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(A) In the steady state,the rate of heat flow $Q$ through a rod of cross-sectional area $A$ and thermal conductivity $K$ is given by Fourier's law of

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(A) In the steady state,the rate of heat flow $Q$ through a rod of cross-sectional area $A$ and thermal conductivity $K$ is given by Fourier's law of heat conduction:$Q = -KA \frac{dT}{dx}$Since the rod is uniform and in a steady state,$Q$,$K$,and $A$ are constants.Therefore,$\frac{dT}{dx} = -\frac{Q}{KA} = 	ext{constant}$.This implies that the temperature gradient is constant throughout the rod.$A$ constant negative slope indicates a straight line starting from $T = 100\,^{\circ}C$ at $x = 0$ and decreasing linearly to $T = 0\,^{\circ}C$ at the other end $B$.

End $A$ of a copper rod is maintained in a steam chamber at $100\,^{\circ}C$ and the other end $B$ is maintained at $0\,^{\circ}C$. Assuming $x = 0$ at $A$,the $T-x$ graph in steady state will be:

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