Question

A rod of negligible heat capacity has length 20 cm, area of cross section 1.0 cm² and thermal conductivity 200 W m⁻¹°C⁻¹. The temperature of one end is maintained at 0°C and that of the other end is slowly and linearly varied from 0°C to 60°C in 10 minutes. Assuming no loss of heat through the sides, find the total heat transmitted through the rod in these 10 minutes.

Answer 1

Given:
Length of the rod, l = 20 cm = 0.2 m
Area of cross section of the rod, A = 1.0 cm²= 1.0$\times$10^$-$4 m²
Thermal conductivity of the material of the rod, k = 200 W m^$-$1$°$C^$-$1
The temperature of one end of the rod is increased uniformly by 60$°$C within 10 minutes.
This mean that the rate of increase of the temperature of one end is 0.1$°$C per second.
$\Rightarrow \frac{60}{10\times 60}°\mathrm{C}/\mathrm{s}$

So, total heat flow can be found by adding heat flow every second.
$$\begin{gathered} \mathrm{Rate}\mathrm{of}\mathrm{flow}\mathrm{of}\mathrm{heat}=\frac{d\mathrm{Q}}{dt} \\ {Q}_{\mathrm{net}}=\sum \frac{KA}{d}\left(T_2-T_1\right)\times ∆t \end{gathered}$$

For each interval,
$∆t=1$

$$\begin{gathered} Q_{\mathrm{net}}=\frac{KA}{d}\times 0.1+\frac{KA}{d}\times 0.2+\frac{KA}{d}\times 0.3+......+\frac{KA}{d}\times 60.0 \\ {Q}_{\mathrm{net}}=\frac{KA}{d}\left(0.1+0.2+.....+60.0\right) \end{gathered}$$

Sum of n terms of an AP is given by
$S_n=\frac{n}{2}\left(a+a_n\right)$
$$\begin{gathered} \Rightarrow Q_{\mathrm{net}}=\frac{KA}{d}\times \frac{600}{2}\left(0.1+60\right) \\ \Rightarrow Q_{\mathrm{net}}=\frac{200\times {10}^{-4}}{20\times {10}^{-2}}\times \frac{600}{2}\times 60.1 \\ \Rightarrow Q_{\mathrm{net}}=1800\mathrm{J}\left(\mathrm{approximately}\right) \end{gathered}$$