Animals in cold climates often depend on two layers of insulation: a layer of body fat [of thermal conductivity 0.20 W/(m⋅K) ] surrounded by a layer of air trapped inside fur or down. We can model a black bear (Ursus americanus) as a sphere 1.5 m in diameter having a layer of fat 4.0 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 2.7∘C during hibernation, while the inner surface of the fat layer is at 31.0∘C. What should the temperature at the fat-inner fur boundary be so that the bear loses heat at a rate of 50.8 ?

a) 29.6° C

Explanation:

a) Our goal is to find the temperature of the outer surface of the fat layer, and that could be calculated using the following equation  

H=Q/t=kA[T_hot-T_cold]/L                            (1)

Where T_cold is the temperature of the outer surface of the fat layer and we can just call it T_f , and (A) is the surface area of the fat layer, and since we are modelling the bear as a sphere with a radius of (r = 1.5 m/2 = 0.75 m), then the surface area of the spherical fat layer is [4 x (0.75 m)^2]. Let's rearrange eq.(1) to solve for (T_f) or T_cold

T_f =T_hot-HL/kA                                          (2)

Now, substitute into eq.(2) with (50.8 W) for the rate of heat lost by the bear which is (H), (4 x 10-2 m) for the thickness (L), (0.2 W/m • K) and (31° C) for T_hot

T_f = (31° C)-(50.8 W)*(4*10^-2 m)/(0.2 W/m K)*(4*(0.75 m)^2)

T_f = 29.6° C

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