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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

An excess of zinc metal is added to \(50.0 \mathrm{~mL}\) of a \(0.100 \mathrm{M} \mathrm{AgNO}_{3}\) solution in a constant-pressure calorimeter like the one pictured in Figure 5.8 . As a result of the reaction \(\mathrm{Zn}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+2 \mathrm{Ag}(s)\) the temperature rises from \(19.25^{\circ} \mathrm{C}\) to \(22.17^{\circ} \mathrm{C}\). If the heat capacity of the calorimeter is \(98.6 \mathrm{~J} /{ }^{\circ} \mathrm{C},\) calculate the enthalpy change for the given reaction on a molar basis. Assume that the density and specific heat of the solution are the same as those for water, and ignore the specific heats of the metals.

Define calorimetry and describe two commonly used calorimeters. In a calorimetric measurement, why is it important that we know the heat capacity of the calorimeter? How is this value determined?

Consider the reaction: \(2 \mathrm{Na}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g)\) When 2 moles of Na react with water at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), the volume of \(\mathrm{H}_{2}\) formed is \(24.5 \mathrm{~L}\). Calculate the work done in joules when \(0.34 \mathrm{~g}\) of Na reacts with water under the same conditions. (The conversion factor is \(1 \mathrm{~L} \cdot \mathrm{atm}=101.3 \mathrm{~J} .)\)

Calculate the standard enthalpy change for the reaction $$ 2 \mathrm{Al}(s)+\mathrm{Fe}_{2} \mathrm{O}_{3}(s) \longrightarrow 2 \mathrm{Fe}(s)+\mathrm{Al}_{2} \mathrm{O}_{3}(s) $$ given that $$\begin{aligned} 2 \mathrm{Al}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(s) & \\ \Delta H_{\mathrm{rxn}}^{\circ}=&-1669.8 \mathrm{~kJ} / \mathrm{mol} \\ 2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s) & \\ \Delta H_{\mathrm{rxn}}^{\circ}=-822.2 \mathrm{~kJ} / \mathrm{mol} \end{aligned}$$

A quantity of \(50.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) is mixed with \(50.0 \mathrm{~mL}\) of \(0.400 \mathrm{M} \mathrm{HNO}_{3}\) in a constant-pressure calorimeter having a heat capacity of \(496 \mathrm{~J} /{ }^{\circ} \mathrm{C}\). The initial temperature of both solutions is the same at \(22.4^{\circ} \mathrm{C}\). What is the final temperature of the mixed solution? Assume that the specific heat of the solutions is the same as that of water and the molar heat of neutralization is \(-56.2 \mathrm{~kJ} / \mathrm{mol}\).

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