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(I) A heat engine exhausts 8200 J of heat while performing 2600 J of useful work.What is the efficiency of this engine?

Short Answer

Expert verified
The efficiency of the engine is approximately 24.07%.

Step by step solution

01

Understanding Heat Engine Efficiency

The efficiency of a heat engine (\( \eta \)) is calculated using the formula:\[\eta = \frac{W}{Q_{in}}\]where \( W \) is the work done by the engine and \( Q_{in} \) is the heat absorbed by the engine.
02

Given Information

From the problem, we know:- The work done (\( W \)) is 2600 J.- The heat exhausted (\( Q_{out} \)) is 8200 J.
03

Finding Heat Input

Since the heat input (\( Q_{in} \)) is the sum of the work done and the heat exhausted, calculate it using:\[Q_{in} = W + Q_{out}\]Substitute the given values:\[Q_{in} = 2600 \text{ J} + 8200 \text{ J} = 10800 \text{ J}\]
04

Calculating Efficiency

Substitute the values of \( W \) and \( Q_{in} \) into the efficiency formula:\[\eta = \frac{2600 \text{ J}}{10800 \text{ J}}\]Calculate the efficiency:\[\eta = 0.2407\]
05

Express Efficiency as a Percentage

Convert the efficiency to a percentage by multiplying by 100:\[\eta = 0.2407 \times 100 \approx 24.07\%\]

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Key Concepts

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

Understanding the Concept of Work Done
In the context of heat engines, the term "work done" represents the useful energy output by the engine. This energy is what allows the engine to perform tasks, such as powering a machine or generating electricity. The work done by an engine is often denoted by the symbol \( W \). In our example, the engine performed 2600 J (joules) of work. Joules are a standard unit of energy, and they are critical in measuring how much useful energy a machine is utilizing for the task at hand. Understanding this concept helps us analyze engine efficiency and how effectively energy is being converted into work.
The Role of Heat Exhausted
Heat exhausted, denoted as \( Q_{out} \), refers to the amount of heat energy expelled by the engine into the surroundings. This hot exhaust is wasted energy that isn't converted into useful work. In our example, the heat exhausted by the engine amounts to 8200 J. One should remember that not all energy in a heat engine is converted to useful work. A significant portion is often lost as heat loss to the environment. This concept is crucial for understanding why no heat engine can be 100% efficient; there is always some energy loss to the surroundings.
Efficiency Formula in Heat Engines
The efficiency formula of a heat engine helps us determine how well an engine converts heat energy into useful work. The formula is expressed as:\[\eta = \frac{W}{Q_{in}}\]where \( \eta \) is the efficiency, \( W \) is the work done, and \( Q_{in} \) is the total heat absorbed by the engine. In our example, \( Q_{in} \) is the total of the work done (2600 J) and the heat exhausted (8200 J), making \( Q_{in} = 10800 \text{ J} \). Thus, the efficiency can be calculated by dividing the work done by the heat input. This results in an efficiency of about 24.07%. From this, it's evident that only about a quarter of the input energy is used for work.
Energy Conversion in Heat Engines
Energy conversion is the process through which heat energy is transformed into mechanical energy in heat engines. The primary goal of a heat engine is to maximize this conversion to perform useful work efficiently. However, due to the second law of thermodynamics, some energy is always wasted, often as heat. In the context of our problem, the engine converts a part of the total absorbed energy (heat input) into useful work (2600 J) while other parts are lost as heat exhausted (8200 J). This process's effectiveness is captured by the engine's efficiency, which is calculated using the efficiency formula. Understanding energy conversion is essential for investigating how efficiently an engine operates and the potential for improving its performance.

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Most popular questions from this chapter

(II) A four-cylinder gasoline engine has an efficiency of 0.22 and delivers 180 J of work per cycle per cylinder. If the engine runs at 25 cycles per second (1500 rpm), determine (\(a\)) the work done per second, and (\(b\)) the total heat input per second from the gasoline. (\(c\)) If the energy content of gasoline is 130 MJ per gallon, how long does one gallon last?

(II) 1.0 kg of water at 35\(^\circ\)C is mixed with 1.0 kg of water at 45\(^\circ\)C in a well-insulated container. Estimate the net change in entropy of the system.

(I) A heat engine's high temperature \(T_H\) could be ambient temperature, because liquid nitrogen at 77 K could be \(T_L\) and is cheap. What would be the efficiency of a Carnot engine that made use of heat transferred from air at room temperature (293 K) to the liquid nitrogen "fuel"(Fig.15\(-\)25)?

Suppose a power plant delivers energy at 880 MW using steam turbines. The steam goes into the turbines superheated at 625 K and deposits its unused heat in river water at 285 K. Assume that the turbine operates as an ideal Carnot engine. (\(a\)) If the river flow rate is 37 m\(^3\)/s, estimate the average temperature increase of the river water immediately downstream from the power plant. (\(b\)) What is the entropy increase per kilogram of the downstream river water in J/kg\(\cdot\)K?

(II) Use the conservation of energy to explain why the temperature of a well- insulated gas increases when it is compressed\(-\)say, by pushing down on a piston\(-\)whereas the temperature decreases when the gas expands. Show your reasoning.

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