An iron bar $(L_{1}=0.1 \; m, A_{1}=0.02 \; m^{2}, K_{1}=79 \; W m^{-1} K^{-1})$ and a brass bar $(L_{2}=0.1 \; m, A_{2}=0.02 \; m^{2}, K_{2}=109 \; W m^{-1} K^{-1})$ are soldered end to end as shown in the figure. The free ends of the iron bar and brass bar are maintained at $373 \; K$ and $273 \; K$ respectively. Obtain expressions for and hence compute:
$(i)$ the temperature of the junction of the two bars,
$(ii)$ the equivalent thermal conductivity of the compound bar,and
$(iii)$ the heat current through the compound bar.

(N/A) Given: $L_{1}=L_{2}=L=0.1 \; m$,$A_{1}=A_{2}=A=0.02 \; m^{2}$,$K_{1}=79 \; W m^{-1} K^{-1}$,$K_{2}=109 \; W m^{-1} K^{-1}$,$T_{1}=373 \; K$,$T_{2}=273 \; K$.
Under steady-state conditions,the heat current $(H_{1})$ through the iron bar is equal to the heat current $(H_{2})$ through the brass bar.
$H = H_{1} = H_{2} = \frac{K_{1} A (T_{1}-T_{0})}{L} = \frac{K_{2} A (T_{0}-T_{2})}{L}$
$(i)$ Junction temperature $T_{0}$:
$K_{1}(T_{1}-T_{0}) = K_{2}(T_{0}-T_{2})$
$T_{0} = \frac{K_{1} T_{1} + K_{2} T_{2}}{K_{1} + K_{2}} = \frac{(79 \times 373) + (109 \times 273)}{79 + 109} = \frac{29467 + 29757}{188} = \frac{59224}{188} \approx 315 \; K$.
$(ii)$ Equivalent thermal conductivity $K'$:
For series combination,$\frac{2L}{K'} = \frac{L}{K_{1}} + \frac{L}{K_{2}} \implies \frac{2}{K'} = \frac{1}{K_{1}} + \frac{1}{K_{2}} = \frac{K_{1}+K_{2}}{K_{1} K_{2}}$
$K' = \frac{2 K_{1} K_{2}}{K_{1}+K_{2}} = \frac{2 \times 79 \times 109}{79 + 109} = \frac{17222}{188} \approx 91.6 \; W m^{-1} K^{-1}$.
$(iii)$ Heat current $H$:
$H = \frac{K' A (T_{1}-T_{2})}{2L} = \frac{91.6 \times 0.02 \times (373 - 273)}{2 \times 0.1} = \frac{91.6 \times 0.02 \times 100}{0.2} = \frac{183.2}{0.2} = 916 \; W$.