91Ó°ÊÓ

Given Data:

The surface area of skier is\(A = 1.60\;{{\rm{m}}^2}\)

The temperature of skier is\({T_1} = 10^\circ {\rm{C}} = 283\;{\rm{K}}\)

The temperature of surrounding is\({T_2} = - 15^\circ {\rm{C}} = 258\;{\rm{K}}\)

The emissivity of skier clothes is\(\varepsilon = 0.200\)

The net rate of heat transfer is calculated by using the Stefan’s law. This law gives the heat transfer without any medium and difference in temperature between surrounding and skier.

The net rate of heat transfer by radiation from skier 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.

\(\begin{aligned}{}Q &= \left( {0.200} \right)\left( {5.67 \times {{10}^{ - 8}}\;{\rm{W}} \cdot {{\rm{m}}^{ - 2}} \cdot {{\rm{K}}^4}} \right)\left( {1.60\;{{\rm{m}}^2}} \right)\left[ {{{\left( {283\;{\rm{K}}} \right)}^4} - {{\left( {258\;{\rm{K}}} \right)}^4}} \right]\\Q &= 36\;{\rm{W}}\\Q &= \left( {36\;{\rm{W}}} \right)\left( {\frac{{1\;{\rm{J}}/{\rm{s}}}}{{1\;{\rm{W}}}}} \right)\\Q &= 36\;{\rm{J}}/{\rm{s}}\end{aligned}\)

Therefore, the net rate of heat transfer from skier by radiation is \(36\;{\rm{J}}/{\rm{s}}\).