A long metallic bar is carrying heat from one | vedclass

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Question

The heat flow rate is given by

$\frac{{dQ}}{{dt}} = \frac{{kA\left( {{	heta _1} - 	heta } ight)}}{x}$

$ \Rightarrow {	heta _1} - 	heta&nbs

Vedclass · Accepted Answer

Vedclass · Answer

The heat flow rate is given by

$\frac{{dQ}}{{dt}} = \frac{{kA\left( {{	heta _1} - 	heta } ight)}}{x}$

$ \Rightarrow {	heta _1} - 	heta  = \frac{x}{{kA}}\frac{{dQ}}{{dt}} \Rightarrow 	heta  = {	heta _1} - \frac{x}{{kA}}\frac{{dQ}}{{dt}}$

Where $	heta _1$ is the temperature of hot and $	heta $ is temperature at a distance $x$ from hot end.

The above equation can be graphically represented by option $(a)$.

A long metallic bar is carrying heat from one of its ends to the other end under steady-state. The variation of temperature $\theta$ along the length $x$ of the bar from its hot end is best described by which of the following figures?

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