Answer:
What is the rate of heat conduction through the 3.00-cm-thick fur of a large animal having a 1.40 m2 surface area? Assume that the animal’s skin temperature is 32.0°C, that the air temperature is −5.00∘C, and that fur has the same thermal conductivity as air. (b) What food intake will the animal need in one day to replace this heat transfer?
a) = 39.7kW
b) = 819kcal
Explanation:
Given that
d = 3.00cm = 3.0 × 10⁻²m
k(fur)=k(air) = 0.023 w/m °C
A = 1.4m²
T₁(animal) = 32 °C
T₂(air) = -5 °C
Δt = 1day =86400s
The rate of conduction heat transfer through a material
= [tex]\frac{kA(T_1 - T_2)}{d}[/tex]
= [tex]\frac{0.023 * 1.40 * (32 (-5))}{3.0 * 10^-^2}[/tex]
= 39.7W
b) P = Q / Δt
Q = P × Δt
= 39.7 × 86400
Q = 3.43 × 10⁶J
1kcal = 4186J
Q(intake) = 3.43 × 10⁶J × (1kcal / 4186J)
= 819kcal
