Question

Three bars of equal lengths and equal area of cross-sections are connected in series. Their thermal conductivities are in the ratio of $2:4:3$. If the open ends of the first and the last bars are at temperature $200^{\circ }C$ and $18^{\circ }C$, respectively in the steady state, then calculate the temperatures of both the junctions.

• A. $116^{\circ }C,74^{\circ }C$
• B. $120^{\circ }C,180^{\circ }C$
• C. $125^{\circ }C,50^{\circ }C$
• D. $130^{\circ }C,40^{\circ }C$

Answer 1

Answer: (A)

Suppose ${\theta }_1$ and ${\theta }_2$ be the temperatures of junctions $B$ and $C$, respectively.

In the steady state, the rate of flow of heat through each bar will be same.
$\frac{Q}{t}=\frac{2K\times A\left(200-{\theta }_1\right)}{x}=\frac{4K\times A\left({\theta }_1-{\theta }_2\right)}{x}$
$=\frac{3K\times A\left({\theta }_2-18\right)}{x}$
$2\left(200-{\theta }_1\right)=4\left({\theta }_1-{\theta }_2\right)=3\left({\theta }_2-18\right)$
$200-{\theta }_1=2{\theta }_1-2{\theta }_2$
and $4{\theta }_1-4{\theta }_2=3{\theta }_2-54$
$\Rightarrow 3{\theta }_1-2{\theta }_2=200$
and $4{\theta }_1-7{\theta }_2=-54$
Solving the given expressions for ${\theta }_1$ and ${\theta }_2,$ we get
$\Rightarrow {\theta }_2=74^{\circ }C$
${\theta }_1=116^{\circ }C$