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Heat and Thermodynamics question

2022 · Q175
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Heat and Thermodynamics question

2022 · Q175

NEETPhysicsHeat and ThermodynamicsMCQ+4 / −1

Two rods one made of copper and other made of steel of same length and same cross sectional area are joined together. The thermal conductivity of copper and steel are 385 J s−-−1 K−-−1 m−-−1 and 50 J s−-−1 K−-−1 m−-−1 respectively. The free ends of copper and steel are held at 100∘^\circ∘C and 0∘^\circ∘C respectively. The temperature at the junction is, nearly :

  1. A
    88.5∘C88.5^\circ C88.5∘C
  2. B
    12∘C12^\circ C12∘C
  3. C
    50∘C50^\circ C50∘C
  4. D
    73∘C73^\circ C73∘C
View written solutionFree

Correct answer: A

We know in conduction, rate of flow of heat $$q = {{ - K.A.\Delta T} \over {\Delta x}}$$

As it is a case of steady state heat transfer

$$ \Rightarrow {{ - {K_{cu}} \times A \times [100 - {T_j}]} \over l} = {{ - {K_{steel}} \times A \times [{T_j} - 0]} \over l}$$

Where Tj is temperature of the junction

$$ \Rightarrow 385(100 - {T_j}) = 50({T_j})$$

$$ \Rightarrow {{100} \over {{T_j}}} - 1 = {{50} \over {385}} \Rightarrow {{100} \over {{T_j}}} = 1 + {{50} \over {385}} \Rightarrow {{100} \over {{T_j}}} = 1.1298$$

$$ \Rightarrow {T_j} = 88.5^\circ C$$

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