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A Carnot engine has an efficiency of \(0.40 .\) The Kelvin temperature of its hot reservoir is quadrupled, and the Kelvin temperature of its cold reservoir is doubled. What is the efficiency that results from these changes?

Short Answer

Expert verified
The new efficiency is 0.72.

Step by step solution

01

Understanding Initial Efficiency

The efficiency of a Carnot engine is given by the formula \( \eta = 1 - \frac{T_C}{T_H} \), where \( T_C \) is the temperature of the cold reservoir and \( T_H \) is the temperature of the hot reservoir. Initially, the efficiency is \( \eta = 0.40 \).

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Key Concepts

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

Carnot cycle
The Carnot cycle is a theoretical concept in thermodynamics that describes an idealized heat engine. The cycle consists of two isothermal processes and two adiabatic processes. During each cycle, the engine absorbs heat from a high-temperature reservoir and converts part of it into work, rejecting the rest to a low-temperature reservoir. It's important to note that the Carnot cycle is an ideal model, which means it provides a benchmark for the maximum possible efficiency of any real heat engine.
This cycle defines the Carnot efficiency, which depends on the temperatures of the two reservoirs. The larger the temperature difference between the hot and cold reservoirs, the more efficient the engine can potentially be. Understanding the Carnot cycle helps in appreciating why no real engine can be 100% efficient, as there will always be some heat loss in real scenarios.
thermodynamics
Thermodynamics is a branch of physics that deals with the relationships and conversions between heat and other forms of energy. There are four laws of thermodynamics, which form the backbone of the conversation around energy conversion.
Key points include:
  • First Law: Energy cannot be created or destroyed, only transformed from one form to another.
  • Second Law: Heat will always flow spontaneously from hotter to colder bodies, and not the other way around unless external work is applied.
  • Third Law: As temperature approaches absolute zero, the entropy of a perfect crystal approaches zero.
In the context of the Carnot engine, the laws of thermodynamics ensure that efficiency is capped by the temperature differences between reservoirs. These laws dictate how energy efficiency is calculated and why no engine or cycle can surpass the Carnot efficiency.
Kelvin temperature scale
The Kelvin temperature scale is the standard unit of measuring thermodynamic temperature used in scientific communities. Unlike Celsius or Fahrenheit, the Kelvin scale begins at absolute zero, which is the theoretical point where all thermal motion ceases.
Some key features include:
  • Absolute zero is defined as 0 Kelvin (K).
  • The scale is linear, which means each increment represents the same change in thermal energy.
  • 0°C equals 273.15 K.
This temperature scale is crucial in calculations involving Carnot engines because it ensures that temperature differences are absolute and directly comparable. In the efficiency formula for a Carnot engine, using Kelvin allows us to accurately model the behavior of heat flows and energy transformations.

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Most popular questions from this chapter

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 rescrvoir at \(298 \mathrm{K}\) and a rescrvoir at \(248 \mathrm{K}\). Find \(Q\).

Multiple-Concept Example 6 deals with the same concepts as this problem does. What is the efficiency of a heat engine that uses an input heat of \(5.6 \times 10^{4} \mathrm{J}\) and rejects \(1.8 \times 10^{4} \mathrm{J}\) of heat?

Due to design changes, the efficiency of an engine increases from 0.23 to \(0.42 .\) For the same input heat \(\left|Q_{\mathrm{H}}\right|,\) these changes increase the work done by the more efficient engine and reduce the amount of heat rejected to the cold reservoir. Find the ratio of the heat rejected to the cold reservoir for the improved engine to that for the original engine.

A Carnot engine operates between temperatures of 650 and \(350 \mathrm{K}\). To improve the efficiency of the engine, it is decided either to raise the temperature of the hot reservoir by \(40 \mathrm{K}\) or to lower the temperature of the cold reservoir by \(40 \mathrm{K}\). Which change gives the greater improvement? Justify your answer by calculating the efficiency in each case.

A diesel engine does not use spark plugs to ignite the fuel and air in the cylinders. Instead, the temperature required to ignite the fuel occurs because the pistons compress the air in the cylinders. Suppose that air at an initial temperature of \(21{ }^{\circ} \mathrm{C}\) is compressed adiabatically to a temperature of \(688^{\circ} \mathrm{C}\). Assume the air to be an ideal gas for which \(\gamma=\frac{7}{5}\). Find the compression ratio, which is the ratio of the initial volume to the final volume.

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