Heat is conducted through a copper plate at the rate of 460 cal/s・cm

1.	Heat is conducted through a copper plate at the rate of 460 cal/s・cm2. Calculate the temperature gradient when the steady-state is reached.(Kcopper= 92 cal/ms℃)
Answer» ANSWER: Temperature gradient of the rod is given as $\frac{dT}{dx} = 5 deg C/ m$ Explanation: As we KNOW that THERMAL energy will flow through the metal by CONDUCTION method given as $\frac{dQ}{dt} = kA\frac{dT}{dx}$ so we know that $\frac{1]{A}\frac{dQ}{dt} = k\frac{dT}{dx}$ now we know that $\frac{1}{A}\frac{dQ}{dt} = 460 cal/ s m^2$ now we have $460 cal/s m^2 = 92 cal/ m s C \frac{dT}{dx}$ now we have $\frac{dT}{dx} = 5 deg C/ m$ #LEarn Topic : thermal conduction brainly.in/question/12602371