Question

One advantage to being small is that it is easier to cool off after exertion than if you are large. For example, hummingbirds generate a lot of heat as their wings beat quickly, but they can cool off very quickly in response. (The flip side is, it is harder to stay warm when the air temperature is low.) In this problem we will develop a simple model of the smallest shrew and a polar bear cooling off. We will model both mammals as spheres (Volume = 43?r3 ; surface area = 4?r2 ) with a 0.50 cm barrier of air next to their skin (k= .026 W/(mK)). In both cases we will take the internal temperature to be 40 degrees C, and the outside temperature to be 22 degrees C. The shrew has a radius of 0.0175 m, while the polar bear has a radius of 1.5 m

Part A. What is the rate of heat loss through conduction for the shrew? Give all answers in this question to three sig figs.

Part B. What is the rate of heat loss through conduction for the bear?

Part C. Let's assume that the shrew needs to loose 100J of thermal energy in order to cool off to a safer internal temperature. We want to find out how much thermal energy the bear needs to loose. We will do this by assuming that the amount of thermal energy each has to loose is proportional to their volume: Thermal needed to loose = Constant * Volume of mammal and that the constant is the same for both the shrew and the bear. There are two ways (at least) to solve this. One is to find the value of the constant for shrew and use that value for the bear. The other is to use proportions, knowing that the constant is the same for both mammals.

Part D. How long will it take the shrew to loose this 100 J of heat through conduction? Hint: if the rate of heat loss via conduction were 50 Watts, the shrew would be losing 50 Joules every second (Watt=Joule/second), so it would take 2 seconds to cool off.

Part E. How long would it take the bear to cool off?

Answer 1

For conduction, we use Fourier's Law, which in this case is given by

where is the rate of heat transfer, is the thermal conductivity, is the cross-sectional surface area, is the temperature difference between the ends and is the distance between the ends. In our problem, is the surface area of the animal, is the temperature difference between the outside temperature (22 deg.) and the body temperature (40 deg.). This means = 18 degrees. Since the size of one degree is the same as the size of one Kelvin, this also means = 18 Kelvin. is the thickness of the air barrier (0.5 cm.)

PART A

Known values-
= 0.026 W/mK (given)
= 18 Kelvin. (40-22 deg.)
= 0.5 cm = 0.005m

Calculating area, = = = 0.003848451.

Putting all these values in the right hand side of Fourier's law, we get

(negative means heat is being lost)

So, up to 3 significant figures, the rate of heat loss for the shrew is 0.360 Watt.   

PART B

Using the same values as earlier, we just need to recalculate area as = = = 28.2743.

Putting all these values in the right hand side of Fourier's law, we get   

(negative means heat is being lost)

So, up to 3 significant figures, the rate of heat loss for the bear is 2650 Watt.

PART C

Since the constant is the same for both animals, we can equate the ratios as

We know the heat needed to be lost by the shrew = 100J. So the heat needed to be lost by the bear is given by the above ratio (using volume = ) as

So, up to 3 significant figures, the bear needs to lose 6.30*10^7 Joules.  

PART D

Assuming the rate of heat loss to be constant as 0.360 Watt, then this is equal to the total heat loss divided by the time taken.

Solving this give the time taken = 277.777 seconds.

So, up to 3 significant figures, the shrew will take 278 seconds.  

PART E

Assuming the rate of heat loss to be constant as 2650 Watt, then this is equal to the total heat loss divided by the time taken.

Solving this give the time taken = 23773.5849057 seconds.

So, up to 3 significant figures, the bear will take 23800 seconds.