Thermodynamics Part 5

Concept:Steady-state heat conduction can be modelled using thermal resistances in series and parallel, analogous to Ohm's law.Explanation:Let rod $y$ have thermal conductivity $k$, then rod $x$ has $3k$. Both have same length $L$ and area $A$.Thermal resistance $R_{th} = \frac{L}{kA}$. So $R_y = R$ and $R_x = \frac{R}{3}$.From $A$ to $F$: $R_{AB} = \frac{R}{3}$ (rod $x$), $R_{EF} = R$ (rod $y$).Between $B$ and $E$: path $BCE$ gives $R+R=2R$, path $BDE$ gives $\frac{R}{3}+\frac{R}{3}=\frac{2R}{3}$. These are in parallel.Parallel combination: $R_{BE} = \frac{(2R)(2R/3)}{2R+2R/3} = \frac{4R^2/3}{8R/3}= \frac{R}{2}$.Total resistance: $R_{\text{total}} = R_{AB}+R_{BE}+R_{EF} = \frac{R}{3}+\frac{R}{2}+R = \frac{11R}{6}$.Overall temperature difference: $\Delta T = 100-40 = 60^\circ \text{C}$.Heat current: $H = \frac{60}{11R/6} = \frac{360}{11R}$.For segment $A$ to $B$: $H = \frac{100-T_B}{R/3}$ $\Rightarrow \frac{360}{11R} = \frac{3(100-T_B)}{R}$ $\Rightarrow 100-T_B = \frac{120}{11}$ $\Rightarrow T_B = 100 - \frac{120}{11} = \frac{980}{11} \approx 89.09^\circ \text{C}$.For segment $E$ to $F$: $H = \frac{T_E-40}{R}$ $\Rightarrow \frac{360}{11R} = \frac{T_E-40}{R}$ $\Rightarrow T_E-40 = \frac{360}{11}$ $\Rightarrow T_E = 40 + \frac{360}{11} = \frac{800}{11} \approx 72.72^\circ \text{C}$.Answer:Temperatures at $B$ and $E$ are approximately $89^\circ \text{C}$ and $73^\circ \text{C}$, respectively, matching option C.