Answer:
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[tex]\pi[/tex] 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[tex]\pi[/tex]*(0.75 m)^2)
T_f = 29.6° C