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(D) Let $K_b$ and $K_c$ be the thermal conductivities of brass and copper respectively. Given $K_b : K_c = 1 : 4$. Let $T$ be the temperature of the i

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$80$

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(D) Let $K_b$ and $K_c$ be the thermal conductivities of brass and copper respectively. Given $K_b : K_c = 1 : 4$. Let $T$ be the temperature of the interface. Since the plates are in series,the rate of heat flow $H$ through both plates must be the same in steady state. $H = \frac{K_c A(100 - T)}{x} = \frac{K_b A(T - 0)}{x}$ Where $A$ is the area and $x$ is the thickness of each plate. Canceling $A$ and $x$ from both sides,we get: $K_c(100 - T) = K_b(T)$ $\frac{K_c}{K_b} = \frac{T}{100 - T}$ Since $\frac{K_b}{K_c} = \frac{1}{4}$,we have $\frac{K_c}{K_b} = 4$. Substituting this value: $4 = \frac{T}{100 - T}$ $4(100 - T) = T$ $400 - 4T = T$ $5T = 400$ $T = 80^{\circ} C$

$A$ slab consists of two identical plates of copper and brass. The free face of the brass is at $0^{\circ} C$ and that of copper at $100^{\circ} C$. If the thermal conductivities of brass and copper are in the ratio $1: 4$,then the temperature of the interface is (in $^{\circ} C$)

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