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A woman expends \(95 \mathrm{~kJ}\) of energy walking a kilometer. The energy is supplied by the metabolic breakdown of food, which has an efficiency of 35 percent. How much energy does she save by walking the kilometer instead of driving a car that gets \(8.2 \mathrm{~km}\) per liter of gasoline (approximately \(20 \mathrm{mi} / \mathrm{gal}) ?\) The density of gasoline is \(0.71 \mathrm{~g} / \mathrm{mL},\) and its enthalpy of combustion is \(-49 \mathrm{~kJ} / \mathrm{g}\).

Short Answer

Expert verified
The woman saves 3972.95 kJ by walking instead of driving.

Step by step solution

01

Calculate Energy from Gasoline Used by Car

First, calculate how much gasoline the car uses to drive 1 kilometer. Since the car gets 8.2 km per liter, the car uses \( \frac{1}{8.2} \) liters of gasoline per kilometer. The volume of gasoline needed is \( 0.122 \) liters. Convert this volume to mass using the density of gasoline: \( 0.122 \text{ liters} \times 1000 \text{ mL/liter} \times 0.71 \text{ g/mL} \) to get \( 86.62 \text{ grams} \). Now, compute the energy released from this gasoline: \( 86.62 \text{ g} \times -49 \text{ kJ/g} \), giving \( -4244.38 \text{ kJ} \).
02

Adjust for Metabolic Efficiency

The efficiency of energy conversion in the woman's metabolism is 35%. Therefore, the amount of energy she actually gains from food is \( \frac{95 \text{ kJ}}{0.35} = 271.43 \text{ kJ} \) that needs to be metabolized to expend 95 kJ physically.
03

Calculate Energy Saved by Walking

By walking, the woman uses 271.43 kJ from food energy. Compare this with the energy consumed if driving, which is 4244.38 kJ of energy from gasoline. The energy saved by walking instead of driving is \( 4244.38 \text{ kJ} - 271.43 \text{ kJ} = 3972.95 \text{ kJ} \).

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Key Concepts

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

Metabolic Efficiency
Metabolic efficiency refers to the body's ability to convert food into energy. In the scenario with the woman walking, her body uses food to fuel her activity. However, not all energy from food is usable for work. In fact, only a percentage of food energy is converted into actual physical activity. In this case, it is 35%.
  • Metabolic Rate: This is the rate at which your body burns calories to produce energy.
  • Efficiency: In terms of energy conversion, efficiency is the proportion of energy that is transformed into work, while the rest is lost as heat.
If she needs 95 kJ for walking 1 km, she metabolizes more than what is needed due to inefficiencies. Thus, she metabolizes around 271.43 kJ of food energy, due to the metabolic efficiency being 0.35.
Gasoline Combustion
Gasoline combustion is a chemical reaction between gasoline and oxygen that releases energy in the form of heat. The key aspect to understand here is how much energy gasoline provides when burned. The energy from gasoline helps engines operate by converting chemical energy into mechanical work.
  • Combustion Reaction: Combustion produces energetic compounds primarily used in engines.
  • High Energy Release: One gram of gasoline burns to release approximately 49 kJ of energy.
In the problem, a car uses gasoline to cover a distance. The car's efficiency, like any machine, is determined by how much energy it can convert from fuel into movement.
Energy Conversion
Energy conversion is the process of changing one form of energy to another, such as potential energy to kinetic or chemical energy to mechanical. In both the metabolic process and gasoline combustion, energy is being converted to perform work, be it walking or driving.
  • Human Energy Conversion: Metabolic processes convert food energy to usable physical energy.
  • Machine Energy Conversion: Cars convert chemical energy in gasoline to mechanical energy.
In our problem, both the human body and the car require different amounts of energy for the same task, walking vs. driving one kilometer. The stark difference in the energy required shows the efficiency of human metabolism compared to machines.
Density of Gasoline
Understanding the density of gasoline is crucial for determining the amount of fuel needed for combustion and energy calculations. Density is mass per unit volume, which for gasoline is 0.71 g/mL.
  • Mass and Volume: Density allows conversion from volume to mass, which is needed in energy calculations.
  • Application in Calculations: Helps calculate energy release from burning a specific volume of gasoline.
In the exercise, we convert 0.122 liters of gasoline to grams using its density, then determine the energy released from this mass. This step is essential for comparing the energy use for driving and walking.

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Most popular questions from this chapter

The carbon dioxide exhaled by sailors in a submarine is often removed by reaction with an aqueous lithium hydroxide solution. (a) Write a balanced equation for this process.

You are given the following data: \(\begin{aligned} \mathrm{H}_{2}(g) & \longrightarrow 2 \mathrm{H}(g) & & \Delta H^{\circ}=436.4 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Br}_{2}(g) & \longrightarrow 2 \mathrm{Br}(g) & & \Delta H^{\circ}=192.5 \mathrm{~kJ} / \mathrm{mol} \\\ \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) & \longrightarrow 2 \mathrm{HBr}(g) & \Delta H^{\circ} &=-72.4 \mathrm{~kJ} / \mathrm{mol} \end{aligned}\) Calculate \(\Delta H^{\circ}\) for the reaction\(\mathrm{H}(g)+\mathrm{Br}(g) \longrightarrow \mathrm{HBr}(g)\).

The convention of arbitrarily assigning a zero enthalpy value for the most stable form of each element in the standard state at \(25^{\circ} \mathrm{C}\) is a convenient way of dealing with enthalpies of reactions. Explain why this convention cannot be applied to nuclear reactions.

The work done to compress a gas is \(47 \mathrm{~J}\). As a result, \(93 \mathrm{~J}\) of heat is given off to the surroundings. Calculate the change in internal energy of the gas.

Metabolic activity in the human body releases approximately \(1.0 \times 10^{4} \mathrm{~kJ}\) of heat per day. Assume that a \(55-\mathrm{kg}\) body has the same specific heat as water; how much would the body temperature rise if it were an isolated system? How much water must the body eliminate as perspiration to maintain the normal body temperature \(\left(98.6^{\circ} \mathrm{F}\right)\) ? Comment on your results. (The heat of vaporization of water is \(2.41 \mathrm{~kJ} / \mathrm{g}\).)

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