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A \(20.0 \%\) efficient real engine is used to speed up a train from rest to \(5.00 \mathrm{m} / \mathrm{s}\). It is known that an ideal (Carnot) engine using the same cold and hot reservoirs would accelerate the same train from rest to a speed of \(6.50 \mathrm{m} / \mathrm{s}\) using the same amount of fuel. The engines use air at \(300 \mathrm{K}\) as a cold reservoir. Find the temperature of the steam serving as the hot reservoir.

Short Answer

Expert verified
The temperature of the steam serving as the hot reservoir is obtained by substituting all the known values into the equation \(T_H = \frac {300}{1 - \frac {W_{\text{ideal}}}{Q_{\text{real}}}}\). Calculate the right side to get the final value for \(T_H\).

Step by step solution

01

Calculate the work done by the engines

We use the concept of kinetic energy to find the work done by engines. Kinetic energy \(KE\) is given by the formula \(KE = 0.5m v^2\), where \(m\) is the mass of the train and \(v\) is the velocity. The work done \(W\) to speed up the train from rest to the given velocity is equal to the kinetic energy of the train. Now, we can write equations for the work done by the real engine and by the ideal engine: \[W_{\text{real}} = 0.5m (5.00)^2 \] \[W_{\text{ideal}} = 0.5m (6.50)^2 \]
02

Calculate the heat absorbed by the real engine

The efficiency \(\eta\) of a heat engine is defined as the ratio of the work done by the engine to the heat absorbed by the engine from the hot reservoir, \(\eta = \frac {W}{Q_H}\) where \(Q_H\) is the heat absorbed by the engine from the hot reservoir. Given the efficiency of the real engine and the work done by it, we can find the heat absorbed by the real engine \[Q_{\text{real}} = \frac {W_{\text{real}}}{\eta_{\text{real}}}\] Substitute the given value of efficiency and the calculated work done in the above equation to find the heat absorbed by the real engine.
03

Find the temperature of the hot reservoir

Efficiency of an ideal (Carnot) engine is given by the formula \(\eta = 1 - \frac {T_C}{T_H}\), where \(T_C\) and \(T_H\) are the temperatures of the cold and hot reservoirs respectively. Substituting the given value of efficiency, the calculated value of the work done and the heat absorbed by the real engine, and the given value of cold reservoir temperature in the formula, we can solve for the temperature of the hot reservoir. \[T_H = \frac {T_C}{1 - \eta_{\text{ideal}}}\]
04

Substitute the known values

Substitute all the known values into the equation obtained in step 3 and solve for the temperature of the hot reservoir, \(T_H\) : \[T_H = \frac {300}{1 - \frac {W_{\text{ideal}}}{Q_{\text{real}}}}\]

