Question

The two rods shown in figure (28-E6) have identical geometrical dimensions. They are in contact with two heat baths at temperatures 100°C and 0°C. The temperature of the junction is 70°C. Find the temperature of the junction if the rods are interchanged.

Answer 1

As the rods are connected in series, the rate of flow of heat will be same in both the cases.

Case 1:
Rate of flow of heat is given by
$\frac{dQ}{dt}=\frac{KA∆T}{l}$
Rate of heat flow in rod P will be same as that in rod Q.

$$\begin{gathered} \therefore \frac{K_{\mathrm{p}}\mathit{\times }A\mathit{\times }\left(100-70\right)}{l}=\frac{K_{\mathrm{Q}}\times A\times \left(70-0\right)}{l} \\ \Rightarrow 30K_{\mathrm{P}}\mathit{}=70K_{\mathrm{Q}} \\ \Rightarrow K_{\mathrm{Q}}=\frac{3}{7}{K}_{\mathrm{P}}......\left(\mathrm{i}\right) \end{gathered}$$

Case 2:
Again, the rate of flow of heat will be same in rod P and Q.

$$\begin{gathered} \therefore \frac{K_{\mathrm{Q}}\times A\times \left(100-\mathrm{T}\right)}{l}=\frac{K_{\mathrm{P}}\times A\times \left(\mathrm{T}-0\right)}{l} \\ 100K_{\mathrm{Q}}-K_{\mathrm{Q}}\mathrm{T}=K_{\mathrm{P}}\mathrm{T} \\ 100K_{\mathrm{Q}}-K_{\mathrm{Q}}T=\frac{70}{30}{K}_{\mathrm{Q}}T[\mathrm{using}(\mathrm{i}\left)\right] \\ 100-T=\frac{7}{3}T \\ 100=\frac{10}{3}T \\ \Rightarrow T=30°\mathrm{C} \end{gathered}$$