Question

A metal rod of cross sectional area 1.0 cm² is being heated at one end. At one time, the temperatures gradient is 5.0°C cm⁻¹ at cross section A and is 2.5°C cm⁻¹ at cross section B. Calculate the rate at which the temperature is increasing in the part AB of the rod. The heat capacity of the part AB = 0.40 J°C⁻¹, thermal conductivity of the material of the rod = 200 W m⁻¹°C⁻¹. Neglect any loss of heat to the atmosphere

Answer 1

Let the temperatures at the ends A and B be T_A and T_B, respectively.
Rate of flow of heat at end A of the rod is given by
$\frac{{\mathrm{dQ}}_{\mathrm{A}}}{\mathrm{d}t}=K\mathrm{A}.\frac{\mathrm{d}}{\mathrm{d}l}\left(T_{\mathrm{A}}\right)$
Rate of flow of heat at end B of the rod is given by
$\frac{d{Q}_{\mathrm{B}}}{\mathrm{d}t}=K\mathrm{A}.\frac{\mathrm{d}}{\mathrm{d}l}\left(T_{\mathrm{B}}\right)$
Heat absorbed by the rod = ms∆T
Here, s is the specific heat of the rod and ∆T is the temperature difference between ends A and B.

Rate of heat absorption by the rod is given by
$$\begin{gathered} \frac{\mathrm{d}Q}{\mathrm{d}t}=ms\frac{\mathrm{d}T}{\mathrm{d}t} \\ \therefore ms\frac{\mathrm{dT}}{\mathrm{d}t}=K\mathrm{A}.\frac{{\mathrm{dT}}_{\mathrm{A}}}{\mathrm{d}l}-K\mathrm{A}.\frac{{\mathrm{dT}}_{\mathrm{B}}}{\mathrm{d}l} \\ \Rightarrow \left(0.4\right).\frac{\mathrm{dT}}{\mathrm{d}t}=200\times 1\times {10}^{-4}\left(5-2.5\right) \\ \frac{\mathrm{dT}}{\mathrm{d}t}=12.5°\mathrm{C}/\sec \end{gathered}$$