(II) A 100-W light bulb generates 95 W of heat, which is dissipated through a glass bulb that has a radius of 3.0 cm and is 0.50 mm thick. What is the difference in temperature between the inner and outer surfaces of the glass?

In the conduction heat transfer process, the medium's molecules do not leave their mean position. They transfer heat energy by oscillating about their mean position.

The rate of heat generated by the light bulb is\(P = 95\;{\rm{W}}\).

The radius of the bulb is \(r = 3.0\;{\rm{cm}}\).

The thickness of the bulb is \(l = 0.50\;{\rm{mm}}\).

From table 14-4:

The thermal conductivity of the glass 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 expression for the area of the surface area of the light bulb is given as:

\(A = 4\pi {r^2}\)

The expression for the conduction heat transfer rate is given as:

\(\begin{array}{c}P = \frac{{kA\Delta T}}{l}\\\Delta T = \frac{{Pl}}{{kA}}\\ = \frac{{Pl}}{{k\left( {4\pi {r^2}} \right)}}\end{array}\)

Substitute the values in the above equation.

\(\begin{array}{c}\Delta T = \frac{{\left( {95\;{\rm{W}}} \right)\left[ {\left( {{\rm{0}}{\rm{.5}}\;{\rm{mm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{mm}}}}} \right)} \right]}}{{\left( {0.84\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{s}} \cdot {\rm{m}} \cdot {\rm{^\circ C}}}}} \right.\\} {{\rm{s}} \cdot {\rm{m}} \cdot {\rm{^\circ C}}}}} \right)\left[ {4\pi {{\left\{ {\left( {{\rm{3}}\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)} \right\}}^2}} \right]}}\\ = 5{\rm{^\circ C}}\end{array}\)

Thus, the difference in temperatures between the inner and outer surfaces of the glass is \(5{\rm{^\circ C}}\).