A heat source at T = 10^3, K is connected to another heat reservoir at T = 10^2, K by a copper slab

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10-2.Transmission of Heat

hard

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

$200$

(JEE MAIN-2019)

Solution

$\frac{{\Delta Q}}{{\Delta t}} = \frac{{kA}}{\ell }\left( {{T_2} – {T_1}} \right)$

$\frac{1}{A}\left( {\frac{{\Delta Q}}{{\Delta t}}} \right) = \frac{k}{\ell }\left( {{T_2} – {T_1}} \right)$

Standard 11

Physics

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(IIT-1988)