Question

(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?

Short answer

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

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

Step-by-step solution

Understanding the heat transfer rate by the radiation process

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.

Given data

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 \({T_{\rm{s}}} = - 5{\rm{^\circ C}}\).

(a) Evaluation of the area of the sphere

The area of the sphere can be calculated as:

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

Evaluation of the power radiated by the tungsten sphere

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 \(71\;{\rm{W}}\).

(b) Evaluation of the net flow rate of energy out of the sphere

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\({\rm{25}}\;{\rm{W}}\).