JEE Mains · Physics · STD 11 - 10.2 transmission of heat
A heat source at \(T = 10^3\, K\) is connected to another heat reservoir at \(T = 10^2\, K\) by a copper slab which is \(1\, m\) thick. Given that the thermal conductivity of copper is \(0.1\, WK^{-1}\, m^{-1}\), the energy flux through it in the steady state is ........... \(Wm^{-2}\)
- A \(90\)
- B \(120\)
- C \(65\)
- D \(200\)
Answer & Solution
Correct Answer
(A) \(90\)
Step-by-step Solution
Detailed explanation
\(\frac{{\Delta Q}}{{\Delta t}} = \frac{{kA}}{\ell }\left( {{T_2} - {T_1}} \right)\) \(\frac{1}{A}\left( {\frac{{\Delta Q}}{{\Delta t}}} \right) = \frac{k}{\ell }\left( {{T_2} - {T_1}} \right)\)
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