A picture window has the dimensions of \({\bf{1}}{\bf{.40}}\;{\bf{m \times 2}}{\bf{.50}}\;{\bf{m}}\) and is made of glass 5.20 mm thick. On a winter day, the temperature of the outside surface of the glass is \({\bf{ - 20}}{\bf{.0}}\;{\bf{^\circ C}}\), while the temperature inside is comfortable \({\bf{19}}{\bf{.5}}\;{\bf{^\circ C}}\).

(a) At what rate is heat being lost through the window by conduction?

(b)At what rate would heat be lost through the window if you covered it with a \({\bf{0}}{\bf{.750}}\;{\bf{mm}}\) thick layer of paper (thermal conductivity \({\bf{0}}{\bf{.0500}}\;{{\bf{W}} \mathord{\left/{\vphantom {{\bf{W}} {\bf{m}}}} \right.\\} {\bf{m}}} \cdot {\bf{K}}\))

The cross-sectional area of the window is

\(\begin{array}{c}{\rm{A}} = 1.40\;{\rm{m}} \times {\rm{2}}{\rm{.50}}\;{\rm{m}}\\ = 3.5\;{{\rm{m}}^{\rm{2}}}\end{array}\)

The temperature difference between the two ends of the glass window:

\(\begin{array}{c}\Delta T = \left( {19.5 - \left( { - 20} \right)} \right)\;{\rm{^\circ C}}\\ = {\rm{39}}{\rm{.5}}\;{\rm{^\circ C}}\end{array}\).

The width of glass:\(L = 5.20\;{\rm{mm}}\).

The glass has thermal conductivity: \(\kappa = 0.8\;{{\rm{W}} \mathord{\left/{\vphantom {{\rm{W}} {{\rm{m}} \cdot {\rm{K}}}}} \right.\\} {{\rm{m}} \cdot {\rm{K}}}}\).

When a hot and a cold body are brought in contact, the transfer of heat takes place between them from the hot body to the cold body. This phenomenon is called thermal conduction. The transfer of heat will continue until the temperature of both bodies becomes equal. The thermal conductivity of an object is given as

\(\kappa {\bf{ = }}\frac{{\bf{L}}}{{{\bf{A\Delta T}}}}\frac{{{\bf{dQ}}}}{{{\bf{dt}}}}\;\; \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)\)

Using equation (1), the rate of heat transfer through the glass can be calculated as-

\(\frac{{dQ}}{{dt}} = \kappa A\frac{{\Delta T}}{L}\)

For the given values, the equation becomes-

\(\begin{array}{c}\frac{{dQ}}{{dt}} = \left( {0.8\;{{\rm{W}} \mathord{\left/{\vphantom {{\rm{W}} {{\rm{m}}{\rm{.K}}}}} \right.\\} {{\rm{m}}{\rm{.K}}}}} \right)\left( {3.5\;{{\rm{m}}^2}} \right)\frac{{39.5\;^\circ {\rm{C}}}}{{5.20 \times {{10}^{ - 3}}\;{\rm{m}}}}\\ = 2.13 \times {10^4}\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {\rm{s}}}} \right.\\} {\rm{s}}}\end{array}\)

the rate of transfer of heat through the glass is \( = 2.13 \times {10^4}\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {\rm{s}}}} \right.} {\rm{s}}}\) .

the rate of heat transfer from inner surface to outer surface of the glass window is \(2.13 \times {10^4}\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {\rm{s}}}} \right.\\} {\rm{s}}}\).