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(D) The thermal resistance $R$ of a slab is given by $R = \frac{L}{KA}$,where $L$ is the thickness,$K$ is the thermal conductivity,and $A$ is the area

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(D) The thermal resistance $R$ of a slab is given by $R = \frac{L}{KA}$,where $L$ is the thickness,$K$ is the thermal conductivity,and $A$ is the area.For the first material: $R_1 = \frac{x}{KA}$.For the second material: $R_2 = \frac{4x}{(2K)A} = \frac{2x}{KA}$.Since the slabs are in series,the equivalent thermal resistance $R_{eq}$ is:$R_{eq} = R_1 + R_2 = \frac{x}{KA} + \frac{2x}{KA} = \frac{3x}{KA}$.The rate of heat transfer $\frac{dQ}{dt}$ is given by:$\frac{dQ}{dt} = \frac{T_2 - T_1}{R_{eq}} = \frac{T_2 - T_1}{\frac{3x}{KA}} = \frac{KA(T_2 - T_1)}{3x}$.Comparing this with the given expression $\left[\frac{A(T_2 - T_1)K}{x}\right] \cdot f$,we get:$f = \frac{1}{3}$.

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