/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 A 52-kg mountain climber, starti... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(e=|W| /\left|Q_{\mathrm{H}}\right|,\) 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 9.06%.

Step by step solution

01

Identify Known Values

The mountain climber weighs 52 kg, and she climbs a vertical distance of 730 m. The energy generated through metabolic processes is given as \(4.1 \times 10^6 \; \text{J}\). We need to find the efficiency \( e \) using the formula \( e = \frac{|W|}{|Q_{H}|} \).
02

Calculate Work Done (W)

The work done by the mountain climber is equal to the gravitational potential energy gained, which can be calculated as: \[ |W| = mgh \] where \( m = 52 \; \text{kg} \), \( g = 9.8 \; \text{m/s}^2 \), and \( h = 730 \; \text{m} \). Thus, \[ |W| = 52 \times 9.8 \times 730 = 371,384 \; \text{J} \]
03

Apply the Efficiency Formula

The efficiency as a heat engine is given by the formula: \[ e = \frac{|W|}{|Q_{H}|} \] where \(|W| = 371,384 \; \text{J}\) and \(|Q_{H}| = 4.1 \times 10^6 \; \text{J}\). Plug these values into the formula: \[ e = \frac{371,384}{4.1 \times 10^6} \approx 0.0906 \]
04

Express Efficiency as a Percentage

To express the efficiency as a percentage, multiply the efficiency by 100: \[ e \approx 0.0906 \times 100 = 9.06\% \]

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Key Concepts

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

Understanding Heat Engines
A heat engine is a simple yet fascinating concept in thermodynamics. It operates on the principle of converting heat energy from a high temperature source into mechanical work. Every heat engine must expel some unused energy to a lower temperature sink, meaning not all input energy can be transformed directly into work. This inevitable loss accounts for the less-than-perfect efficiency measured in any heat engine.
In the context of our mountain climber, her body acts as a heat engine. By burning calories, she generates energy that helps her ascend the mountain. The efficiency of this biological engine is calculated by assessing how much of the input energy (from food) results in meaningful work (climbing). This is symbolized as \( e = \frac{|W|}{|Q_{H}|} \), where \(|W|\) is the work done, and \(|Q_{H}|\) is the input energy. In our problem, her efficiency is found to be around 9.06%, showing that while her body does real work, a majority of the input heat is lost as heat.
The Role of Gravitational Potential Energy
Gravitational potential energy is a critical factor when discussing activities that involve a change in height, like climbing. This form of energy depends on an object's mass, the height it ascends, and the gravitational field strength, represented by the equation \( |W| = mgh \).
For our mountain climber, ascending 730 meters gives her a gravitational potential energy rise of 371,384 J. This value explicitly represents the work done against gravity, which is vital in understanding the amount of mechanical work required to achieve her climb. Thus, this value of work done is a key part of calculating her heat engine efficiency as it determines the effective energy use from the total energy intake. Understanding this concept helps gauge how physical efforts relate to energy transformations in various contexts.
Examining Metabolic Processes
Metabolic processes refer to the complex biochemical reactions within our bodies that convert food into energy. This energy then supports cellular activities and physical exertions. During climbing, the body utilizes stored chemical energy to fuel muscles, generating significant amounts of heat as a by-product.
When considering our climber, her total metabolic energy output during the climb is given as \(4.1 \times 10^6 \; \text{J} \). However, not all this energy is transformed into climbing work; instead, much is released as thermal energy, or waste, underlining the principle that metabolic processes are not perfectly efficient.
By analyzing the fractional efficiency, or how much of the metabolic energy translates into actual work, we understand the limitations of biological systems. This insight is not only important for calculating efficiencies but also in improving physical training and dietary regimens to maximize energy use during intense activities.

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Most popular questions from this chapter

One mole of neon, a monatomic gas, starts out at conditions of standard temperature and pressure. The gas is heated at constant volume until its pressure is tripled, then further heated at constant pressure until its volume is doubled. Assume that neon behaves as an ideal gas. For the entire process, find the heat added to the gas.

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