100 - 	heta)$.$	heta = 400 - 4	heta$.$5	heta = 400$.$	heta = 80^\circ C$.

$A$ slab consists of two parallel layers of copper and brass of the same thickness and having thermal conductivities in the ratio $1 : 4$. If the free face of brass is at $100^\circ C$ and that of copper at $0^\circ C$,the temperature of the interface is ........ $^\circ C$.

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