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(D) Let the temperature of the common interface be $T$. In a steady state,the rate of heat flow $H$ is the same through both slabs.The rate of heat fl

Vedclass · Accepted Answer

$0.33$

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(D) Let the temperature of the common interface be $T$. In a steady state,the rate of heat flow $H$ is the same through both slabs.The rate of heat flow is given by $H = \frac{KA \Delta T}{l}$.For the first slab (conductivity $K$,thickness $x$): $H_1 = \frac{KA(T_2 - T)}{x}$.For the second slab (conductivity $2K$,thickness $4x$): $H_2 = \frac{(2K)A(T - T_1)}{4x} = \frac{KA(T - T_1)}{2x}$.Since $H_1 = H_2$:$\frac{KA(T_2 - T)}{x} = \frac{KA(T - T_1)}{2x}$$T_2 - T = \frac{T - T_1}{2}$$2T_2 - 2T = T - T_1$$3T = 2T_2 + T_1 \implies T = \frac{2T_2 + T_1}{3}$.Substituting $T$ back into the expression for $H_1$:$H = \frac{KA}{x} \left( T_2 - \frac{2T_2 + T_1}{3} \right)$$H = \frac{KA}{x} \left( \frac{3T_2 - 2T_2 - T_1}{3} \right)$$H = \frac{KA}{x} \left( \frac{T_2 - T_1}{3} \right) = \frac{1}{3} \left( \frac{KA(T_2 - T_1)}{x} \right)$.Comparing this with the given expression $\left( \frac{A(T_2 - T_1)K}{x} \right)f$,we get $f = \frac{1}{3} \approx 0.33$.

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