Constants Part A What should the temperature at the fat-inner fur boundary be so

Question

Constants Part A What should the temperature at the fat-inner fur boundary be so that the bear loses heat at a rate of 52.0 W? IVO ALD * oo ? 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 1.60 m in diameter having a layer of fat 3.90 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 2.60 C during hibernation, while the inner surface of the fat layer is at 31.2°C TE °C Submit Request Answer Assume the surface area of each layer is the same and given by the surface area of the spherical model constructed for the black bear. Also, assume a steady state of the heat current through the layers. - Part B How thick should the air layer (contained within the fur) be so that the bear loses heat at a rate of 52.0 W? IVO ALO * oo ? Lair = Lair = cm Submit Request Answer

Explanation / Answer

Given,

Conductivity, k = 0.200 W/m K

Thickness of fat layer, t = 3.9 cm

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

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

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

then 52 = 0.2*4*pi*(0.8)^2*(31.2 - T)/0.039

52 = 41.24 * (31.2 - T)

T = 29.94 degree celcius

part B:

temperature at inner surface is 29.94 degree celcius

temperature at outer surface is 2.6 degree celcius

let thickness be x m

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

then 52 = 0.025*4*pi*(0.8)^2*(29.94-2.6)/x

Thickness, x = 0.1057 m = 10.57 cm

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