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A Carnot engine operates with a large hot reservoir and a much smaller cold reservoir. As a result, the temperature of the hot reservoir remains constant while the temperature of the cold reservoir slowly increases. This temperature change decreases the efficiency of the engine to 0.70 from \(0.75 .\) Find the ratio of the final temperature of the cold reservoir to its initial temperature.

Short Answer

Expert verified
The ratio of the final temperature to the initial temperature of the cold reservoir is 1.2.

Step by step solution

01

Understanding Efficiency Change

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. The efficiency decreases from 0.75 to 0.70, indicating an increase in \(T_c\).
02

Express Initial Efficiency

Initially, the efficiency is 0.75. So, \( 0.75 = 1 - \frac{T_{c1}}{T_h} \). This implies \( \frac{T_{c1}}{T_h} = 0.25 \).
03

Express Final Efficiency

Finally, the efficiency becomes 0.70. Hence, \( 0.70 = 1 - \frac{T_{c2}}{T_h} \). This implies \( \frac{T_{c2}}{T_h} = 0.30 \).
04

Find Ratio of Temperatures

To find the ratio of the final to the initial temperature of the cold reservoir, we use: \[ \frac{T_{c2}}{T_{c1}} = \frac{(T_{c2}/T_h)}{(T_{c1}/T_h)} = \frac{0.30}{0.25} = 1.2. \] This tells us how much the cold reservoir's temperature has increased.

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Key Concepts

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

Thermodynamic Cycles
In the world of thermodynamics, a Carnot engine is often used as an idealized model to understand how engines work. A thermodynamic cycle in this context is a series of processes that involve heat transfer and work done on or by a system. The goal is typically to convert heat energy into mechanical work efficiently. The Carnot cycle is considered the most efficient cycle possible between two heat reservoirs, as it assumes no loss in heat due to friction or other inefficiencies.
The cycle consists of four steps: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression. During these steps, the system absorbs heat from a hot reservoir, performs work, releases heat to a cold reservoir, and then returns to its original state. This cycle demonstrates how temperature difference between the reservoirs influences the efficiency of energy conversion. If we can closely mimic the Carnot cycle, we can achieve maximum efficiency in real-world engines.
Temperature Ratio
The concept of temperature ratio is crucial when analyzing the efficiency of a Carnot engine. The formula for efficiency is given by: \[ \eta = 1 - \frac{T_c}{T_h} \] where \( T_c \) is the temperature of the cold reservoir and \( T_h \) is the temperature of the hot reservoir (in Kelvin).
As you can see from the formula, the efficiency directly depends on the ratio of these temperatures. A lower temperature ratio means higher efficiency. For example, if \( T_c \) increases, the ratio \( \frac{T_c}{T_h} \) increases, and hence, the efficiency \( \eta \) decreases as observed in the original exercise.
The change in the temperature ratio can tell us how the performance of the engine varies over time. Tracking this ratio is crucial for ensuring the optimal operation of any Carnot engine or similar thermodynamic system.
Heat Reservoirs
Heat reservoirs are essential components in thermodynamic cycles, particularly in the Carnot engine. A heat reservoir is essentially a large system that can absorb or provide heat without undergoing any significant changes in its own temperature.
In a Carnot engine, there are two key reservoirs:
  • The hot reservoir provides heat to the engine system, kept at a constant high temperature.
  • The cold reservoir absorbs the engine's output heat, gradually increasing in temperature as seen in the exercise.
The purpose of these reservoirs is to facilitate the flow of energy necessary for performing mechanical work. Ideally, the hot reservoir remains constant to maintain consistent energy input, while the cold reservoir can absorb all residual heat without disrupting the cycle.
Understanding how these reservoirs interact with the thermodynamic cycle helps us grasp why certain temperature changes in either reservoir lead to variations in engine efficiency. This understanding is imperative for improving thermal efficiency in practical applications where ensuring optimum temperature stability can lead to better performance.

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

A system undergoes a two-step process. In the first step, the internal energy of the system increases by 228 \(\mathrm{J}\) when 166 \(\mathrm{J}\) of work is done on the system. In the second step, the internal energy of the system increases by 115 \(\mathrm{J}\) when 177 \(\mathrm{J}\) of work is done on the system. For the overall process, find the heat. What type of process is the overall process? Explain.

An irreversible engine operates between temperatures of 852 and 314 \(\mathrm{K}\) . It absorbs 1285 \(\mathrm{J}\) of heat from the hot reservoir and does 264 \(\mathrm{J}\) of work. \(\quad\) (a) What is the change \(\Delta S_{\text { universe }}\) in the entropy of the universe associated with the operation of this engine? \((\mathbf{b})\) If the engine were reversible, what would be the magnitude \(|W|\) of the work it would have done, assuming that it operated between the same temperatures and absorbed the same heat as the irreversible engine? (c) Using the results of parts (a) and (b), find the difference between the work produced by the reversible and irreversible engines.

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.

Suppose that the gasoline in a car engine burns at \(631^{\circ} \mathrm{C}\) , while the exhaust temperature (the temperature of the cold reservoir) is \(139^{\circ} \mathrm{C}\) and the outdoor temperature is \(27^{\circ} \mathrm{C}\) . Assume that the engine can be treated as a Carnot engine (a gross oversimplification). In an attempt to increase mileage performance, an inventor builds a second engine that functions between the exhaust and outdoor temperatures and uses the exhaust heat to produce additional work. Assume that the inventor's engine can also be treated as a Carnot engine. Determine the ratio of the total work produced by both engines to that produced by the first engine alone.

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