Question

The emissivity of tungsten is \({\bf{0}}{\bf{.350}}\). A tungsten sphere with radius \({\bf{1}}.{\bf{50}}{\rm{ }}{\bf{cm}}\) is suspended within a large evacuated enclosure whose walls are at\({\bf{290}}.{\bf{0}}{\rm{ }}{\bf{K}}\). What power input is required to maintain the sphere at \({\bf{3000}}.{\bf{0}}{\rm{ }}{\bf{K}}\). if heat conduction along the supports is ignored?

Short answer

The input power required to maintain the sphere at \(3000\;{\rm{K}}\) is \(4.55\;{\rm{kW}}\).

Step-by-step solution

Given Data

The temperature of the sphere is\(T = 3000\;{\rm{K}}\).

The temperature of the large evacuated enclosure \(T = 290\;{\rm{K}}\)

The emissivity of tungsten is \(0.350\).

Radius of tungsten sphere is \(r = 1.50\;{\rm{cm}} = 0.015\;{\rm{m}}\).

Stefan-Boltzmann constant: \(\sigma = 5.67 \times {10^{ - 8}}\;{{\rm{W}} \mathord{\left/{\vphantom {{\rm{W}} {{{\rm{m}}^{\rm{2}}}{{\rm{K}}^{\rm{4}}}}}} \right.\\} {{{\rm{m}}^{\rm{2}}}{{\rm{K}}^{\rm{4}}}}}\)

Concept

In daily life, objects simultaneously absorbs and radiate energy to the surrounding. For such a body, Stefan’s law is given as-

\(\frac{{{\bf{dQ}}}}{{{\bf{dt}}}}{\bf{ = A\sigma e}}\left( {{{\bf{T}}^4}{\bf{ - T}}_o^4} \right)\;\; \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left( 2 \right)\)

Here,\({{\bf{T}}_{\bf{o}}}\)is the absolute temperature of the surrounding.

Calculation

The surface area of the spherical pot is given as-

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

For \(r = 0.015\;{\rm{m}}\), the surface area of the sphere will be-

\(\begin{array}{c}A = 4\pi {\left( {0.015\;{\rm{m}}} \right)^2}\\ = 2.83 \times {10^{ - 3}}\;{{\rm{m}}^2}\end{array}\)

From equation (2), the heat radiated per second through the sphere is given as-

\(\frac{{dQ}}{{dt}} = A\sigma e\left( {{T^4} - T_o^4} \right)\)

For the given values the above equation becomes-

\(\begin{array}{c}\frac{{dQ}}{{dt}} = 2.83 \times {10^{ - 3}}\;{{\rm{m}}^2} \times 5.67 \times {10^{ - 8}}\;{{\rm{W}} \mathord{\left/{\vphantom {{\rm{W}} {{{\rm{m}}^{\rm{2}}}{{\rm{K}}^{\rm{4}}}}}}\right.\\}{{{\rm{m}}^{\rm{2}}}{{\rm{K}}^{\rm{4}}}}} \times 0.350 \times \left\{ {{{\left( {3000\;{\rm{K}}} \right)}^4} - {{\left( {290\;{\rm{K}}} \right)}^4}} \right\}\\ = \left( {4.55 \times {{10}^3}} \right)\;{\rm{W}}\\ ={\rm{4}}{\rm{.55}}\;{\rm{kW}}\end{array}\)

The input power of the sphere is \(4.55\;{\rm{kW}}\).