Energy for our activities is provided by the chemical energy of the foods we eat
ID: 1460129 • Letter: E
Question
Energy for our activities is provided by the chemical energy of the foods we eat. The absolute value of the rate of conversion of this chemical energy into other forms of energy (E/t) is called the metabolic rate. The metabolic rate depends on many factors - a person's weight, physical activity, the efficiency of bodily processes, and the fat-muscle ratio. The table lists the metabolic rates of people under several different conditions and in several different units of measure: 1kcal=1000calories=4186J. Dieticians call a kcal simply a Cal. A piece of bread provides about 70 kcal of metabolic energy.
In 1 hour of heavy exercise a 68-kg person metabolizes 600kca90kcal=510kcal more energy than when at rest. Typically, reducing kilocalorie intake by 3500 kcal(either by burning it in exercise or not consuming it in the first place) results in a loss of 0.45 kg of body mass (the mass is lost through exhaling carbon dioxide - the product of metabolism).
Part A
A 68-kg person wishes to lose 3.6 kg in 4 months. Estimate the time that this person should spend in moderate exercise each day to achieve this goal (without altering her food consumption).
Express your answer to two significant figures.
Table. Energy usage rate during various activities. Type of activity E/t (watts) E/t (kcal/h) E/t (kcal/day) 45-kg person at rest 80 70 1600 68-kg person at rest 100 90 2100 90-kg person at rest 120 110 2600 68-kg person walking 3 mph 280 240 5800 68-kg person moderate exercise 470 400 10,000 68-kg person heavy exercise 700 600 14,000Explanation / Answer
reducing 3500 Kcal cause in loss of 0.45 kg weight.
amount of kcal burn to reduce 3.6 kg for 68 kg weighted person : 3.6 *3500/0.45 = 280000 Kcal
so total hour need to burn 31500 Kcal for 68 kg weighted person by moderate exercise : 280000/400
= 700 hour
so in four month he have to work daily by ; 700/120
=5.833 hours/day
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