MHT CET · Physics · Thermal Properties of Matter
Two metal slabs of same cross-sectional area have thicknesses \(d_1\) and ; and thermal conductivities \(K_1\) and \(K_2\) respectively, are connected in series. The free ends of the two slabs are kept at temperatures \(T_1\) and \(T_2\left(T_1>T_2\right)\). The temperature \(T\) of their common junction is
- A \(\frac{K_1 T_1 d_2+K_2 T_2 d_1}{K_1 d_2+K_2 d_1}\)
- B \(\frac{K_1 T_1+K_2 T_2}{K_1+K_2}\)
- C \(\frac{K_1 T_1+K_2 T_2}{T_1+T_2}\)
- D \(\frac{K_1 T_1 d_1+K_2 T_2 d_2}{K_1 d_2+K_2 d_1}\)
Answer & Solution
Correct Answer
(A) \(\frac{K_1 T_1 d_2+K_2 T_2 d_1}{K_1 d_2+K_2 d_1}\)
Step-by-step Solution
Detailed explanation
Heat current, \(\dot{Q}_1=\frac{K_1\left(T_1-T\right) A}{d_1}\)
For second slab,
Heat current, \(\dot{Q}_2=\frac{K_2\left(T-T_2\right) A}{d_2}\)
As slabs are in series same heat current flows through them
\(\dot{Q}_1=\dot{Q}_2\)
\(\therefore \frac{K_1\left(T_1-T\right) A}{d_1}=\frac{K_2\left(T-T_2\right) A}{d_2}\)
Therefore, temperature \(T\) of their common junction is
\(T=\frac{K_1 T_1 d_2+K_2 T_2 d_1}{K_2 d_1+K_1 d_2}\)
For second slab,
Heat current, \(\dot{Q}_2=\frac{K_2\left(T-T_2\right) A}{d_2}\)
As slabs are in series same heat current flows through them
\(\dot{Q}_1=\dot{Q}_2\)
\(\therefore \frac{K_1\left(T_1-T\right) A}{d_1}=\frac{K_2\left(T-T_2\right) A}{d_2}\)
Therefore, temperature \(T\) of their common junction is
\(T=\frac{K_1 T_1 d_2+K_2 T_2 d_1}{K_2 d_1+K_1 d_2}\)
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