91Ó°ÊÓ

Heat is defined as the energy transit due to temperature gradient. It always flows from the region of high temperature to the region of low temperature and measured in Joules. The amount of heat which is transferred per unit of time is known as the rate of heat transfer.

A healthy man of height \({\rm{1}}{\rm{.6 m}}\) and width \({\rm{40 cm}}\) is walking sitting outdoors on a cold night. The temperature of the body of the man is \({\rm{21^\circ C}}\) and the temperature outdoors is \({\rm{ - 4^\circ C}}\). Assuming the thickness of the body is \({\rm{2 cm}}\), its thermal conductivity is \({\rm{0}}{\rm{.27 W/m}} \cdot {\rm{^\circ C}}\), calculate the rate of heat transfer from his to outer surrounding.

The area of the man is,

\(A = hw\)

Here, \(h\)is the height of the man, and \(w\) is the width.

Substitute \({\rm{1}}{\rm{.6 m}}\) for \(w\), and \({\rm{40 cm}}\) for \(w\),

\(\begin{align}A &= \left( {1.6{\rm{ m}}} \right) \times \left( {40{\rm{ cm}}} \right)\\ &= \left( {1.6{\rm{ m}}} \right) \times \left( {40{\rm{ cm}}} \right) \times \left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)\\ &= 0.64{\rm{ }}{{\rm{m}}^2}\end{align}\)

The rate of heat transfer is,

\(\dot Q = \frac{{kA\left( {{T_b} - {T_s}} \right)}}{d}\)

Here, \(k\)is the thermal conductivity of the body, \(A\) is the area of the body, \({T_b}\) is the temperature of the body, \({T_s}\) is the temperature of the surrounding, and \(d\) is the thickness of the body.

Substitute \({\rm{0}}{\rm{.27 W/m}} \cdot {\rm{^\circ C}}\) for \(k\), \({\rm{0}}{\rm{.64 }}{{\rm{m}}^{\rm{2}}}\) for \(A\), \({\rm{21^\circ C}}\)for \({T_b}\), \({\rm{ - 4^\circ C}}\) for \({T_s}\), and \({\rm{2 cm}}\) for \(d\).

\(\begin{align}\dot Q &= \frac{{\left( {0.27{\rm{ W}}/{\rm{m}} \cdot ^\circ {\rm{C}}} \right) \times \left( {0.64{\rm{ }}{{\rm{m}}^2}} \right) \times \left( {\left( {21^\circ {\rm{C}}} \right) - \left( { - 4^\circ {\rm{C}}} \right)} \right)}}{{\left( {2{\rm{ cm}}} \right)}}\\ &= \frac{{\left( {0.27{\rm{ W}}/{\rm{m}} \cdot ^\circ {\rm{C}}} \right) \times \left( {0.64{\rm{ }}{{\rm{m}}^2}} \right) \times \left( {\left( {21^\circ {\rm{C}}} \right) - \left( { - 4^\circ {\rm{C}}} \right)} \right)}}{{\left( {2{\rm{ cm}}} \right) \times \left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)}}\\ &= 216{\rm{ W}}\end{align}\)

Hence, the rate of heat transfer is \({\rm{216 W}}\).