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The heat that goes into a particular Carnot engine is 4.00 times greater than the work it performs. What is the engine's efficiency?

Short Answer

Expert verified
The engine's efficiency is 25%.

Step by step solution

01

Understand Carnot Efficiency Formula

The efficiency (\(\eta\)) of a Carnot engine is defined as the ratio of work output (\(W\)) to the heat input (\(Q_H\)) from the hot reservoir. This is given by the formula: \[\eta = \frac{W}{Q_H}\]
02

Determine Given Ratio

We are given that the heat input (\(Q_H\)) is 4 times the work output (\(W\)). This implies \(Q_H = 4W\).
03

Substitute into Efficiency Formula

Substitute the ratio \(Q_H = 4W\) into the efficiency formula: \[\eta = \frac{W}{4W}\].
04

Simplify the Efficiency Expression

Simplify the expression \(\frac{W}{4W}\) to get \[\eta = \frac{1}{4}\].
05

Convert to Percentage

Convert the efficiency into percentage by multiplying by 100. Thus, \(\eta = \frac{1}{4} \times 100 = 25\%\).

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Key Concepts

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

Understanding Thermodynamic Cycles
Thermodynamic cycles are a fundamental concept in understanding how heat engines, like the Carnot engine, operate. A thermodynamic cycle consists of a series of processes that involve transferring energy as heat and work into and out of a system. The system returns to its initial state at the end of the cycle, which means that the parameters defining the system (such as volume, pressure, and temperature) are the same at the beginning and at the end.
A popular example of a thermodynamic cycle is the Carnot cycle, which is an idealized cycle that provides the upper limit on the efficiency any real engine can achieve. It involves four steps:
  • Isothermal expansion - the system absorbs heat from a hot reservoir at a constant temperature.
  • Adiabatic expansion - the system expands without transferring heat to or from the environment, causing a drop in temperature.
  • Isothermal compression - the system releases heat to a cold reservoir at a constant temperature.
  • Adiabatic compression - the system is compressed, which raises its temperature without any heat exchange.
The Carnot cycle is crucial for understanding the theoretical limits of engine efficiency, as it considers both heat transfer and work done, along with the critical aspect of temperature differences between reservoirs.
Exploring Heat Engines
Heat engines are fascinating machines that convert thermal energy (or heat) into mechanical energy or work. They operate on the principles of thermodynamics, absorbing heat from a high-temperature source, converting part of the absorbed heat into work, and dumping the remaining heat to a low-temperature sink. An important point to note is that it is impossible for a heat engine to convert all the absorbed heat into work due to the second law of thermodynamics.
A Carnot engine is a hypothetical example of a perfect heat engine. It operates on the previously mentioned Carnot cycle and defines the maximum possible efficiency a heat engine can achieve. In the real world, no engine can achieve Carnot efficiency due to practical limitations such as friction, heat losses, and material properties. However, studying Carnot engines helps engineers set a benchmark to strive towards when designing practical heat engines.
By understanding the operation of heat engines, we learn how energy resources like fossil fuels, nuclear energy, and solar power are converted into work in various forms, whether it's powering car engines, generating electricity, or even in refrigeration processes.
Calculating Efficiency of Heat Engines
The efficiency of any heat engine, including the Carnot engine, gives us an idea of how well it can convert heat into work. This efficiency is calculated using the formula \[ \eta = \frac{W}{Q_H} \]where \(\eta\) is the efficiency, \(W\) is the work output, and \(Q_H\) is the heat input from the hot reservoir. It is crucial to realize that the efficiency will always be less than 100% due to inevitable energy losses.
In the exercise provided, you calculate the efficiency by determining the ratio of work to heat input. Since the given condition is that the heat input is four times the work, the efficiency turns out to be \(\frac{1}{4}\), which, when converted to a percentage, is 25%.
Efficiency calculation is essential in improving energy systems as it points out how much of the energy input doesn’t go to wasteful heat and can instead be utilized for productive work. It also provides insight into optimizing energy usage, thereby allowing us to develop more eco-friendly and sustainable technologies.

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Most popular questions from this chapter

An ideal gas is held in an insulated container at the temperature \(T .\) All the gas is initially in one-half of the container, with a partition separating the gas from the other half of the container, which is a vacuum. If the partition ruptures, and the gas expands to fill the entire container, is the final temperature greater than, less than, or equal to \(T\) ? Explain.

(a) A gas is expanded reversibly and adiabatically. Does its entropy increase, decrease, or stay the same? (b) Choose the best explanation from among the following: I. The process is reversible, and no heat is added to the gas. Therefore, the entropy of the gas remains the same. II. Expanding the gas gives it more volume to occupy, and this increases its entropy. III. The gas is expanded with no heat added to it, and hence its temperature will decrease. This, in turn, will lower its entropy.

A basketball player does \(2.43 \times 10^{5}\) J of work during her time in the game, and evaporates \(0.110 \mathrm{kg}\) of water. Assuming a latent heat of \(2.26 \times 10^{6} \mathrm{J} / \mathrm{kg}\) for the perspiration (the same as for water), determine (a) the change in the player's internal energy and (b) the number of nutritional calories the player has converted to work and heat.

A heat engine operates between a high-temperature reservoir at \(610 \mathrm{K}\) and a low-temperature reservoir at \(320 \mathrm{K}\). In one cycle, the engine absorbs 6400 J of heat from the high-temperature reservoir and does 2200 J of work. What is the net change in entropy as a result of this cycle?

(a) If 535 J of heat are added to 45 moles of a monatomic gas at constant volume, how much does the temperature of the gas increase? (b) Repeat part (a), this time for a constant-pressure process.

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