91Ó°ÊÓ

In the conduction heat transfer process, heat energy transfers from warm to cool molecules when they are in contact. It does not occur in vacuum, and it is a slower process compared to the radiation process.

The length of each side is \(a = 4.0\;{\rm{m}}\).

The thickness of the wall is \(l = 14\;{\rm{cm}}\).

The power of each bulb is \({P_{\rm{e}}} = 100\;{\rm{W}}\).

The temperature of the air in one room is \({T_1} = 30{\rm{^\circ C}}\).

The temperature of the air in the other room is \({T_2} = 10{\rm{^\circ C}}\).

From table 14-4:

The thermal conductivity of the brick is \(k = 0.84\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{s}} \cdot {\rm{m}} \cdot {\rm{^\circ C}}}}} \right.\\} {{\rm{s}} \cdot {\rm{m}} \cdot {\rm{^\circ C}}}}\).

The surface area of the wall can be calculated as:

\(\begin{array}{c}A = {a^2}\\ = {\left( {4\;{\rm{m}}} \right)^2}\\ = 16\;{{\rm{m}}^{\rm{2}}}\end{array}\)

The heat conduction rate can be calculated as:

\(\begin{array}{c}P = \frac{{kA\left( {{T_1} - {T_2}} \right)}}{l}\\ = \frac{{\left( {0.84\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{s}} \cdot {\rm{m}} \cdot {\rm{^\circ C}}}}} \right.\\} {{\rm{s}} \cdot {\rm{m}} \cdot {\rm{^\circ C}}}}} \right)\left( {16\;{{\rm{m}}^{\rm{2}}}} \right)\left[ {\left( {30{\rm{^\circ C}}} \right) - \left( {10{\rm{^\circ C}}} \right)} \right]}}{{\left( {{\rm{14}}\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)}}\\ = 1920\;{\rm{W}}\end{array}\)

The number of bulbs needed to maintain the temperature difference can be calculated as:

\(\begin{array}{c}n = \frac{P}{{{P_{\rm{e}}}}}\\ = \frac{{1920\;{\rm{W}}}}{{100\;{\rm{W}}}}\\ = 19.2\\ \approx 20\end{array}\)

Thus, the number of bulbs needed to maintain the temperature difference is \(20\).