A spherical pot contains \({\bf{0}}{\bf{.75}}\;{\bf{L}}\) of hot coffee (essentially water) at an initial temperature of \({\bf{95}}\;{\bf{^\circ C}}\). The pot has an emissivity of \({\bf{0}}{\bf{.60}}\), and the surroundings are at \({\bf{20}}{\bf{.0}}\;{\bf{^\circ C}}\). Calculate the coffee’s rate of heat loss by radiation.

The temperature of the coffee is\(T = 95\;^\circ {\rm{C}} = 368\;{\rm{K}}\).

The temperature of the surroundings are\(T = 20\;^\circ {\rm{C}} = 293\;{\rm{K}}\)

The emissivity of the pot is \(0.60\).

Volume of coffee is \(V = 0.75\;{\rm{L}}\).

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

Stefan-Boltzmann Law states that the rate at which heat is radiated through a black body is directly proportional to the fourth power of the absolute temperature of the black body. The rate of heat radiation through a perfect black body, on heating, is given as-

\(\frac{{{\bf{dQ}}}}{{{\bf{dt}}}}{\bf{ = A\sigma e}}{{\bf{T}}^{\bf{4}}}\;\; \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)\)

Any other body except the black body simultaneously emits the energy and absorbs it from the surrounding. For such a body, the rate of heat transfer 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.

The volume of the spherical pot is given as-

\(V = \frac{4}{3}\pi {r^3}\)

Here, \(r\) is the radius of the sphere.

For \(V = 0.75\;{\rm{L}}\), the radius of sphere is-

\(\begin{array}{c}r = \sqrt[3]{{0.75\;{\rm{L}} \times \frac{{\rm{3}}}{{4\pi }}}}\\ = 0.56\;{\rm{m}}\end{array}\)

The surface area of the spherical pot is given as-

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

For \(r = 0.56\;{\rm{m}}\), the surface area of the sphere will be-

\(\begin{array}{c}A = 4\pi {\left( {0.56\;{\rm{m}}} \right)^2}\\ = 3.94\;{{\rm{m}}^2}\end{array}\)

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

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

For the given values the above equation becomes-

\(\begin{array}{c}\frac{{dQ}}{{dt}} = 3.94\;{{\rm{m}}^2} \times 5.67 \times {10^{ - 8}}\;{{\rm{W}} \mathord{\left/{\vphantom {{\rm{W}} {{{\rm{m}}^{\rm{2}}}{{\rm{K}}^{\rm{4}}}}}} \right.\\} {{{\rm{m}}^{\rm{2}}}{{\rm{K}}^{\rm{4}}}}} \times 0.60 \times \left\{ {{{\left( {368\;{\rm{K}}} \right)}^4} - {{\left( {293\;{\rm{K}}} \right)}^4}} \right\}\\ = \left( {1.47 \times {{10}^3}} \right)\;{\rm{W}}\\ = {\rm{1}}{\rm{.47}}\;{\rm{kW}}\end{array}\)

The rate at which heat is lost from the spherical pot is \(1.47\;{\rm{kW}}\).