Question

In figure shows an aluminium rod joined to a copper rod. Each of the rods has a length of 20 cm and area of cross section $0.20c{m}^2$. The junction is maintained at a constant temperature ${40}^{\circ }$ and the two ends are maintained at ${80}^{\circ }$. Calculate the amount of heat taken out from the cold junction in one minute after the steady state is reached. The conductivities are $K_{Al}=200W{m}^{-1\circ }{C}^{-1}$ and $K_{Cu}=400W{m}^{-1\circ }{C}^{-1}$

${40}^{\circ }C{80}^{\circ }C\begin{array}{cc}\text{Aluminium}& \text{Copper}\end{array}{80}^{\circ }C$

Answer 1

$K_{Al}=200W{m}^{-1\circ }{C}^{-1}$

$K_{Cu}=400W{m}^{-1\circ }{C}^{-1}$

A = 0.2 $c{m}^2=2\times {10}^{-5}{m}^2$

l = 20 cm =$2\times {10}^{-1}m$

Heat drawn per sec.

$Q_{Al}+Q_{Cu}=\frac{K_{Al}\times A\times (80-40)}{l}+\frac{K_{Cu}\times A\times (80-40)}{l}$

$=\frac{2\times {10}^{-5}\times 40}{2\times {10}^{-1}}[200+400]$

= 2.4 J

Heat drawn per minute = $2.4\times 60=144J$