Animals in cold climates often depend on two layers of insulation: a layer of body fat [of thermal conductivity 0.200 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 1.40 m in diameter having a layer of fat 4.00 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 2.90 ∘C during hibernation, while the inner surface of the fat layer is at 30.9 ∘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.
What should the temperature at the fat-inner fur boundary be so that the bear loses heat at a rate of 50.8?
How thick should the air layer (contained within the fur) be so that the bear loses heat at a rate of 50.8?

Respuesta :

Answer: a) 29.25 ∘C b) 7.66 cm

Explanation:

a) We need to find the temperature at the fat-inner fur boundary so bear loses heat at rate of 50.8

Rate of heat is related to temperature as:

Rate= Thermal conductivity of fat * Area * Change in temperature / Fat layer length

So,

50.8= 0.2 * [tex]4\pi *(\frac{1.4}{2} )^{2}[/tex] * ( 30.9 + 273 - T ) / 0.040

T= 302.25 K

T= 29.25 ∘C

b)

We have

Thermal conductivity of air= 0.024 W/(m⋅K)

Rate= Thermal conductivity of air * Area * Change in temperature / Air layer length

50.8= 0.024  * [tex]4\pi *(\frac{1.4}{2} )^{2}[/tex] * ( 29.25- 2.9 ) / L

L= 0.076 m

L= 7.66 cm