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A 200-lb man decides to add to his exercise routine by walking up three flights of stairs ( \(45 \mathrm{ft}\) ) 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 weighs 90.7184 kg and climbs a total height of 45 ft (13.716 m per flight). The total work done is found by calculating the potential energy: \(potential\_energy = (90.7184\;kg) \times (9.81\;m/s^2) \times (13.716\;m)\), then multiplying by 3 flights and 20 times per day. Finally, convert the work done to Calories and compare it to the 245 Cal of a serving of French fries. If the total Calories expended is greater than or equal to 245 Cal, the man's assumption is correct.

Step by step solution

01

Convert the man's weight to mass

First, we need to convert the man's weight (in pounds) to mass (in kilograms) using the conversion factor 1 lb = 0.453592 kg. \(mass = \frac{200\;lb}{1} \times \frac{0.453592\;kg}{1\;lb} = 90.7184\;kg\) The man's mass is approximately 90.7184 kg.
02

Calculate the potential energy per flight

To find the work done in climbing one flight of stairs, we will find the change in potential energy. The gravitational potential energy is given by the equation: \(potential\_energy = mgh\) where m is the mass, g is the acceleration due to gravity (approximately 9.81 m/sĀ²), and h is the height climbed. \(potential\_energy = (90.7184\;kg) \times (9.81\;m/s^2) \times (45 \times 0.3048\;m)\) The height traveled is multiplied by the conversion factor 1 ft = 0.3048 m.
03

Calculate the work done for climbing three flights of stairs 20 times

To find the total work done, we will multiply the potential energy per flight by the number of flights and the number of times the man climbs them each day. \(total\_work = potential\_energy \times 3 \times 20\)
04

Convert the work done to Calories

Now, we need to convert the total work done in Joules to Calories. The conversion factor is 1 Cal = 4184 J. \(total\_energy \; (in \; Calories) = \frac{total\_work}{4184}\)
05

Compare the total energy to the energy content of the French fries

Finally, we will compare the total energy expended by the man in climbing the stairs to the energy content of an order of French fries (which is given as 245 Calories in the problem statement). If the total energy expended is equal to or greater than the energy content of the French fries, the man's assumption is correct. In conclusion, calculate the total work done while climbing the stairs and compare it to the energy content of the French fries using the steps provided. This will help us determine if the man can consume an extra 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.

Work-Energy Principle
The work-energy principle is a critical concept in understanding how forces acting on an object result in a change in its energy. In the context of the exercise, as the man walks up the stairs, he does work against the force of gravity. The work done by the man in climbing the stairs is equal to the increase in his gravitational potential energy.

This relationship is mathematically expressed as: \( Work = \triangle Kinetic\text{ }Energy + \triangle Potential\text{ }Energy \) where \( \triangle \) stands for 'change in'. Since the man starts and ends at rest, his kinetic energy change is zero, so all the work he does is converted into gravitational potential energy.

It's important to note that work, and hence energy, are scalar quantities, meaning they only have magnitude and not a direction. This principle allows us to calculate how much energy a person needs to exert to perform a certain task, such as climbing stairs, which ties in perfectly to our exercise scenario.
Caloric Energy Conversion
When we speak of 'calories,' we often refer to the energy content of food. One Calorie, with a capital 'C', as used in food energy content, is actually equivalent to 1 kilocalorie (kcal) or 1000 calories (with a lowercase 'c'), which is a unit of energy in physics.

To understand how physical work relates to food consumption, we use the conversion factor between mechanical energy (in Joules) and caloric energy (in Calories). As the solution illustrates, \( 1 \text{ Cal} = 4184 \text{ J} \). This is vital in everyday life, particularly when balancing energy intake from food with energy expenditure through physical activity, such as climbing the stairs in our exercise.

Using this conversion, we can determine if the calories burned through exercise, in this case climbing stairs, are sufficient to offset the caloric intake from an extra order of French fries, thus allowing an individual to maintain or manage their weight effectively.
Mass to Weight Conversion
Weight is the force that gravity exerts on an object with mass, and it varies depending on the gravitational field strength of the planet or celestial body you're on. In contrast, mass is a measure of how much matter is in an object and remains constant regardless of location.

In the exercise, the man's weight in pounds must be converted to mass in kilograms before we can calculate the potential energy. The conversion factor is \( 1 \text{ lb} = 0.453592 \text{ kg} \) as shown in the solution. This step is critical because the equations for potential energy and work require the mass of the object, not its weight. Understanding the difference between mass and weight is crucial in physics and allows for accurate calculations, whether you're assessing the potential energy on Earth or any other environment with a different gravitational field.

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Most popular questions from this chapter

Consider the combustion of liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH}(I)\) : \(\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) \(\Delta H=-726.5 \mathrm{~kJ}\) (a) What is the enthalpy change for the reverse reaction? (b) Balance the forward reaction with whole-number coefficients. What is \(\Delta H\) for the reaction represented by this equation? (c) Which is more likely to be thermodynamically favored, the forward reaction or the reverse reaction? (d) If the reaction were written to produce \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) instead of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\), would you expect the magni- tude of \(\Delta H\) to increase, decrease, or stay the same? Explain.

Using values from Appendix \(C\), calculate the value of \(\Delta H^{\circ}\) for each of the following reactions: (a) \(4 \mathrm{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Br}_{2}(l)\) (b) \(2 \mathrm{Na}(\mathrm{OH})(s)+\mathrm{SO}_{3}(g) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{CH}_{4}(g)+4 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(l)+4 \mathrm{HCl}(g)\) (d) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+6 \mathrm{HCl}(g) \longrightarrow 2 \mathrm{FeCl}_{3}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)\)

The specific heat of iron metal is \(0.450 \mathrm{~J} / \mathrm{g}-\mathrm{K}\). How many J of heat are necessary to raise the temperature of a 1.05-kg block of iron from \(25.0{ }^{\circ} \mathrm{C}\) to \(88.5^{\circ} \mathrm{C}\) ?

Ammonia \(\left(\mathrm{NH}_{3}\right)\) boils at \(-33^{\circ} \mathrm{C} ;\) at this temperature it has a density of \(0.81 \mathrm{~g} / \mathrm{cm}^{3}\). The enthalpy of formation of \(\mathrm{NH}_{3}(g)\) is \(-46.2 \mathrm{~kJ} / \mathrm{mol}\), and the enthalpy of vaporization of \(\mathrm{NH}_{3}(l)\) is \(23.2 \mathrm{~kJ} / \mathrm{mol}\). Calculate the enthalpy change when \(1 \mathrm{~L}\) of liquid \(\mathrm{NH}_{3}\) is burned in air to give \(\mathrm{N}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\). How does this compare with \(\Delta H\) for the complete combustion of \(1 \mathrm{~L}\) of liquid methanol \(\mathrm{CH}_{3} \mathrm{OH}(l) ?\) For \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{l})\), the density at \(25^{\circ} \mathrm{C}\) is \(0.792 \mathrm{~g} / \mathrm{cm}^{3}\), and \(\Delta H_{f}^{\circ}\) equals \(-239 \mathrm{~kJ} / \mathrm{mol}\).

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is used as a fuel in race cars. (a) Write a balanced equation for the combustion of liquid methanol in air. (b) Calculate the standard enthalpy change for the reaction, assuming \(\mathrm{H}_{2} \mathrm{O}(g)\) as a product. (c) Calculate the heat produced by combustion per liter of methanol. Methanol has a density of \(0.791 \mathrm{~g} / \mathrm{mL}\) (d) Calculate the mass of \(\mathrm{CO}_{2}\) produced per \(\mathrm{kJ}\) of heat emitted.

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