Small birds like that in (Figure 1) can migrate over long distances without feed

Question

Small birds like that in (Figure 1) can migrate over long distances without feeding, storing energy mostly as fat rather than carbohydrate. Fat is a good form of energy storage because it provides the most energy per unit mass: 1.00 grams of fat provides about 9.40 (food) Calories, compared to 4.20 (food) Calories per 1.00 grams of carbohydrate. Remember that Calories associated with food, which are always capitalized, are not exactly the same as calories used in physics or chemistry, even though they have the same name. More specifically, one food Calorie is equal to 1000 calories of mechanical work or 4186 joules. Therefore, in this problem use the conversion factor 1Cal=4186J.

A)Consider a bird that flies at an average speed of 10.7 m/s and releases energy from its body fat reserves at an average rate of 3.70 W (this rate represents the power consumption of the bird). Assume that the bird consumes 4.00g of fat to fly over a distance db without stopping for feeding. How far will the bird fly before feeding again?

B)How many grams of carbohydrate mcarb would the bird have to consume to travel the same distance db?

C) Field observations suggest that a migrating ruby-throated hummingbird can fly across the Gulf of Mexico on a nonstop flight traveling a distance of about 800 . Assuming that the bird has an average speed of 40.0 and an average power consumption of 1.70 , how many grams of fat does a ruby-throated hummingbird need to accomplish the nonstop flight across the Gulf of Mexico?

Explanation / Answer

here,

v = 10.7 m/s = 38.52 km/hr
p = 3.70 W

A)

4g of fat = 37.6 Cal = 157393.6 J

Power = work done /time
time = work done /power
t = 157393.6 / 3.70
t = 42538.81 /3600
t = 11.816 hr

D = v * t
D = 38.52 * 11.816
D = 455.152 km

B)

as 1g of fat = 9.4 Cal as compared to 4.2 of gram of carb,
9.4 / 4.2 = 2

so, if the bird consume 4g of fat during the flight :
It would consume 2.2381 * 4 = 8.952 g

C)

d = 800 km
v = 40 km/hr
p = 1.70 watts

t = d/v = 800/40 = 20 hr = 72000 s

Power = Energy /time
Energy = 1.70 * 72000
E = 122400 J

as 1 gram of fat = 9.4 * 4186 = 39348 J

Bird requirement = 122400 / 39348
               = 3.1107 g of fat

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.