Two rectangular blocks, having identical dimensions, can be arranged : Heat Conduction
By Problem: Two rectangular blocks, having identical dimensions, can be arranged either in configuration I or in configuration II as shown in the figure. One of the blocks has thermal conductivity κ and the other 2κ. The temperature difference between the ends along the x-axis is the same in both the configurations. It takes 9 s to transport a certain amount of heat from the hot end to the cold end in the configuration I. The time to transport the same amount of heat in the configuration II is? (IIT JEE 2013)
- 2 s
- 3 s
- 4.5 s
- 6.0 s
Answer: The answer is (A) i.e., 2 s.
Solution: Let $T_h$ and $T_l$ be the temperatures of hot and cold ends and $T_m$ be the temperature at the middle point of configuration I. The rate of heat flow is equal in both the blocks of configuration I because the blocks are connected in series, i.e., \begin{align} \frac{\Delta Q_{\text{I}}}{\Delta t_{\text{I}}}&=\kappa A \frac{(T_h-T_m)}{x}\nonumber \\ & =2\kappa A\frac{(T_m-T_l)}{x}. \end{align} Solve above equation to get $T_m={(T_h+2T_l)}/{3}$. Thus, above equation becomes \begin{align} &\frac{\Delta Q_{\text{I}}}{\Delta t_\text{I}}=2\kappa A\frac{(T_h-T_l)}{(3x)}. \end{align} In the configuration II, total rate of heat flow is the sum of heat flows through the two blocks because the blocks are joined in parallel i.e., \begin{align} \frac{\Delta Q_{\text{II}}}{\Delta t_{\text{II}}}&=\kappa A \frac{(T_h-T_l)}{x}+2\kappa A\frac{(T_h-T_l)}{x}\nonumber\\ &=3\kappa A\frac{(T_h-T_l)}{x}. \end{align} Divide first equation by the second. Substitute $\Delta Q_{\text{I}}=\Delta Q_{\text{II}}$ and $\Delta t_{\text{I}}=9$ s to get $\Delta t_\text{II}=2$ s.
Alternately, the thermal resistance ($\frac{x}{\kappa A}$) of series combination in configuration I is $\frac{3}{2}\frac{x}{\kappa A}$ and that of parallel combination in configuration II is $R_\text{II}=\frac{1}{3}\frac{x}{\kappa A}$.
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