91Ó°ÊÓ

In the conduction process, heat energy is conveyed through a substance by means of collisions between hotter (faster) molecules or electrons with their slower-moving neighbors.

The length of the rod is \(l = 56\;{\rm{cm}}\).

The diameter of the rod is \(d = 2.0\;{\rm{cm}}\).

The temperature of the rod at one end is \({T_1} = 460{\rm{\circ C}}\).

The temperature of the rod at the other end is \({T_2} = 22{\rm{\circ C}}\).

From table 14-4,

The thermal conductivity of the copper is \(k = 380\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{s}} \cdot {\rm{m}} \cdot {\rm{\circ C}}}}} \right.\\} {{\rm{s}} \cdot {\rm{m}} \cdot {\rm{\circ C}}}}\).

The cross-section area of the rod can be calculated as:

\(\begin{array}{c}A = \frac{\pi }{4}{d2}\\ = \frac{\pi }{4}{\left[ {\left( {2\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)} \right]2}\\ = 3.14 \times {10{ - 4}}\;{{\rm{m}}{\rm{2}}}\end{array}\)

The heat conduction rate along the rod can be calculated as:

\(\frac{Q}{t} = kA\frac{{{T_1} - {T_2}}}{l}\)

Here, k is the thermal conductivity, A is the cross-section area, and l is the length,

Substitute the values in the above expression.

\(\begin{array}{c}\frac{Q}{t} = \left( {380\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{s}} \cdot {\rm{m}} \cdot {\rm{\circ C}}}}} \right.\\} {{\rm{s}} \cdot {\rm{m}} \cdot {\rm{\circ C}}}}} \right)\left( {3.14 \times {{10}{ - 4}}\;{{\rm{m}}{\rm{2}}}} \right)\frac{{\left[ {\left( {460{\rm{\circ C}}} \right) - \left( {22{\rm{\circ C}}} \right)} \right]}}{{\left[ {\left( {56\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)} \right]}}\\ = 93.3\;{\rm{W}}\\ \approx {\rm{93}}\;{\rm{W}}\end{array}\)

Thus, the heat conduction rate along the rod is \({\rm{93}}\;{\rm{W}}\).