Question

Consider a person outdoors on a cold night. Construct a problem in which you calculate the rate of heat transfer from the person by all three heat transfer methods. Make the initial circumstances such that at rest the person will have a net heat transfer and then decide how much physical activity of a chosen type is necessary to balance the rate of heat transfer. Among the things to consider are the size of the person, type of clothing, initial metabolic rate, sky conditions, amount of water evaporated, and volume of air breathed. Of course, there are many other factors to consider and your instructor may wish to guide you in the assumptions made as well as the detail of analysis and method of presenting your results.

Short answer

The rate of heat transfer is \(216{\rm{ W}}\).

Step-by-step solution

Defining heat

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.

Construction of problem

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.

Calculating of area of the man

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 \(w\),

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

Calculating of rate of heat transfer

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 \({\rm{216 W}}\).