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A 200-lh man decides to add to his exercise routine hy walking up three flights of stairs (45 ft) 20 times per day. He figures that the work required to increase his potential energy in this way will permit him to eat an extra order of French fries, at 245 Cal, without adding to his weight. Is he correct in this assumption?

Short Answer

Expert verified
The man's assumption is incorrect. By walking up three flights of stairs 20 times per day, he burns approximately 58.3 Calories, which is less than the 245 Calories in an extra order of French fries. Therefore, this additional exercise routine will not permit him to eat an extra order of French fries without adding to his weight.

Step by step solution

01

Calculate Work Done

We must first calculate the work done by the man when he walks up the stairs. The formula for work is: Work = force x distance In this case, force is equal to man's weight (since force acting on a body = mass x acceleration due to gravity), and distance is equal to the height of the stairs. Let's convert the weight into appropriate units. Here, 1 lb = 0.4536 kg.
02

Convert weight into kilograms

Convert 200 lb to kg: \( 200 lb \times 0.4536 = 90.72 kg \)
03

Calculate force

Calculate the force exerted by the man: Force = mass x gravitational acceleration Force = 90.72 kg x 9.8 m/s² Force ≈ 889.06 N
04

Convert height into meters

We must also convert the height of the stairs (45 ft) into meters. 1 ft = 0.3048 meters Height = 45 ft x 0.3048 = 13.716 meters
05

Calculate work done for one flight of stairs

Calculate the work done by the man for one flight of stairs: Work = force x distance Work = 889.06 N x 13.716 meters Work ≈ 12,196.4 J (Joules)
06

Calculate work done for 20 flights of stairs

Calculate the work done by the man after walking up 20 flights of stairs: Total work = work for one flight of stairs x number of flights Total work = 12,196.4 J x 20 Total work ≈ 243,928 J
07

Convert work done into calories

Convert the total work done into calories. 1 calorie (cal) = 4.184 Joules (J). Total work in calories = total work (J) / 4.184 Total work in calories ≈ 243,928 J / 4.184 Total work in calories ≈ 58,321.7 cal We need to convert this to nutritional Calories (which are actually kilocalories). To do this, divide the value by 1000. Total work in Calories = 58,321.7 cal / 1000 Total work in Calories ≈ 58.3 Calories
08

Compare with French fries Calories

Finally, let's compare the Calories burnt by the man with the Calories in an extra order of French fries (245 Calories). As 58.3 Calories < 245 Calories, the man's assumption is incorrect. The work required to increase his potential energy by walking up three flights of stairs 20 times per day is not enough to compensate for the extra Calories from an order of French fries without adding to his weight.

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Key Concepts

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

Potential Energy
Potential energy is the stored energy in an object due to its position relative to the earth. In the case of the man walking up the stairs, his body gains potential energy because it is being elevated to a higher altitude. The potential energy gained is proportional to both his weight and the height of the stairs.
  • Formula: Potential energy can be expressed as \( PE = mgh \), where \( m \) is mass, \( g \) is the acceleration due to gravity (9.8 m/s²), and \( h \) is height.
  • Application: When the man climbs a set of stairs, his potential energy increases because he is lifting his body weight against gravity.
The essence of this exercise is to determine whether the gained potential energy, converted from calories burned, offsets the calories consumed.
Work and Energy
Work and energy are closely connected concepts in physics, representing the transfer and transformation of energy.
  • Work Definition: Work is the process of using force to move an object over a distance. It can be calculated using the formula \( W = Fd \), where \( F \) is the force exerted and \( d \) is the distance moved in the direction of the force.
  • Energy Transformation: As the man climbs stairs, chemical energy from his body is transformed into mechanical energy (work done) and potential energy.
  • Unit Conversion: Noteworthy is the conversion of Joules (the unit for work and energy in the metric system) to calories, given \( 1 \, \text{calorie} = 4.184 \, \text{Joules} \).
In this real-world example, calculating the work done gives insight into how physical activity translates into energy expenditure, helping understand daily caloric needs.
Caloric Expenditure
Caloric expenditure refers to the amount of energy, measured in calories, that a person uses during physical activity. It is a vital part of managing body weight and ensuring energy balance.
  • Burning Calories: Physical activities, such as climbing stairs, require calories because the muscles need energy to function. These calories come from the body's stores of fats, carbohydrates, and proteins.
  • Exercise vs. Caloric Intake: In this exercise, the man's assumption is that his physical activity (caloric expenditure from climbing stairs) will allow him to consume "extra" calories without weight gain. However, as calculations show, the calories expended during his exercise routine are insufficient to cover an extra caloric intake of 245 Calories from French fries.
Understanding caloric expenditure helps individuals align their dietary choices with their fitness goals, ultimately promoting a healthier lifestyle.

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Most popular questions from this chapter

(a) When a 0.235-g sample of benzoic acid is combusted in a bomb calorimeter (Figure 5.18), the temperature rises \(1.642^{\circ} \mathrm{C}\). When a \(0.265-\mathrm{g}\) sample of caffeine, \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{O}_{2} \mathrm{~N}_{4}\), is burned, the temperature rises \(1.525^{\circ} \mathrm{C}\). Using the value \(26.38 \mathrm{~kJ} / \mathrm{g}\) for the heat of combustion of benzoic acid, calculate the heat of combustion per mole of caffeine at constant volume. (b) Assuming that there is an uncertainty of \(0.002^{\circ} \mathrm{C}\) in each temperature reading and that the masses of samples are measured to \(0.001 \mathrm{~g}\), what is the estimated uncertainty in the value calculated for the heat of combustion per mole of caffeine?

When a \(6.50\)-g sample of solid sodium hydroxide dissolves in \(100.0 \mathrm{~g}\) of water in a coffee-cup calorimeter (Figure 5.17), the temperature rises from \(21.6\) to \(37.8^{\circ} \mathrm{C}\). (a) Calculate the quantity of heat (in \(\mathrm{kJ}\) ) released in the reaction. (b) Using your result from part (a), calculate \(\Delta H\) (in \(\mathrm{kJ} / \mathrm{mol} \mathrm{NaOH}\) ) for the solution process. Assume that the specific heat of the solution is the same as that of pure water.

The heat of combustion of fructose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), is \(-2812 \mathrm{~kJ} / \mathrm{mol}\). If a fresh golden delicious apple weighing \(4.23 \mathrm{oz}(120 \mathrm{~g})\) contains \(16.0 \mathrm{~g}\) of fructose, what caloric content does the fructose contribute to the apple?

The specific heat of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\), is \(2.22 \mathrm{~J} / \mathrm{g}-\mathrm{K}\). (a) How many J of heat are needed to raise the temperature of \(80.0 \mathrm{~g}\) of octane from \(10.0\) to \(25.0^{\circ} \mathrm{C}\) ? (b) Which will require more heat, increasing the temperature of \(1 \mathrm{~mol}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\) by a certain amount or increasing the temperature of \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}(l)\) by the same amount?

The complete combustion of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\), to form \(\mathrm{H}_{2} \mathrm{O}(g)\) and \(\mathrm{CO}_{2}(g)\) at constant pressure releases \(1235 \mathrm{~kJ}\) of heat per mole of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\). (a) Write a balanced thermochemical equation for this reaction. (b) Draw an enthalpy diagram for the reaction.

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