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 =  43πr3 ; surface area =  4πr2 ) 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

A. What is the rate of heat loss through conduction for the shrew?

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

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.

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.

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

Answer 1