JEE PYQ: Thermodynamics Question 39
Question 39 - 2019 (10 Jan Shift 1)
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:
(1) 90 Wm$^{-2}$
(2) 120 Wm$^{-2}$
(3) 65 Wm$^{-2}$
(4) 200 Wm$^{-2}$
Show Answer
Answer: (1)
Solution
Energy flux, $\frac{1}{A}\left(\frac{dQ}{dt}\right) = \frac{k\Delta T}{\ell}$
$= \frac{(0.1)(900)}{1} = 90$ W/m$^{2}$
BETA
