Question

The left end of a copper rod (length = 20 cm, area of cross section = 0.20 cm²) is maintained at 20°C and the right end is maintained at 80°C. Neglecting any loss of heat through radiation, find (a) the temperature at a point 11 cm from the left end and (b) the heat current through the rod. Thermal conductivity of copper = 385 W m⁻¹°C⁻¹.

Answer 1

Area of cross section, A = 0.2 cm² = 0.2 × 10⁻⁴ m²
Thermal conductivity, k = 385 W m^–1 °C^–1

$$\begin{gathered} \mathrm{Rate}\mathrm{of}\mathrm{flow}\mathrm{of}\mathrm{heat}=\frac{\mathrm{Temperature}\mathrm{difference}}{\mathrm{Thermal}\mathrm{resistance}} \\ \frac{\mathrm{\Delta Q}}{\mathrm{\Delta }t}=\frac{kA\mathit{(}{T}_1-T_2\mathit{)}}{{\displaystyle l}} \\ \frac{\mathrm{\Delta Q}}{\mathrm{\Delta }t}=\left(\frac{80-20}{0.2}\right)\times 385\times 0.2\times {10}^{-4} \\ =2310\times {10}^{-3} \\ =2.31\mathrm{J}/\sec \end{gathered}$$

Let te temperature of point C be T, which is at a distance of 11 cm from the left end.
Rate of flow of heat is given by

$$\begin{gathered} \frac{\mathrm{\Delta Q}}{\mathrm{\Delta }t}=\frac{\mathrm{k}A\mathrm{\Delta }T}{{\displaystyle l}} \\ \Rightarrow \frac{\mathrm{\Delta }T}{l}=\left(\frac{\mathrm{\Delta Q}}{\mathrm{\Delta }t}\right)\times \frac{1}{\mathrm{kA}} \\ \frac{T-20}{11\times {10}^{-2}}=\frac{2.31}{383\times 0.2\times {10}^{-4}} \\ \Rightarrow T=33+20 \\ \Rightarrow T=53°\mathrm{C} \end{gathered}$$