Question

Animals in cold climates often depend on two layers of insulation: a layer of body fat [of thermal conductivity 0.200W/(m⋅K) ] surrounded by a layer of air trapped inside fur or down. We can model a black bear (Ursus americanus) as a sphere 1.60m in diameter having a layer of fat 3.90cm thick. (Actually, the thickness varies with the season, but we are interested in hibernation, when the fat layer is thickest.) In studies of bear hibernation, it was found that the outer surface layer of the fur is at 2.80∘C during hibernation, while the inner surface of the fat layer is at 30.9∘C a) What should the temperature at the fat-inner fur boundary be so that the bear loses heat at a rate of 51.4W ? b) How thick should the air layer (contained within the fur) be so that the bear loses heat at a rate of 51.4W ?

Answer 1

Answer:

Explanation:

Using the equation

H = Q/t = k A ( T hot - T cold) / L

where H is the rate of heat loss = 51.4 W, T cold be temperature of the outer surface, A is the surface area of the fat layer which is a model of sphere ( surface area of a sphere ) = 4πr² where diameter = 1.60 m

radius = 1.60 m / 2 = 0.80 m

A = 4 × 3.142 × ( 0.8²) = 8.04352 m²

making T cold subject of the formula

T cold =  T hot -   $\frac{HL}{KA}$  = 30.9° C - ( 51.4 W × 3.9 × 10⁻² m) / ( 0.2 W/mK × 8.04352 m² ) =  30.9° C - 1.25 ° C = 29.65° C

b) The thickness of air layer for the bear to lose heat t a rate of 51.4 W

thermal conductivity of air is 0.024 W/mK and rearranging the earlier formula

L = $\frac{kA( T HOT - T COLD )}H}$ = (0.024 W/ m K × 8.04352 m²) ( 29.65° C - 2.8°C) / 51.4 W = 0.101 m = 10.1 m