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Chapter 15: Problem 48

A 52 -kg mountain climber, starting from rest, climbs a vertical distance of 730 \(\mathrm{m}\) . At the top, she is again at rest. In the process, her body generates \(4.1 \times 10^{6} \mathrm{J}\) of energy via metabolic processes. In fact, her body acts like a heat engine, the efficiency of which is given by Equation 15.11 as \(e=|W| / Q_{\mathrm{H}} |,\) where \(|W|\) is the magnitude of the work she does and \(\left|Q_{\mathrm{H}}\right|\) is the magnitude of the input heat. Find her efficiency as a heat engine.

Short Answer

Expert verified
The climber's efficiency as a heat engine is approximately 9.07%.

Step by step solution

01

Determine the Work Done

The work done by the mountain climber can be calculated by using the formula for gravitational potential energy, which is \[ W = mgh \]where:- \( m \) is the mass of the climber (52 kg),- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)), and- \( h \) is the height climbed (730 m).So,\[ W = 52 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 730 \, \text{m} = 371,672 \, \text{J} \]
02

Find the Input Energy

The input energy, \( Q_H \), generated by the climber during the metabolic process is given as \[ Q_H = 4.1 \times 10^6 \, \text{J} \].
03

Calculate Efficiency

Efficiency \( e \) is calculated using the formula \[ e = \frac{|W|}{Q_H} \], where \(|W|\) is the work done and \(Q_H\) is the input energy.Substituting the known values:\[ e = \frac{371,672 \, \text{J}}{4.1 \times 10^6 \, \text{J}} \approx 0.09065 \]. To express this as a percentage, multiply by 100:\[ e \approx 9.07 \% \].

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Key Concepts

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

Work Done
When discussing the term "work done" in physics, it specifically refers to the energy transferred when a force moves an object over a distance. In the case of our mountain climber, the force is her own weight and the distance is the vertical height she climbs. The formula used to calculate work done here is closely related to gravitational potential energy, as both involve height and mass. To determine the work done, we use the formula:
  • \[ W = mgh \]
  • Where \( m \) is the mass (in kilograms), \( g \) is the gravitational acceleration (approximately \( 9.8 \, \text{m/s}^2 \)), and \( h \) is the height climbed (in meters).
In the mountain climber's scenario, all these factors come together to show how much work her body did in lifting itself by 730 meters vertically, resulting in a calculated work of 371,672 Joules. This shows how energy is expended to overcome gravitational pull when climbing.
Gravitational Potential Energy
Gravitational potential energy (GPE) is a form of energy related to an object's position in a gravitational field, often Earth. When an object like a mountain climber ascends, it gains gravitational potential energy because it is moved against the force of gravity. This energy is stored due to its elevated position and can be calculated using the same formula as the work done: \[ GPE = mgh \]This highlights the fundamental relationship between work done in lifting an object and the gravitational potential energy it gains. For our climber, climbing up a height of 730 meters while weighing 52 kilograms equates to an increase in gravitational potential energy by 371,672 Joules. This mirrors the work done because the entire effort exerted goes into overcoming gravity and elevating her to a higher potential energy state at the top of the climb.
Metabolic Processes
Metabolic processes in the context of our mountain climber refers to the biochemical reactions and pathways that enable her body to convert food into energy. This includes all the intricate processes that happen within her body to both produce and use energy effectively. During the climb, her metabolism generates a total of \(4.1 \times 10^6 \text{ J}\) of energy. Not all of this energy converts directly into work done (lifting herself up), but rather a substantial portion is lost as heat. The climber's body functions similarly to a heat engine, with some energy being used for mechanical work while the rest dissipates as heat. This efficiency is characterized by how well her body converts stored chemical energy into work. The calculation of efficiency in this exercise illustrates this concept, showing that only about 9.07% of the energy produced was used for the climbing work, emphasizing the challenge of converting biological energy into mechanical output in human activity.

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Most popular questions from this chapter

When a .22 -caliber rifle is fired, the expanding gas from the burning gunpowder creates a pressure behind the bullet. This pressure causes the force that pushes the bullet through the barrel. The barrel has a length of 0.61 \(\mathrm{m}\) and an opening whose radius is \(2.8 \times 10^{-3} \mathrm{m} . \mathrm{A}\) bullet (mass \(=2.6 \times 10^{-3} \mathrm{kg}\) ) has a speed of 370 \(\mathrm{m} / \mathrm{s}\) after passing through this barrel. Ignore friction and determine the average pressure of the expanding gas.

Heat \(Q\) flows spontaneously from a reservoir at 394 \(\mathrm{K}\) into a reservoir at 298 \(\mathrm{K}\) . Because of the spontaneous flow, 2800 \(\mathrm{J}\) of energy is rendered unavailable for work when a Carnot engine operates between the reservoir at 298 \(\mathrm{K}\) and a reservoir at 248 \(\mathrm{K}\) . Find \(Q\) .

Heat is added to two identical samples of a monatomic ideal gas. In the first sample the heat is added while the volume of the gas is kept constant, and the heat causes the temperature to rise by 75 \(\mathrm{K}\) . In the second sample, an identical amount of heat is added while the pressure (but not the volume) of the gas is kept constant. By how much does the temperature of this sample increase?

A system undergoes a two-step process. In the first step, the internal energy of the system increases by 228 \(\mathrm{J}\) when 166 \(\mathrm{J}\) of work is done on the system. In the second step, the internal energy of the system increases by 115 \(\mathrm{J}\) when 177 \(\mathrm{J}\) of work is done on the system. For the overall process, find the heat. What type of process is the overall process? Explain.

Three moles of an ideal monatomic gas are at a temperature of 345 \(\mathrm{K}\) . Then, 2438 \(\mathrm{J}\) of heat is added to the gas, and 962 \(\mathrm{J}\) of work is done on it. What is the final temperature of the gas?
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