Question

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 1

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

$\frac{dT}{dx} = 5 deg C/ m$

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Topic : thermal conduction

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