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  • Two rods of same length are arranged as shown. The ends are maintained at temperature 100∘c and 70∘c. The temperature at the junction point

Two rods of the same length but different thermal conductivities K1 ​ and  K2 ​ are joined end to end (in series) are arranged. The ends are maintained at temperatures of 100∘C and 70∘C. The temperature T at the junction point?

Answers (1)

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Given- 

  • Lengths of rods = same
  • Thermal conductivities: K1, K2 ?
  • One end of rod 1 = 100°C
  • One end of rod 2 = 70°C
  • The temperature at the junction = T

To find- T

Solution- In steady-state heat conduction, the rate of heat flow (Q) through both rods must be equal = $$
\frac{K_{1}A(100−T)} {L}= \frac{K_{2}A(T−70)} {L}
$$

As A and L are the same. So, $$
{K_{1}(100−T)}= {K_{2}(T−70)} 
$$ $$
100K_{1}+ 70 K_{2}= T{(K_{1}+ K_{2})} 
$$  $$
T= \frac {100K_{1}+ 70 K_{2}}{(K_{1}+ K_{2})}
$$

Posted by

Saniya Khatri

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