Question

The ends of copper rod of length $1\text{m}$ and area of cross section $1{\text{cm}}^2$ are maintained at $0^{\circ }C$ and ${100}^{\circ }C$. At the centre of the rod there is a source of heat of power $25\text{W}$. Thermal conductivity of copper is $400\text{W/m-K}$. In steady state, the temperature at the section of rod at which source is supplying heat, will be

• A. ${150.50}^{\circ }C$
• B. ${325.75}^{\circ }C$
• C. ${206.25}^{\circ }C$
• D. ${126.25}^{\circ }C$

Answer 1

The correct option is C ${206.25}^{\circ }C$

According to the energy balance, the rate of heat generated would be equal to the rate of heat conducted through the rod at steady state.



Hence we can write,
$25=H_1+H_2$
$25=\frac{kA}{0.5}(T-100)+\frac{kA}{0.5}(T-0)$
$\frac{25\times 0.5}{400\times 1\times {10}^{-4}}=2T-100$
$\therefore 2T=\frac{1250}{4}+100$
$T={206.25}^{\circ }C$