Question
A wall has two layers $$A$$ and $$B,$$ each made of different material. Both the layers have the same thickness. The thermal conductivity of the meterial of $$A$$ is twice that of $$B.$$ Under thermal equilibrium, the temperature difference across the wall is $${36^ \circ }C.$$ The temperature difference across the layer $$A$$ is
A.
$${6^ \circ }C$$
B.
$${12^ \circ }C$$
C.
$${18^ \circ }C$$
D.
$${24^ \circ }C$$
Answer :
$${12^ \circ }C$$
Solution :
$$\eqalign{ & {\theta _A} - {\theta _B} = {36^ \circ }C\,\,\left( {{\text{Given}}} \right) \cr & {K_A} = 2\,{K_B}\,\,\left( {{\text{Given}}} \right) \cr & {\theta _C} = \frac{{\frac{{{K_A}}}{\ell }{\theta _A} + \frac{{{K_B}}}{\ell }{\theta _B}}}{{\frac{{{K_A}}}{\ell } + \frac{{{K_B}}}{\ell }}} \cr} $$
$$\eqalign{ & \therefore \,\,{\theta _C} = \frac{{2\,{\theta _A} + {\theta _B}}}{3} \cr & = \frac{{2\,{\theta _A} + {\theta _A} - 36}}{3} \cr & = \frac{{3\left( {{\theta _A} - 12} \right)}}{3} \cr & \therefore \,\,{\theta _A} - {\theta _C} = 12 \cr} $$
$$\eqalign{ & {\theta _A} - {\theta _B} = {36^ \circ }C\,\,\left( {{\text{Given}}} \right) \cr & {K_A} = 2\,{K_B}\,\,\left( {{\text{Given}}} \right) \cr & {\theta _C} = \frac{{\frac{{{K_A}}}{\ell }{\theta _A} + \frac{{{K_B}}}{\ell }{\theta _B}}}{{\frac{{{K_A}}}{\ell } + \frac{{{K_B}}}{\ell }}} \cr} $$
$$\eqalign{ & \therefore \,\,{\theta _C} = \frac{{2\,{\theta _A} + {\theta _B}}}{3} \cr & = \frac{{2\,{\theta _A} + {\theta _A} - 36}}{3} \cr & = \frac{{3\left( {{\theta _A} - 12} \right)}}{3} \cr & \therefore \,\,{\theta _A} - {\theta _C} = 12 \cr} $$
