A composite metal bar of uniform section is made up of length $25 cm$ of copper, $10 cm$ of nickel and $15 cm$ of aluminium. Each part being in perfect thermal contact with the adjoining part. The copper end of the composite rod is maintained at ${100^o}C$ and the aluminium end at ${0^o}C$. The whole rod is covered with belt so that there is no heat loss occurs at the sides. If ${K_{{\rm{Cu}}}} = 2{K_{Al}}$ and ${K_{Al}} = 3{K_{{\rm{Ni}}}}$, then what will be the temperatures of $Cu - Ni$ and $Ni - Al$ junctions respectively
Diffcult
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(b) If suppose ${K_{Ni}} = K \Rightarrow {K_{Al}} = 3K$ and ${K_{Cu}} = 6K.$
Since all metal bars are connected in series
Since all metal bars are connected in series
So ${\left( {\frac{Q}{t}} \right)_{Combination}} = {\left( {\frac{Q}{t}} \right)_{Cu}} = {\left( {\frac{Q}{t}} \right)_{Al}} = {\left( {\frac{Q}{t}} \right)_{Ni}}$
and $\frac{3}{{{K_{eq}}}} = \frac{1}{{{K_{Cu}}}} + \frac{1}{{{K_{Al}}}} + \frac{1}{{{K_{Ni}}}} = \frac{1}{{6K}} + \frac{1}{{3K}} + \frac{1}{K} = \frac{9}{{6K}}$
==> ${K_{eq}} = 2K$
Hence, if ${\left( {\frac{Q}{t}} \right)_{Combination}} = {\left( {\frac{Q}{t}} \right)_{Cu}}$
==> $\frac{{{K_{eq}}\,A(100 - 0)}}{{{l_{Combination}}}} = \frac{{{K_{Cu}}A(100 - {\theta _1})}}{{{l_{Cu}}}}$
==> $\frac{{2K\,A\,(100 - 0)}}{{(25 + 10 + 15)}} = \frac{{6K\,A\,(100 - {\theta _1})}}{{25}}$
==> ${\theta _1} = 83.33^\circ C$
Similar if ${\left( {\frac{Q}{t}} \right)_{Combination}} = {\left( {\frac{Q}{t}} \right)_{Al}}$
==> $\frac{{2K\,A(100 - 0)}}{{50}} = \frac{{3K\,A({\theta _2} - 0)}}{{15}}$
==> ${\theta _2} = {20^o}C$
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