Consider the situation as shown in the figure. Find the amount of heat flowing per second through a cross-section of the bent part,if the total heat taken out per second from the end at $200^{\circ} C$ is $150 \ J/s$.

Two slabs $A$ and $B$ of different materials but of the same thickness are joined end to end to form a composite slab. The thermal conductivities of $A$ and $B$ are $k_1$ and $k_2$ respectively. $A$ steady temperature difference of $12^{\circ} C$ is maintained across the composite slab. If $k_1=\frac{k_2}{2}$, the temperature difference across slab $A$ is (in $^{\circ} C$)

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 $f$.

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

Two rods having the same cross-sectional area are used to connect two reservoirs at temperatures $100\,^{\circ}C$ and $0\,^{\circ}C$ as shown. The temperature of the junction is $70\,^{\circ}C$. If the rods are now interchanged,the temperature of the junction will be ......... $^{\circ}C$.

Two identical plates of different metals are joined to form a single plate whose thickness is double the thickness of each plate. If the coefficients of thermal conductivity of each plate are $2$ and $3$ respectively,then the thermal conductivity of the composite plate will be: