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A heat engine takes in 360 J of energy from a hot reservoir and performs 25.0 J of work in each cycle. Find (a) the efficiency of the engine and (b) the energy expelled to the cold reservoir in each cycle.

Short Answer

Expert verified
The efficiency of the engine is 6.94%, and the energy expelled to the cold reservoir in each cycle is 335 J.

Step by step solution

01

Understand Efficiency

The efficiency of a heat engine is defined as the ratio of the work it performs to the energy it takes in. Mathematically, efficiency (e) is given by the formula: \( e = \frac{W}{Q_h} \) where \( W \) is the work done by the engine and \( Q_h \) is the heat energy absorbed from the hot reservoir.
02

Calculate the Efficiency

Given that the heat engine performs \( 25.0 \, J \) of work (W) and takes in \( 360 \, J \) of energy from a hot reservoir (Q_h), we can substitute these values into the efficiency formula to calculate the efficiency of the engine: \( e = \frac{25.0 \, J}{360 \, J} = 0.0694 \). To express this as a percentage, we multiply by 100, giving us an efficiency of 6.94%.
03

Calculate the Energy Expelled to the Cold Reservoir

The energy expelled to the cold reservoir (Q_c) in each cycle can be determined using the first law of thermodynamics, which states that the energy absorbed from the hot reservoir is equal to the work done plus the energy expelled to the cold reservoir: \( Q_h = W + Q_c \). We can rearrange this to solve for \( Q_c \): \( Q_c = Q_h - W = 360 \, J - 25.0 \, J = 335 \, J \).

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Key Concepts

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

Understanding Thermodynamics
Thermodynamics is a branch of physics that deals with the relationships between heat and other forms of energy. In essence, thermodynamics looks at how energy changes from one form into another and how these transformations affect the physical properties of the substances involved.

The subject is vast and includes concepts like temperature, entropy, and the very heart of our topic – heat engines. A heat engine is a system that converts heat or thermal energy—and often, but not always, this involves the use of a 'working substance'—into mechanical work. Think of steam engines or car engines; these are everyday examples of heat engines.

In explaining the textbook problem at hand, the engine in question receives a certain amount of thermal energy from a hot source and converts part of this energy into work, with the remainder expelled to a cooler reservoir. This process is a classic scenario studied in thermodynamics.
Delving into the First Law of Thermodynamics
The first law of thermodynamics, also known as the law of energy conservation, establishes that energy cannot be created or destroyed in an isolated system. Instead, energy can only be transformed from one form to another. This fundamental principle forms the backbone of our understanding of how heat engines operate.

In the context of the heat engine in our exercise, the first law implies that the energy supplied to the engine from the hot reservoir must equal the sum of the work done by the engine and the heat expelled to the cold reservoir. When we demonstrate this concept with numbers \(Q_h = W + Q_c\), we cement our comprehension of how energy balance plays a vital role in engineering and environmental considerations.
Exploring the Work-Energy Principle
The work-energy principle is a fundamental concept in physics that states that the work done on an object is equal to the change in its energy. This principle is closely related to the first law of thermodynamics because it is essentially an application of energy conservation to mechanical systems.

In the heat engine example, the work done by the engine \( W \) is a direct representation of the energy converted from the thermal energy taken in \( Q_h \) to mechanical work. The principle allows us to predict and understand how much work can be extracted from a certain amount of energy, just as we calculated in the exercise. Understanding this relationship not only helps with homework but also lays the foundation for more complex applications in areas such as mechanical design and energy resource management.

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Most popular questions from this chapter

A Carnot engine has a power output of 150 \(\mathrm{kW}\) . The engine operates between two reservoirs at \(20.0^{\circ} \mathrm{C}\) and \(500^{\circ} \mathrm{C}\). (a) How much energy does it take in per hour? (b) How much energy is lost per hour in its exhaust?

A firebox is at 750 K, and the ambient temperature is 300 K. The efficiency of a Carnot engine doing 150 J of work as it transports energy between these constant-temperature baths is 60.0%. The Carnot engine must take in energy 150 J/0.600 ! 250 J from the hot reservoir and must put out 100 J of energy by heat into the environment. To follow Carnot’s reasoning, suppose that some other heat engine S could have efficiency 70.0%. (a) Find the energy input and wasted energy output of engine S as it does 150 J of work. (b) Let engine S operate as in part (a) and run the Carnot engine in reverse. Find the total energy the firebox puts out as both engines operate together, and the total energy transferred to the environment. Show that the Clausius statement of the second law of thermodynamics is violated. (c) Find the energy input and work output of engine S as it puts out exhaust energy of 100 J. (d) Let engine S operate as in (c) and contribute 150 J of its work output to running the Carnot engine in reverse. Find the total energy the firebox puts out as both engines operate together, the total work output, and the total energy transferred to the environment. Show that the Kelvin–Planck statement of the second law is violated. Thus our assumption about the efficiency of engine S must be false. (e) Let the engines operate together through one cycle as in part (d). Find the change in entropy of the Universe. Show that the entropy statement of the second law is violated.

One of the most efficient heat engines ever built is a steam turbine in the Ohio valley, operating between \(430^{\circ} \mathrm{C}\) and \(1870^{\circ} \mathrm{C}\) on energy from West Virginia coal to produce electricity for the Midwest. (a) What is its maximum theoretical efficiency? (b) The actual efficiency of the engine is 42.0\(\%\) . How much useful power does the engine deliver if it takes in \(1.40 \times 10^{5} \mathrm{J}\) of energy each second from its hot reservoir?

An ideal refrigerator or ideal heat pump is equivalent to a Carnot engine running in reverse. That is, energy \(Q_{c}\) is taken in from a cold reservoir and energy \(Q_{h}\) is rejected to a hot reservoir. (a) Show that the work that must be supplied to run the refrigerator or heat pump is $$W=\frac{T_{h}-T_{c}}{T_{c}} Q_{c}$$ (b) Show that the coefficient of performance of the ideal refrigerator is $$\operatorname{COP}=\frac{T_{c}}{T_{h}-T_{c}}$$

A biology laboratory is maintained at a constant temperature of \(7.00^{\circ} \mathrm{C}\) by an air conditioner, which is vented to the air outside. On a typical hot summer day the outside temperature is \(27.0^{\circ} \mathrm{C}\) and the air conditioning unit emits energy to the outside at a rate of 10.0 \(\mathrm{kW}\) . Model the unit as having a coefficient of performance equal to 40.0\(\%\) of the coefficient of performance of an ideal Carnot device. (a) At what rate does the air conditioner remove energy from the laboratory? (b) Calculate the power required for the work input. (c) Find the change in entropy produced by the air conditioner in 1.00 h. (d) What If? The outside temperature increases to \(32.0^{\circ} \mathrm{C} .\) Find the fractional change in the coefficient of performance of the air conditioner.

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