JEE Mains · Physics · STD 11 - Transmission of heat
- JEE Mains 2023
- Medium
- Numerical
Two plates \(A\) and \(B\) have thermal conductivities \(84\,Wm ^{-1}\,K ^{-1}\) and \(126\,Wm ^{-1}\,K ^{-1}\) respectively. They have same surface area and same thickness. They are placed in contact along their surfaces. If the temperatures of the outer surfaces of \(A\) and \(B\) are kept at \(100^{\circ}\,C\) and \(0{ }^{\circ}\,C\) respectively, then the temperature of the surface of contact in steady state is \(..........\,{ }^{\circ} C\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(40\)
Step-by-step Solution
Detailed explanation
Let the temperature of contact surface is \(T\), then
\(H _{ A }= H _{ B }\)
\(\frac{ K _{ A } A\left(T_A-T\right)}{L}=\frac{ K _{ B } A \left( T - T _{ B }\right)}{ L }\)
\(84(100-T)=126( T -0)\)
\(2(100-T)=3\,T\)
\(200-2\,T =3\,T\)
\(T =40^{\circ}\,C\)
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