Question

A metal rod of cross sectional area 1.0 $c{m}^2$ is being heated at one end. At one time, the temperature gradient is ${5.0}^{\circ }Cc{m}^{-1}$ at cross section A and is ${2.5}^{\circ }Cc{m}^{-1}$ at cross section B. Calculate the rate at which the emperature is increasing in the part AB of the rod. The heat capacity of the part $AB=0.40J^{\circ }{C}^{-1}$, thermal conductivity of the material of the rod = $200W{m}^{-1\circ }{C}^{-1}$. Neglect any loss of heat to the atmosphere.

Answer 1

The rate of heat flow per sec

$\frac{d{Q}_A}{dt}=KA\frac{d{\theta }_A}{dl}$

The rate of heat flow per sec

$\frac{d{\theta }_B}{dt}=KA\frac{d{\theta }_B}{dl}$

The part of heat is absorbed by the rod.

$\frac{Q}{t}=\frac{ms\Delta \theta }{dt}$

Where $\frac{d\theta }{dt}$ = Rate of temperature variation

$\Rightarrow \frac{ms\Delta \theta }{dt}=KA\frac{d{\theta }_A}{dl}-Ka\frac{d{\theta }_B}{dl}$

$\Rightarrow ms\frac{d\theta }{dt}=Ka(\frac{d{\theta }_A}{dl}-\frac{d{\theta }_B}{dl})$

$\Rightarrow 0.4\frac{d\theta }{dt}=200\times 1\times {10}^{-4}$ $(5-2.5{)}^{\circ }C/cm$

$\Rightarrow \frac{d\theta }{dt}={\frac{200\times 2.5\times {10}^{-4}}{0.4\times {10}^{-2}}}^{\circ }C/m$

$=\frac{200\times 2.5\times {10}^{-4}}{0.4\times {10}^{-2}}$

=$1250\times {10}^{-2}=12.50C/s.$