(II) (a) How much power is radiated by a tungsten sphere (emissivity \({\bf{\varepsilon = 0}}{\bf{.35}}\)) of radius 19 cm at a temperature of 25°C? (b) If the sphere is enclosed in a room whose walls are kept at \({\bf{ - 5^\circ C}}\), what is the net flow rate of energy out of the sphere?

The radiation process does not require any medium to carry heat from one place to another.The value of the heat transfer rate by the radiation process can be calculated by examining the value of emissivity, the Stefan-Boltzmann constant, the area of the object, and the temperature of the object.

The emissivity is \({\rm{\varepsilon }} = 0.35\).

The radius of the sphere is \(r = 19\;{\rm{cm}}\).

The temperature of the sphere is \(T = 25{\rm{^\circ C}}\).

The temperature of the surrounding is \({T_{\rm{s}}} = - 5{\rm{^\circ C}}\).

The area of the sphere can be calculated as:

\(\begin{array}{c}A = 4\pi {r^2}\\ = 4\pi {\left[ {\left( {{\rm{19}}\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)} \right]^2}\\ = 4\pi {\left( {0.19\;{\rm{m}}} \right)^2}\end{array}\)

The power radiated by the tungsten sphere can be calculated by using the following equation:

\(\frac{Q}{t} = {\rm{\varepsilon }}\sigma A{T^4}\)

Here, \(\sigma \) is the Stefan-Boltzmann constant, and its value is \(5.67 \times {10^{ - 8}}\;{{\rm{W}} \mathord{\left/{\vphantom {{\rm{W}} {{{\rm{m}}^{\rm{2}}} \cdot {{\rm{K}}^{\rm{4}}}}}} \right.\\} {{{\rm{m}}^{\rm{2}}} \cdot {{\rm{K}}^{\rm{4}}}}}\).

Substitute the values in the above equation.

\(\begin{array}{c}\frac{Q}{t} = \left( {0.35} \right)\left( {5.67 \times {{10}{ - 8}}\;{{\rm{W}} \mathord{\left/{\vphantom {{\rm{W}} {{{\rm{m}}{\rm{2}}} \cdot {{\rm{K}}{\rm{4}}}}}} \right.\\} {{{\rm{m}}{\rm{2}}} \cdot {{\rm{K}}{\rm{4}}}}}} \right)\left[ {4\pi {{\left( {0.19\;{\rm{m}}} \right)}2}} \right]{\left[ {\left( {25{\rm{\circ C}} + 273} \right)\;{\rm{K}}} \right]4}\\ = 70.9\;{\rm{W}}\\ \approx 71\;{\rm{W}}\end{array}\)

Thus, the power radiated by the tungsten sphere is \(71\;{\rm{W}}\).

Thus, the power radiated by the tungsten sphere is \(71\;{\rm{W}}\).

The net flow rate of energy out of the sphere can be calculated as:

\(\begin{array}{c}\frac{Q}{t} = {\rm{\varepsilon }}\sigma A\left( {{T^4} - T_{\rm{s}}^{\rm{4}}} \right)\\ = \left( {0.35} \right)\left( {5.67 \times {{10}^{ - 8}}\;{{\rm{W}} \mathord{\left/{\vphantom {{\rm{W}} {{{\rm{m}}^{\rm{2}}} \cdot {{\rm{K}}^{\rm{4}}}}}} \right.\\} {{{\rm{m}}^{\rm{2}}} \cdot {{\rm{K}}^{\rm{4}}}}}} \right)\left[ {4\pi {{\left( {0.19\;{\rm{m}}} \right)}^2}} \right]\left[ {{{\left\{ {\left( {25{\rm{^\circ C}} + 273} \right)\;{\rm{K}}} \right\}}^4} - {{\left\{ {\left( { - 5{\rm{^\circ C}} + 273} \right)\;{\rm{K}}} \right\}}^4}} \right]\\ = 24.5\;{\rm{W}} \approx {\rm{25}}\;{\rm{W}}\end{array}\)

Thus, the net flow rate of energy out of the sphere is\({\rm{25}}\;{\rm{W}}\).