Question

A large body of lava from a volcano has stopped flowing and is slowly cooling. The interior of the lava is at 1200ºC , its surface is at 450ºC , and the surroundings are at 27.0ºC . (a) Calculate the rate at which energy is transferred by radiation from\({\bf{1}}{\bf{.00 }}{{\bf{m}}^{\bf{2}}}\)of surface lava into the surroundings, assuming the emissivity is 1.00. (b) Suppose heat conduction to the surface occurs at the same rate. What is the thickness of the lava between the 450ºC surface and the 1200ºC interior, assuming that the lava’s conductivity is the same as that of brick?

Short answer

(a) The rate of heat transfer by radiation from skin is\({\bf{1}}{\bf{.50 \times 1}}{{\bf{0}}^{\bf{4}}}\;{\bf{W}}\).

(b) The thickness of lava is \({\bf{0}}{\bf{.043}}\;{\bf{m}}\).

Step-by-step solution

Determination of rate of heat transfer by radiation from lava

(a)

Given Data:

The surface area of lava is\(A = 1\;{{\rm{m}}^2}\)

The temperature of interior of lava is\(T = 1200^\circ {\rm{C}} = 1473\;{\rm{K}}\)

The surface temperature of lava is\({T_1} = 450^\circ {\rm{C}} = 723\;{\rm{K}}\)

The temperature of surrounding is\({T_2} = 27^\circ {\rm{C}} = 300\;{\rm{K}}\)

The emissivity of lava is\(\varepsilon = 1\)

The rate of heat transfer by radiation from skin of body is calculated by using the Stefan’s law. The thickness of the lava is found by using Fourier’s formula for heat conduction.

The rate of heat transfer by radiation from lava is given as:

\(Q = \varepsilon \sigma A\left( {T_1^4 - T_2^4} \right)\)

Here,\(\sigma \)is Stefan’s constant and its value is\(5.67 \times {10^{ - 8}}\;{\rm{W}} \cdot {{\rm{m}}^{ - 2}} \cdot {{\rm{K}}^4}\)

Substitute all the values in the above equation.

\(\begin{aligned}{}Q &= \left( 1 \right)\left( {5.67 \times {{10}^{ - 8}}\;{\rm{W}} \cdot {{\rm{m}}^{ - 2}} \cdot {{\rm{K}}^4}} \right)\left( {1\;{{\rm{m}}^2}} \right)\left( {{{\left( {723\;{\rm{K}}} \right)}^4} - {{\left( {300\;{\rm{K}}} \right)}^4}} \right)\\Q &= 1.50 \times {10^4}\;{\rm{W}}\end{aligned}\)

Therefore, the rate of heat transfer by radiation from skin is \(1.50 \times {10^4}\;{\rm{W}}\).

Determination of thickness of lava

(b)

The thickness of lava is given as:

\(Q = kA\frac{{\left( {T - {T_1}} \right)}}{{2t}}\)

Here,\(k\)is the thermal conductivity of lava and its value is\(1.7\;{\rm{W}}/{\rm{m}} \cdot {\rm{K}}\).

Substitute all the values in the above equation.

\(\begin{aligned}{}1.50 \times {10^4}\;{\rm{W}} &= \left( {1.7\;{\rm{W}}/{\rm{m}} \cdot {\rm{K}}} \right)\left( {1\;{{\rm{m}}^2}} \right)\left( {\frac{{1473\;{\rm{K}} - 723\;{\rm{K}}}}{{2t}}} \right)\\t &= 0.043\;{\rm{m}}\end{aligned}\)

Therefore, the thickness of lava is \(0.043\;{\rm{m}}\).