Animals in cold climates often depend on two layers of insulation: a layer of bo

Question

Animals in cold climates often depend on two layers of insulation: a layer of body fat [of thermal conductivity 0.200 W/(mK) ] surrounded by a layer of air trapped inside fur or down. We can model a black bear (Ursus americanus) as a sphere 1.60 m in diameter having a layer of fat 4.20 cm 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 30.8 C.

Assume the surface area of each layer is constant and given by the surface area of the spherical model constructed for the black bear.

Part A

What should the temperature at the fat-inner fur boundary be so that the bear loses heat at a rate of 50.7 W ? (in degrees celsius)

Part B

How thick should the air layer (contained within the fur) be so that the bear loses heat at a rate of 50.7 W ? (in cm)

Explanation / Answer

rate of heat loss=conductivity*area*change in temperature/change in length

let temperature at fat-iiner fur boundary be T degree celcius.

surface area of a sphere=4*pi*radius^2

then 50.7=0.2*4*pi*(1.6/2)^2*(30.8-T)/0.042

==>50.7=38.298*(30.8-T)

==>30.8-T=1.3238

==>T=29.476 degree celcius

hence temperature at fat-inner fur boundary is 29.476 degree celcius

part B:

temperature at inner surface of air layer is 29.476 degree celcius

temperature at outer surface is 2.8 degree celcius

let thick ness be x m

thermal conductivity of air =0.025 W/(m.K

then 50.7=0.025*4*pi*(1.6/2)^2*(29.476-2.8)/x

==>x=0.025*4*pi*(1.6/2)^2*(29.476-2.8)/50.7=0.10579 m

==>thickness has to be 10.579 cm so that it will lose heat at a rate of 50.7 W

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