Question

The ends of a copper rod of length 1 m and area of cross -section $1c{m}^2$ 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 w. Let A be the the temperature gradient in the first halve of the rod in steady state. Thermal conductivity of copper is $400W{m}^{-1}{K}^{-1}$ find the value of A ($inC/m)$.

Answer 1

At steady state

$H_1+H_2=25$

$\frac{\Delta T}{R_1}+\frac{\Delta T}{R_2}=25$

$\frac{T-0^0}{\frac{0.5}{(400\times {10}^{-4})}}+\frac{T-100}{\frac{0.5}{400\times {10}^{-4}}}=25$

$R=\frac{\rho }{KA}$

On solving T=206°C

Temperature gradient of first half $=\frac{{206}^0}{\frac{1}{2}}=\frac{\Delta T}{AC}={412}^{0}C/m$