Question 11.8: Conductive Losses from the Human Body BIO GOAL Apply the con...
Conductive Losses from the Human Body BIO
GOAL Apply the conduction equation to a human being.
PROBLEM In a human being, a layer of fat and muscle lies under the skin having various thicknesses depending on location. In response to a cold environment, capillaries near the surface of the body constrict, reducing blood flow and thereby reducing the conductivity of the tissues. These tissues form a shell up to an inch thick having a thermal conductivity of about 0.21 W/m · K, the same as skin or fat. (a) Estimate the rate of loss of thermal energy due to conduction from the human core region to the skin surface, assuming a shell thickness of 2.0 cm and a skin temperature of 33.0°C. (Skin temperature varies, depending on external conditions.) (b) Calculate the thermal energy lost due to conduction in 1.0 h. (c) Estimate the change in body temperature in 1.0 h if the energy is not replenished. Assume a body mass of 75 kg and a skin surface area of 1.73 m².
STRATEGY The solution to part (a) requires applying Equation 11.7 for the rate of energy transfer due to conduction. Multiplying the power found in part (a) by the elapsed time yields the total thermal energy transfer in the given time. In part (c), an estimate for the change in temperature if the energy is not replenished can be developed using Equation 11.3, Q = mcΔT.
P=k A \frac{\left(T_h – T_c\right)}{L} [11.7]
(a) Estimate the rate of loss of thermal energy due to conduction.
Write the thermal conductivity equation:
P=\frac{k A\left(T_h – T_c\right)}{L}
Substitute values:
P=\frac{(0.21 J / m \cdot K )\left(1.73 m ^2\right)\left(37.0^{\circ} C – 33.0^{\circ} C \right)}{2.0 \times 10^{-2} m }=73 W
(b) Calculate the thermal energy lost due to conduction in 1.0 h.
Multiply the power P by the time Δt:
Q=P \Delta t=(73 W )(3600 s )=2.6 \times 10^5 J
(c) Estimate the change in body temperature in 1.0 h if the energy is not replenished.
Write Equation 11.3 and solve it for ΔT:
Q = mcΔT
\Delta T=\frac{Q}{m c}=\frac{2.6 \times 10^5 J }{(75 kg )(3470 J / kg \cdot K )}=1.0^{\circ} C
REMARKS The calculation doesn’t take into account the thermal gradient, which further reduces the rate of conduction through the shell. Whereas thermal energy transfers through the shell by conduction, other mechanisms remove that energy from the body’s surface because air is a poor conductor of thermal energy. Convection, radiation, and evaporation of sweat are the primary mechanisms that remove thermal energy from the skin. The calculation shows that even under mild conditions the body must constantly replenish its internal energy. It’s possible to die of exposure even in temperatures well above freezing