Answer: amps
Explanation:
A) The temperature at the fat-inner fur boundary be so that the bear loses heat at a rate of 51.4W is; T_i = 38.52°C
B) The thickness of the layer contained within the fur so that the bear loses heat at a rate of 51.4 W is; t = 13.41 cm
We are given;
Diameter of sphere; d = 1.6 m
Radius of sphere; r = d/2
r = 1.6/2
r = 0.8 m
Thickness of bear; t = 3.9 cm cm = 0.039 m
Outer surface Temperature of fur; T_h = 2.8 ∘C
Inner surface Temperature of fat;T_f = 30.9 ∘C
Thermal conductivity of fat; K_f = 0.2 W/m⋅k
Thermal conductivity of air; K_a = 0.024 W/m⋅k
A) To find the temperature at the fat-inner fur boundary when heat loss is 51.4 W, we will use the heat current formula;
H = K_f•A(T_f - T_i)/t
Where;
A is area = 4πr²
A = 4π × 0.8²
A = 8.04 m²
T_i is the temperature we are looking for
H is heat loss = 51.4
t is thickness
Making T_i the subject gives;
T_i = (T_f × H × t)/(K_f × A)
T_i = (30.9 × 51.4 × 0.039)/(0.2 × 8.04)
T_i = 38.52°C
B) We want to find the thickness of the layer contained within the fur. Thus, we will use K_a instead of K_f. Let us make t the subject in the heat current formula to get;
t = (K_a•A(T_i - T_h)/H
t = (0.024 × 8.04 × (38.52 - 2.8))/51.4
t = 0.1341 m
t = 13.41 cm
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Answer:
Explanation:
Using the equation
H = Q/t = k A ( T hot - T cold) / L
where H is the rate of heat loss = 51.4 W, T cold be temperature of the outer surface, A is the surface area of the fat layer which is a model of sphere ( surface area of a sphere ) = 4πr² where diameter = 1.60 m
radius = 1.60 m / 2 = 0.80 m
A = 4 × 3.142 × ( 0.8²) = 8.04352 m²
making T cold subject of the formula
T cold = T hot - = 30.9° C - ( 51.4 W × 3.9 × 10⁻² m) / ( 0.2 W/mK × 8.04352 m² ) = 30.9° C - 1.25 ° C = 29.65° C
b) The thickness of air layer for the bear to lose heat t a rate of 51.4 W
thermal conductivity of air is 0.024 W/mK and rearranging the earlier formula
L = = (0.024 W/ m K × 8.04352 m²) ( 29.65° C - 2.8°C) / 51.4 W = 0.101 m = 10.1 m
Answer:
from the color of the light
Explanation: