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Expert-verified Found in: Page 506 ### College Physics (Urone)

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 $${\bf{2}}{\bf{.664 \times 1}}{{\bf{0}}^{\bf{5}}}\;{\bf{J/s}}$$.

See the step by step solution

## Step 1: Determination of formula for rate of heat transfer from lava by radiationGiven Data:

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 $$5.67 \times {10^{ - 8}}\;{\rm{W}} \cdot {{\rm{m}}^{ - 2}} \cdot {{\rm{K}}^4}$$

## Step 2: Determination of rate of heat transfer from lava by radiation

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 $$2.664 \times {10^5}\;{\rm{J}}/{\rm{s}}$$. ### Want to see more solutions like these? 