The temperatures of the two outer surfaces of a composite slab are $T_2$ and $T_1$ $(T_2 > T_1)$. The thermal conductivities of the materials are $K$ and $2K$,and their thicknesses are $x$ and $4x$,respectively. In the steady state,the rate of heat flow through the slab is given by $f \left( \frac{A(T_2 - T_1)K}{x} \right)$. Find the value of $f$.

In a composite slab, there are two materials having coefficients of thermal conductivity $K$ and $2K$, and thicknesses $x$ and $4x$ respectively. The temperatures of the two outer surfaces of the composite slab are $T_2$ and $T_1$ $(T_2 > T_1)$. $T_2$ is on the side with conductivity $K$ and $T_1$ is on the side with conductivity $2K$. The rate of heat transfer through the slab in a steady state is $\left[ \frac{A(T_2 - T_1)K}{x} \right] \cdot f$, where $f$ is equal to:

Six identical conducting rods are joined as shown in the figure. Points $A$ and $D$ are maintained at $100^{\circ} C$ and $10^{\circ} C$ respectively. The temperature of junction $C$ will be $.... ^{\circ} C$.

Two rods of same length and material transfer a given amount of heat in $12 \ s$,when they are joined end to end. But when they are joined along their lengths (parallel combination),then they will transfer the same amount of heat under the same conditions in time $t = \dots \ s$.

$A$ wall is made of two layers of thickness $1\ cm$ and $4\ cm$ and thermal conductivities $K$ and $3K$ respectively. If the temperature difference across the wall is $50\ ^oC$,then the temperature difference across the thin layer will be

Two walls have thicknesses $d_1$ and $d_2$ and thermal conductivities $k_1$ and $k_2$,respectively. In the steady state,the temperatures of the outer surfaces are $T_1$ and $T_2$. What is the temperature at the interface of the two walls?