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\(\bullet\) Burning fat by exercise. Each pound of fat contains 3500 food calories. When the body metabolizes food, 80\(\%\) of this energy goes to heat. Suppose you decide to run without stopping, an activity that produces 1290 \(\mathrm{W}\) of metabolic power for a typical person. (a) For how many hours must you run to burn up 1 lb of fat? Is this a realistic exercise plan? (b) If you followed your planned exercise program, how much heat would your body produce when you burn up a pound of fat (c) If you needed to get rid of all of this excess heat by evaporating water (i.e., sweating), how many liters would you need to evaporate? The heat of vaporization of water at body temperature is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg}\)

Step by step solution

01

Calculate Total Energy of 1 Pound of Fat

First, determine the total energy in one pound of fat. Since one pound of fat contains 3500 calories and each calorie is equivalent to 4184 joules, the total energy is calculated as follows: \[ \text{Total energy} = 3500 \times 4184 \text{ J} \] which equals 14,644,000 J.

02

Calculate Effective Energy Used for Running

Determine the fraction of energy used for effective work. Given that only 20% of the metabolized energy is used effectively for work (as 80% goes to heat), the useful energy for burning fat is: \[ \text{Useful energy} = 0.2 \times 14,644,000 \text{ J} \] which equals 2,928,800 J.

03

Determine Duration of Running

Use the metabolic power output to find the duration required. The metabolic power during running is 1290 W, where 1 W equals 1 J/s. The time required to burn 1 lb of fat is calculated as: \[ \text{Time} = \frac{\text{Useful energy}}{\text{Power}} = \frac{2,928,800}{1290} \approx 2269 \text{ seconds} \] Convert seconds to hours: \[ 2269 \text{ s} = \frac{2269}{3600} \approx 0.63 \text{ hours} \] Simply put, about 0.63 hours, or approximately 38 minutes, of running is needed.

04

Calculate Heat Produced

Since 80% of the metabolized energy goes to heat, calculate the heat produced when burning 1 lb of fat: \[ \text{Heat} = 0.8 \times 14,644,000 \text{ J} = 11, 715,200 \text{ J} \] The body produces this much heat while metabolizing 1 lb of fat.

05

Calculate Water Evaporation Needed

Using the heat of vaporization to find the water needed to vaporize all the generated heat. The total heat to be evaporated is 11,715,200 J, and the heat of vaporization is given as \(2.42 \times 10^6 \text{ J/kg} \). The mass of water is calculated by: \[ \text{Mass of water} = \frac{11,715,200}{2.42 \times 10^6} \approx 4.84 \text{ kg} \] Since the density of water is 1 kg/L, this means 4.84 liters are needed.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Energy Conversion

Metabolism involves transforming the energy stored in food into forms that our body can use. This is known as energy conversion. When you exercise, your body taps into energy reserves, primarily burning carbohydrates and fats.
Out of the total energy metabolized from food, our bodies typically convert only a fraction into muscular work—the rest becomes heat, helping maintain body temperature. This illustrates the principle of energy conservation, where energy neither created nor destroyed but merely changes form.

• When fat is burned, each pound equates to 3500 calories or 14,644,000 joules.
• Only 20% of this energy is harnessed for physical activity, meaning 2,928,800 joules go into energy for running.
• The majority—80%—becomes heat, used for body warmth.

Understanding these conversions enhances our grasp of how our bodies sustain exercise, using both energy and producing heat.

Heat of Vaporization

The heat of vaporization is the energy required to turn a liquid into a gas. For water, this value is quite high and is essential in regulating body temperature through sweating.
Essentially, when you exercise, your body heats up. To cool down, sweat evaporates from the skin surface, carrying heat away.

• At body temperature, the heat of vaporization of water is approximately \(2.42 \times 10^6 \mathrm{J/kg}\).
• This high energy requirement means a significant amount of heat is removed as sweat evaporates, aiding in cooling.
• To remove the heat generated by burning 1 pound of fat (11,715,200 J), about 4.84 kg or liters of water must evaporate.

Recognizing the heat of vaporization explains why hydration is crucial during exercise, as adequate sweating allows efficient heat dissipation.

Caloric Burn

Caloric burn measures energy expended during physical activities. It accounts for effective energy use, primarily from fat and carbohydrate metabolism.
Running, for instance, is an intensive form of exercise promoting significant caloric burn. In our example, with a power output of 1290 watts, running for about 0.63 hours burns up the energy equivalent of 1 pound of fat.

• This translates to about 38 minutes of sustained running for an average person.
• Understanding caloric burn helps individuals plan exercise routines to achieve weight goals.
• When calculating, remember not all burned calories are visible as energy; some convert to heat!

Accurately estimating caloric burn can leverage exercise's benefits for weight management and cardiovascular health.

Exercise Physiology

Exercise physiology explores how exercise impacts bodily functions, focusing on energy use, metabolism, and fatigue. It delves into how the body optimizes its energy sources during activities like running.
In this context, energy is primarily sourced from metabolizing stored fats and carbohydrates. This is crucial in designing successful and sustainable workout routines.

• Understanding metabolic power, where 1290 watts signify substantial energy output, helps in assessing performance capabilities.
• Concepts such as metabolic rate and energy balance are explored, influencing caloric burn and weight management.
• Moreover, the physiological response to exercise—like sweating to manage heat—is part of this field.

Through exercise physiology, individuals can tailor fitness plans that enhance endurance and muscle efficiency, based on their food energy conversions and caloric needs.