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Multiple-Concept Example 6 deals with the concepts that are important in this problem. In doing \(16600 \mathrm{~J}\) of work, an engine rejects \(9700 \mathrm{~J}\) of heat. What is the efficiency of the engine?

Short Answer

Expert verified
The engine's efficiency is approximately 63.12%.

Step by step solution

01

Understand the Problem

To find the efficiency of the engine, we need to determine how much of the work input is converted into useful output. The engine's efficiency can be calculated by comparing the work done to the heat energy absorbed by the engine.
02

Identify Given Values

From the exercise, we are given that the engine performs work of \(16600 \text{ J}\) and rejects \(9700 \text{ J}\) of heat.
03

Apply the Efficiency Formula

Efficiency (\(\eta\)) can be found using the formula:\[ \eta = \frac{W_{out}}{Q_{in}} \times 100\% \]Where \(W_{out}\) is the work done by the engine and \(Q_{in}\) is the heat absorbed by the engine. We need to find \(Q_{in}\).
04

Calculate Heat Absorbed (Input Heat)

Using the energy conservation principle, the heat absorbed by the engine \(Q_{in}\) can be calculated as:\[ Q_{in} = W_{out} + Q_{rejected} = 16600 \text{ J} + 9700 \text{ J} \]\[ Q_{in} = 26300 \text{ J} \]
05

Calculate Efficiency

Now, plug the values into the efficiency formula:\[ \eta = \frac{16600 \text{ J}}{26300 \text{ J}} \times 100\% \]\[ \eta \approx 63.12\% \]

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Key Concepts

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

Engine Efficiency
Engine efficiency is an important concept that explains how well an engine converts heat from fuel into useful work output. It is a measure of how much of the energy input is actually used for its intended purpose, rather than being lost to other processes.
To calculate engine efficiency, the formula is:
  • \[ \eta = \frac{W_{out}}{Q_{in}} \times 100\% \]
Here, \( W_{out} \) represents the useful work done by the engine, and \( Q_{in} \) is the total heat energy taken in.
The efficiency is usually expressed as a percentage, indicating how much of the input energy is converted into work. If an engine has an efficiency of 100%, it means all the input energy is converted into work with no waste. However, in reality, some energy is always lost as waste heat.
Work and Energy
Work and energy are fundamental concepts in thermodynamics that describe how engines and machines operate. Work is the amount of energy transferred when a force is applied over a distance.
In the context of this exercise, work is the energy output
  • The engine does \( 16600 \, \text{J} \) of work.
Energy is the capacity to do work, and it can exist in many forms, such as kinetic, potential, and thermal energy. In engines, energy is typically input as heat (from burning fuel), which is then converted to work.
Understanding how energy is transferred and converted is vital for calculating efficiency and optimizing engine design. This helps determine how much of the input energy becomes useful work and how much is lost during the process.
Heat Transfer
Heat transfer is a crucial aspect of engine operation, involving the movement of thermal energy from one place to another. It occurs in three primary ways: conduction, convection, and radiation.
In engines, heat is often transferred from the fuel to the engine to produce work. However, some of this heat doesn't contribute to work and is rejected as waste heat. In our exercise,
  • The engine rejects \( 9700 \, \text{J} \) of heat.
Understanding heat transfer allows engineers to improve engine efficiency by minimizing energy losses.
This involves designing systems that maximize the useful transfer of energy while reducing waste. By managing heat transfer effectively, engines can be made more efficient and environmentally friendly. Proper insulation, strategic venting, and material choices are key factors in achieving efficient heat management.

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Most popular questions from this chapter

In moving out of a dormitory at the end of the semester, a student does \(1.6 \times 10^{4} \mathrm{~J}\) of work. In the process, his internal energy decreases by \(4.2 \times 10^{4} \mathrm{~J}\). Determine each of the following quantities (including the algebraic sign): (a) \(W,(\mathrm{~b}) \Delta U\) and \((\mathrm{c}) Q\)

The temperatures indoors and outdoors are 299 and \(312 \mathrm{~K},\) respectively. A Carnot air conditioner deposits \(6.12 \times 10^{5} \mathrm{~J}\) of heat outdoors. How much heat is removed from the house? \(?\)

The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partition. Both sides contain one mole of a monatomic ideal gas \(\left(\gamma=\frac{5}{3}\right),\) with the initial temperature being \(525 \mathrm{~K}\) on the left and \(275 \mathrm{~K}\) on the right. The partition is then allowed to move slowly (i.e., quasi- statically) to the right, until the pressures on each side of the partition are the same. Find the final temperatures on the (a) left and (b) right.

Review Conceptual Conceptual Example 9 before attempting this problem. A window air conditioner has an average coefficient of performance of \(2.0 .\) This unit has been placed on the floor by the bed, in a futile attempt to cool the bedroom. During this attempt \(7.10 \times 10^{4} \mathrm{~J}\) of heat is pulled in the front of the unit. The room is sealed and contains 3800 mol of air. Assuming that the molar specific heat capacity of the air is \(C_{V}=\frac{5}{2} R,\) determine the rise in temperature caused by operating the air conditioner in this manner.

A fifteen-watt heater is used to heat a monatomic ideal gas at a constant pressure of \(7.60 \times 10^{5} \mathrm{~Pa}\) During the process, the \(1.40 \times 10^{-3} \mathrm{~m}^{3}\) volume of the gas increases by \(25.0 \%\). How long was the heater on?

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