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Q57PE

Expert-verifiedFound in: Page 506

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Radiation makes it impossible to stand close to a hot lava flow. Calculate the rate of heat transfer by radiation from **\({\bf{1}}{\bf{.00}}\;{{\bf{m}}^{\bf{2}}}\)** of 1200ºC fresh lava into 30.0ºC surroundings, assuming lava’s emissivity is 1.00.**

The rate of heat transfer from lava by radiation is \({\bf{2}}{\bf{.664 \times 1}}{{\bf{0}}^{\bf{5}}}\;{\bf{J/s}}\).

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

The temperature of fresh lava is \({T_1} = 1200^\circ {\rm{C}} = 1473\;{\rm{K}}\)

The temperature of surrounding is \({T_2} = 30^\circ {\rm{C}} = 303\;{\rm{K}}\)

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

**The radiant energy transfer is calculated by using the Stefan’s law. This law gives the energy transfer without any medium.**

The rate of heat transfer from lava by radiation 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.

\(\begin{aligned}{}Q &= \left( 1 \right)\left( {5.67 \times {{10}^{ - 8}}\;{\rm{W}} \cdot {{\rm{m}}^{ - 2}} \cdot {{\rm{K}}^4}} \right)\left( {1\;{{\rm{m}}^2}} \right)\left( {{{\left( {1473\;{\rm{K}}} \right)}^4} - {{\left( {303\;{\rm{K}}} \right)}^4}} \right)\\Q &= 2.664 \times {10^5}\;{\rm{W}}\\Q &= \left( {2.664 \times {{10}^5}\;{\rm{W}}} \right)\left( {\frac{{1\;{\rm{J}}/{\rm{s}}}}{{1\;{\rm{W}}}}} \right)\\Q &= 2.664 \times {10^5}\;{\rm{J}}/{\rm{s}}\end{aligned}\)

Therefore, the rate of heat transfer from lava by radiation is \(2.664 \times {10^5}\;{\rm{J}}/{\rm{s}}\).

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