Animals in cold climates often depend on two layers of insulation: a layer of body fat [of thermal conductivity 0.200 W/mK] surrounded by a layer of air trapped inside fur or down. We can model a black bear (Ursus americanus) as a sphere 1.50 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 degrees Celsius during hibernation, while the inner surface of the fat layer is at 31.0 degrees Celsius. What should the temperature at the fat-inner fur boundary be so that the bear loses heat at a rate of 51.5 W? How thick should the air layer (contained within the fur) be so that the bear loses heat at a rate of 51.5 W?

To find the temperature at the fat-inner fur boundary, we can use the concept of thermal resistance and apply Fourier's Law of heat conduction.

Let's denote the temperature at the fat-inner fur boundary as Tf, and the thermal conductivity of fat as k_fat (given as 0.200 W/mK). The thermal conductivity of air is much lower and can be considered negligible for this calculation.

The rate of heat transfer (Q) across a material can be calculated using the following formula:

Q = Delta T / R

Where:
- Delta T is the temperature difference across the material
- R is the thermal resistance

In this case, the two resistances we need to consider are the resistance of the fat layer (R_fat) and the resistance of the air layer (R_air).

The thermal resistance (R) of a material is given by its thickness (L) divided by its thermal conductivity (k):

R = L / k

For the fat layer:
R_fat = (L_fat) / k_fat

For the air layer:
R_air = (L_air) / k_air

Since k_air is very small, we can ignore it in the calculation.

Now we can write an equation for the overall thermal resistance of the system:

R_total = R_fat + R_air

Finally, we can find the temperature difference across the fat layer (Delta T_fat) using the given rate of heat transfer:

Delta T_fat = Q * R_total

To find the temperature at the fat-inner fur boundary (Tf), we need to add this temperature difference to the inner surface temperature of the fat layer:

Tf = 31.0°C + Delta T_fat

Now we can proceed with the calculation of the temperature and the thickness of the air layer.

We are given the diameter of the bear (1.50 m), and we can assume it as a sphere. The thickness of the fat layer is given as 3.90 cm, but we need to convert it to meters for consistency:

L_fat = 3.90 cm = 0.039 m

We are also given the rate of heat transfer (Q) as 51.5 W.

Now we can calculate the thermal resistance of the fat layer:

R_fat = (L_fat) / k_fat
= 0.039 m / 0.200 W/mK

Next, we ignore the thermal resistance of the air layer, assuming it's negligible compared to the fat layer.

Therefore, the total thermal resistance:

R_total = R_fat + R_air
= R_fat (since R_air is considered negligible)

Now we can calculate the temperature difference across the fat layer:

Delta T_fat = Q * R_total

Finally, we can find the temperature at the fat-inner fur boundary:

Tf = 31.0°C + Delta T_fat

To calculate the thickness of the air layer, we subtract the thickness of the fat layer from the overall diameter of the bear:

L_air = (Diameter of bear - 2 * L_fat) / 2

Substitute the given values into the equations and calculate to find the respective answers.

Calculate the surface area of the fat-fur boundry, then the surface area of the outer fur boundry.

Outer layer

51.5watts=(31-Tinner)areainner/.0391
solve for Tinner.
Then, on the fur layer...
I don't see the thermal conductivity for the fur layer...

I wasn't given the thermal conductivity of the fur layer, all the info the problem gives is included above...

I believe you have to use 0.200 W/mK = R = (L fat) / k; solve for k and now you should have a new value for R which is (L fat + L fur) / k and assume that the fur you add simply increases the length of the fat insulating layer.

I'll be honest though, this process is not working for me.

It may be that you must also change the Area variable to contain both the Radius to the fat layer and the thickness of the hair

You use the thermal conductivity for air, which is 0.024 W/(m*K), for part B.