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Two rods of same length and cross section are joined along the length. Thermal conductivities of first and second rod are ${K_1}\,\,{\rm{and}}\,\,{K_2}$. The temperature of the free ends of the first and second rods are maintained at ${\theta _1}\,\,{\rm{and }}{\theta _2}$ respectively. The temperature of the common junction is
$\frac{{{\theta _1} + {\theta _2}}}{2}$
$\frac{{{K_2}{K_2}}}{{{K_1} + {K_2}}}({\theta _1} + {\theta _2})$
$\frac{{{K_1}{\theta _1} + {K_2}{\theta _2}}}{{{K_1} + {K_2}}}$
$\frac{{{K_2}{\theta _1} + {K_1}{\theta _2}}}{{{K_1} + {K_2}}}$
Solution
(c) At steady state, rate of heat flow for both blocks will be same i.e., $\frac{{{K_1}A({\theta _1} – \theta )}}{{{l_1}}} = \frac{{{K_2}A(\theta – {\theta _2})}}{{{l_2}}}$(given ${l_1} = {l_2}$)
$ \Rightarrow {K_1}A({\theta _1} – \theta ) = {K_2}A(\theta – {\theta _2})$ $ \Rightarrow \theta = \frac{{{K_1}{\theta _1} + {K_2}{\theta _2}}}{{{K_1} + {K_2}}}$
