An athlete exercising at a moderate rate for 30 minutes performs about 920 kJ of
ID: 201287 • Letter: A
Question
An athlete exercising at a moderate rate for 30 minutes performs about 920 kJ of work. During the first hour of exercise, fats provide about 50% of the energy consumed during work. During the second and third hours of exercise, fats provide 70% and 80% of the energy consumed, respectively. Suppose an athlete exercises for 3 hours at the rate of work just stated. How many grams of fat will be metabolized to provide this work?
[HINT: rate of work= work/time, and that energy provided from fat per unit mass= 34 kJ/g]
Explanation / Answer
Ans. Step 1: 1st Hour:
Given- 920 kJ work done in 30 minutes (= 0.5 hr).
Rate of work = 920 kJ / 0.5 hr = 1840 kJ/hr
# Total work done in 1st hour = 1840 kJ
# Energy provided from fat = 50% of 1840 kJ = 920 kJ
# Amount of fat metabolized = Energy provided by fat / Calorific value of fat
= 920 kJ / (34 kJ/ g)
= 27.0588 g
Step 2: 2nd Hour:
# Total work done in 2nd hour = 1840 kJ (note that work rate is kept constant)
# Energy provided from fat = 70% of 1840 kJ = 1288 kJ
# Amount of fat metabolized = Energy provided by fat / Calorific value of fat
= 1288 kJ / (34 kJ/ g)
= 37.8824 g
Step 3: 3rd Hour:
# Total work done in 3rd hour = 1840 kJ (note that work rate is kept constant)
# Energy provided from fat = 80% of 1840 kJ = 1472 kJ
# Amount of fat metabolized = Energy provided by fat / Calorific value of fat
= 1472 kJ / (34 kJ/ g)
= 43.2942 g
# Total fat metabolized in 3 hours of work = Fat metabolized in (1st + 2nd hr + 3rd hr)
= 27.0588 g + 37.8824 g + 43.2942 g
= 108.2354 g
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