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(D) In a steady state, the rate of heat flow $(H)$ through both slabs connected in series must be the same.The rate of heat flow is given by $H = \fra

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$80$

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(D) In a steady state, the rate of heat flow $(H)$ through both slabs connected in series must be the same.The rate of heat flow is given by $H = \frac{KA(T_1 - T_2)}{L}$, where $K$ is thermal conductivity, $A$ is area, $L$ is thickness, and $(T_1 - T_2)$ is the temperature difference.Let the temperature of the junction $EF$ be $T$.For the first slab (thickness $d$, conductivity $2k$): $H_1 = \frac{(2k)A(100 - T)}{d}$For the second slab (thickness $2d$, conductivity $k$): $H_2 = \frac{kA(T - 0)}{2d}$Since $H_1 = H_2$:$\frac{2kA(100 - T)}{d} = \frac{kA(T - 0)}{2d}$$2(100 - T) = \frac{T}{2}$$4(100 - T) = T$$400 - 4T = T$$5T = 400$$T = 80^oC$.

Heat is conducted across a composite block of two slabs of thickness $d$ and $2d$. Their thermal conductivities are $2k$ and $k$ respectively. All the heat entering the face $AB$ leaves from the face $CD$. The temperature in $^oC$ of the junction $EF$ of the two slabs is:

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