Question

Three identical rods of length $1\text{m}$ each, having cross-sectional area of $1{\text{cm}}^2$ each and made of aluminium, copper and steel respectively are maintained at temperatures of ${12}^{\circ }\text{C},4^{\circ }\text{C}$ and ${50}^{\circ }\text{C}$ respectively at their separate ends. Find the temperature of their common junction.
$[K_{Cu}=400\text{W/m-K},K_{Al}=200\text{W/m-K},K_{\text{Steel}}=50\text{W/m-K}]$

• A. ${10}^{\circ }\text{C}$
• B. ${20}^{\circ }\text{C}$
• C. ${30}^{\circ }\text{C}$
• D. ${40}^{\circ }\text{C}$

Answer 1

The correct option is A ${10}^{\circ }\text{C}$

The thermal resistances of respective rods are,
$R_{Al}=\frac{L}{KA}=\frac{1}{200\times {10}^{-4}}=\frac{{10}^4}{200}$
where area of cross-section of rod is, $A={10}^{-4}{\text{m}}^2$
Similarly $R_S=\frac{{10}^4}{50},R_{Cu}=\frac{{10}^4}{400}$

Let the temperature of common junction $=T$
Thermal current for each rod is given as, $i=\frac{\Delta T}{R}$
Then from Kirchoff's junction law for thermal current (at steady state),
$\Rightarrow i_{Al}+i_S+i_{Cu}=0$
$\Rightarrow \left(\frac{T-12}{R_{Al}}\right)+\left(\frac{T-50}{R_S}\right)+\left(\frac{T-4}{R_{Cu}}\right)=0$
After cancelling common terms,
$\Rightarrow (T-12)200+(T-50)50+(T-4)400=0$
$\Rightarrow 13T=130$
$\Rightarrow T={10}^{\circ }\text{C}$