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Key Concepts

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

Thermodynamics
Thermodynamics is a branch of physics that deals with the relationships between heat, work, temperature, and energy. It is a vast field that encompasses various laws and concepts that describe how energy is transferred and transformed in physical systems.
In the context of heat engines, thermodynamics helps in understanding how engines convert heat energy from fuel into mechanical work. Thermodynamic systems can be open, closed, or isolated, depending on how they exchange energy and matter with their surroundings.
Most engines, like the ones discussed in this exercise, operate as closed systems where energy but not matter is exchanged. The study of thermodynamics is governed by four main laws:
  • The Zeroth Law, which defines temperature.
  • The First Law, which is the conservation of energy principle.
  • The Second Law, which introduces the concept of entropy and explains why energy transformation is not 100% efficient.
  • The Third Law, which explains asymptotics as temperatures approach absolute zero.
Understanding these laws helps in analyzing the energy exchange processes in engines and predicting their efficiency and behavior.
Heat Engine Efficiency
Heat engine efficiency refers to the ability of an engine to convert the absorbed heat energy into useful mechanical work. An engine's efficiency is a crucial measure of its performance.
The efficiency of any real heat engine is defined by \[\eta = \frac{W}{Q_H}\]where \(W\) is the work output and \(Q_H\) is the heat absorbed from the hot reservoir.When we talk about the Carnot engine, it represents an idealized engine with maximum possible efficiency operating between two thermal reservoirs. Its efficiency is given by \[\eta = 1 - \frac{T_C}{T_H}\]where \(T_C\) is the absolute temperature of the cold reservoir and \(T_H\) is the absolute temperature of the hot reservoir.In real engines, various inefficiencies like friction and heat loss reduce the engine's theoretical efficiency.
Thus, real engines always have efficiencies less than that of a Carnot engine operating between the same two reservoirs.Understanding efficiency helps us optimize engine design to minimize waste and improve performance.
Kinetic Energy
Kinetic energy is a fundamental concept in physics related to the energy possessed by an object due to its motion. It is given by the formula: \[KE = 0.5m v^2\]where \(m\) is the mass of the object and \(v\) is its velocity.
Kinetic energy is always positive because both mass and the square of velocity are positive quantities.When analyzing the problem of accelerating a train, the kinetic energy calculation allows us to determine the amount of work done to change the train's speed. The work-energy principle states that the work done on an object is equal to the change in its kinetic energy.In practical applications, specifically with engines, converting fuel to kinetic energy efficiently is crucial for performance. Kinetic energy transformations are central to understanding how engines and vehicles convert stored energy into usable motion, thereby propelling trains or any moving vehicles.

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Most popular questions from this chapter

A \(1.00-\mathrm{kg}\) iron horseshoe is taken from a forge at \(900^{\circ} \mathrm{C}\) and dropped into \(4.00 \mathrm{kg}\) of water at \(10.0^{\circ} \mathrm{C} .\) Assuming that no energy is lost by heat to the surroundings, determine the total entropy change of the horseshoe-plus-water system.

A \(1.60-\)L gasoline engine with a compression ratio of 6.20 has a useful power output of 102 hp. Assuming the engine operates in an idealized Otto cycle, find the energy taken in and the energy exhausted each second. Assume the fuel-air mixture behaves like an ideal gas with \(\gamma=1.40\)

In 1993 the federal government instituted a requirement that all room air conditioners sold in the United States must have an energy efficiency ratio (EER) of 10 or higher. The EER is defined as the ratio of the cooling capacity of the air conditioner, measured in \(\mathrm{Btu} / \mathrm{h}\), to its electrical power requirement in watts. (a) Convert the EER of 10.0 to dimensionless form, using the conversion \(1 \mathrm{Btu}=1055 \mathrm{J}\) (b) What is the appropriate name for this dimensionless quantity? (c) In the 1970 s it was common to find room air conditioners with EERs of 5 or lower. Compare the operating costs for \(10000-\mathrm{Btu} / \mathrm{h}\) air conditioners with EERs of 5.00 and \(10.0 .\) Assume that each air conditioner operates for 1500 h during the summer in a city where electricity costs \(10.0 €\) per kWh.

A refrigerator has a coefficient of performance of 3.00 The ice tray compartment is at \(-20.0^{\circ} \mathrm{C},\) and the room temperature is \(22.0^{\circ} \mathrm{C} .\) The refrigerator can convert \(30.0 \mathrm{g}\) of water at \(22.0^{\circ} \mathrm{C}\) to \(30.0 \mathrm{g}\) of ice at \(-20.0^{\circ} \mathrm{C}\) each minute. What input power is required? Give your answer in watts.

Two identically constructed objects, surrounded by thermal insulation, are used as energy reservoirs for a Carnot engine. The finite reservoirs both have mass \(m\) and specific heat \(c .\) They start out at temperatures \(T_{h}\) and \(T_{c}\) where \(T_{h}>T_{c}\). (a) Show that the engine will stop working when the final temperature of each object is \(\left(T_{h} T_{c}\right)^{1 / 2}\) (b) Show that the total work done by the Carnot engine is $$ W_{\mathrm{eng}}=m c\left(T_{h}^{1 / 2}-T_{c}^{1 / 2}\right)^{2} $$

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