One advantage to being small is that it is easier to cool offafter exertion than if you are large. For example, hummingbirdsgenerate a lot of heat as their wings beat quickly, but they cancool off very quickly in response. (The flip side is, it is harderto stay warm when the air temperature is low.) In this problem wewill develop a simple model of the smallest shrew and a polar bearcooling 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 theirskin (k= .026 W/(mK)). In both cases we will take the internaltemperature to be 40 degrees C, and the outside temperature to be22 degrees C. The shrew has a radius of 0.0175 m, while the polarbear has a radius of 1.5 m

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

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

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

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

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