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Compare two heat engines receiving the same \(Q\), one at \(1200 \mathrm{~K}\) and the other at \(1800 \mathrm{~K},\) both of which reject heat at \(500 \mathrm{~K}\). Which one is better?

Short Answer

Expert verified
The second engine is better, as it has a higher efficiency of 72.22% compared to 58.33% for the first engine.

Step by step solution

01

Understanding Efficiency of Heat Engines

The efficiency of a heat engine depends on the temperatures of the heat source (T_H) and the heat sink (T_C). The efficiency formula is: \[\eta = 1 - \frac{T_C}{T_H}\] where \(T_C\) is the temperature of the cold reservoir and \(T_H\) is the temperature of the hot reservoir.
02

Calculate Efficiency for First Engine

The first heat engine receives heat at \(T_H = 1200\, \mathrm{K}\) and rejects heat at \(T_C = 500\, \mathrm{K}\). Substitute into the efficiency formula: \[\eta_1 = 1 - \frac{500}{1200} = 1 - \frac{5}{12} = \frac{7}{12} \approx 0.5833\]Thus, the efficiency is approximately 58.33%.
03

Calculate Efficiency for Second Engine

The second heat engine receives heat at \(T_H = 1800\, \mathrm{K}\) and also rejects heat at \(T_C = 500\, \mathrm{K}\). Substitute into the efficiency formula:\[\eta_2 = 1 - \frac{500}{1800} = 1 - \frac{5}{18} = \frac{13}{18} \approx 0.7222\]Thus, the efficiency is approximately 72.22%.
04

Compare Efficiencies

The two efficiencies computed are \(\eta_1 \approx 0.5833\) (58.33%) for the first engine and \(\eta_2 \approx 0.7222\) (72.22%) for the second engine. Since \( \eta_2 > \eta_1 \), the second engine, operating at a higher source temperature, is more efficient.

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Key Concepts

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

Heat Engine Efficiency
Heat engines are fascinating devices that convert heat into work. Think of them as machines within a car engine or power plant. The efficiency of a heat engine tells us how well it does this conversion. Higher efficiency means more useful work from a given amount of heat. The efficiency depends crucially on the temperatures involved.

In thermodynamics, we describe efficiency with the formula: - \(\eta = 1 - \frac{T_C}{T_H}\)Here, \(T_C\) is the temperature at which the engine rejects heat to the surroundings (the cold reservoir), and \(T_H\) is the temperature from which the engine absorbs heat (the hot reservoir). Both temperatures are measured in Kelvin, a unit where 0 is absolute zero.

We can see that as \(T_H\) increases or \(T_C\) decreases, the efficiency \(\eta\) improves. This means the engine operates better when the heat source is hotter, or when it releases heat to a cooler space. Efficiency is always less than 1 or 100%, showing that some energy is always lost.
Carnot Cycle
The Carnot Cycle is a fundamental concept that helps us understand the upper limit of heat engine efficiency. Named after the French engineer Sadi Carnot, this cycle details a theoretical engine which achieves maximum efficiency. A Carnot engine undergoes four distinct stages:
  • Isothermal Expansion: Gas expands and performs work on surroundings, absorbing heat at a constant high temperature.
  • Adiabatic Expansion: Expansion continues without heat exchange, causing the gas to cool.
  • Isothermal Compression: Gas is compressed, releasing heat at a constant low temperature.
  • Adiabatic Compression: Compression continues without heat exchange, increasing the gas temperature.
In this cycle, the efficiency is given by:- \(\eta = 1 - \frac{T_C}{T_H}\)This efficiency is the maximum possible for any engine operating between these two temperatures. However, no real engine can achieve Carnot efficiency due to inevitable energy losses, like friction and heat dissipation.
Thermodynamic Processes
Thermodynamic processes describe how a system changes from one state to another. In the context of heat engines, several processes are at work. These include isothermal, adiabatic, isochoric, and isobaric processes.
  • Isothermal Process: The system changes at a constant temperature, typically requiring heat exchange with the surroundings.

  • Adiabatic Process: This change occurs without heat exchange, meaning the system's heat energy remains constant, even if its temperature changes.

  • Isochoric Process: The system's volume stays constant, so any heat added changes the temperature and pressure but not volume.

  • Isobaric Process: The pressure remains constant while the system changes. Heat exchange and work done by the system can alter both temperature and volume in this case.

Each of these processes alter the state of the systems within a heat engine differently. Understanding these changes explains how mechanical work is converted from heat energy. Real engines combine these processes to cycle repeatedly between a hot and cold reservoir, converting as much heat energy into work as possible.

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Most popular questions from this chapter

An air conditioner on a hot summer day removes \(8 \mathrm{~kW}\) of energy from a house at \(21^{\circ} \mathrm{C}\) and pushes energy to the outside, which is at \(31^{\circ} \mathrm{C}\). The house has a mass of \(15000 \mathrm{~kg}\) with an average specific heat of \(0.95 \mathrm{~kJ} / \mathrm{kgK}\). In order to do this, the cold side of the air conditioner is at \(5^{\circ} \mathrm{C}\) and the hot side is at \(40^{\circ} \mathrm{C}\). The air conditioner (refrigerator) has a COP that is \(60 \%\) that of a corresponding Carnot refrigerator. Find the actual COP of the air conditioner and the power required to run it.

A farmer runs a heat pump with a \(2-\mathrm{kW}\) motor. It should keep a chicken hatchery at \(90 \mathrm{~F}\); the hatchery loses energy at a rate of \(10 \mathrm{Btu} / \mathrm{s}\) to the colder ambient \(T_{\mathrm{amb}}\). What is the minimum COP that will be acceptable for the heat pump?

A certain solar-energy collector produces a maximum temperature of \(100^{\circ} \mathrm{C}\). The energy is used in a cycle heat engine that operates in a \(10^{\circ} \mathrm{C}\) environment. What is the maximum thermal efficiency? If the collector is redesigned to focus the incoming light, what should the maximum temperature be to produce a \(25 \%\) improvement in engine efficiency?

A refrigerator should remove \(400 \mathrm{~kJ}\) from some food. Assume the refrigerator works in a Carnot cycle between \(-15^{\circ} \mathrm{C}\) and \(45^{\circ} \mathrm{C}\) with a motorcompressor of \(400 \mathrm{~W}\). How much time does it take if this is the only cooling load?

A house should be heated by a heat pump, \(\beta^{\prime}=2.2\), and maintained at \(20^{\circ} \mathrm{C}\) at all times. It is estimated that it loses \(0.8 \mathrm{~kW}\) for each degree that the ambient is lower than the inside. Assume an outside temperature of \(-10^{\circ} \mathrm{C}\) and find the needed power to drive the heat pump.

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