Assume a goose has a 2.2-cm-thick layer of feather down (on average) and a body surface area of 0.16m2 .What is the rate of heat loss (conduction only) if the goose, with a body temperature of 41 ∘C, is outside on a winter day when the air temperature is 12∘C?

To calculate the rate of heat loss (conduction only) of a goose, we can use the formula:

Q = k * A * ΔT / d

Where:
Q = Heat loss (in watts)
k = Thermal conductivity of air (estimated to be 0.025 W/m·K)
A = Surface area of the goose's body (0.16 m^2)
ΔT = Temperature difference between the goose's body temperature and air temperature (41 - 12 = 29 °C)
d = Thickness of the feather down layer (2.2 cm = 0.022 m)

Plugging in the values, we can calculate the rate of heat loss:

Q = 0.025 * 0.16 * 29 / 0.022

Q = 0.04 * 29 / 0.022

Q = 1.16 / 0.022

Q ≈ 52.73 watts

Therefore, the rate of heat loss (conduction only) for the goose is approximately 52.73 watts.

To determine the rate of heat loss (conduction only) for the goose, we can use the formula:

Q = k * A * ΔT / d

Where:
Q is the rate of heat loss
k is the thermal conductivity of the material (feather down)
A is the surface area of the goose's body
ΔT is the temperature difference between the goose's body and the air
d is the thickness of the feather down layer

First, we need to convert the thickness of the feather down layer from centimeters to meters:

d = 2.2 cm = 0.022 m

Next, we can calculate the temperature difference between the goose's body and the air:

ΔT = 41°C - 12°C = 29°C

Now, we need to find the thermal conductivity of feather down. Unfortunately, the thermal conductivity of feather down is not readily available. However, we can make an estimation using the thermal conductivity of air (0.026 W/m·K) as a rough approximation, since feather down is a good insulator like air.

Substituting the values into the formula, we have:

Q = (0.026 W/m·K) * (0.16 m2) * (29°C) / (0.022 m)

Simplifying the equation:

Q ≈ 0.233 W

Therefore, the rate of heat loss (conduction only) for the goose is approximately 0.233 Watts.