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Chapter 20: Problem 74

A Carnot engine whose high-temperature reservoir is at \(400 \mathrm{~K}\) has an efficiency of \(30.0 \% .\) By how much should the temperature of the low- temperature reservoir be changed to increase the efficiency to \(40.0 \% ?\)

Short Answer

Expert verified
Decrease the low-temperature reservoir by 40 K.

Step by step solution

01

Understand the Efficiency Formula for Carnot Engine

The efficiency of a Carnot engine is given by the formula \( \eta = 1 - \frac{T_L}{T_H} \), where \( \eta \) is the efficiency, \( T_L \) is the temperature of the low-temperature reservoir, and \( T_H \) is the temperature of the high-temperature reservoir.
02

Solve for Initial Low-Temperature Reservoir

Using the initial efficiency \( \eta_1 = 0.30 \), substitute into the efficiency formula: \( 0.30 = 1 - \frac{T_{L1}}{400} \). Rearrange and solve for \( T_{L1} \): \( T_{L1} = 400 (1 - 0.30) = 400 \times 0.70 = 280 \) K.
03

Solve for New Low-Temperature Reservoir

With the new efficiency \( \eta_2 = 0.40 \), substitute into the efficiency formula: \( 0.40 = 1 - \frac{T_{L2}}{400} \). Rearrange and solve for \( T_{L2} \): \( T_{L2} = 400 (1 - 0.40) = 400 \times 0.60 = 240 \) K.
04

Determine the Change in Temperature of the Low-Temperature Reservoir

Find the change in temperature \( \Delta T = T_{L1} - T_{L2} = 280 - 240 = 40 \) K. This means the temperature of the low-temperature reservoir needs to be decreased by 40 K to achieve the higher efficiency.

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Key Concepts

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

Efficiency Formula
The efficiency of a Carnot engine is a fundamental concept in thermodynamics. It tells us how effectively the engine converts heat into work, and it is determined by the efficiency formula: \[ \eta = 1 - \frac{T_L}{T_H} \]where \( \eta \) represents efficiency, \( T_L \) is the temperature of the low-temperature reservoir, and \( T_H \) is the temperature of the high-temperature reservoir.Using this formula:* If the temperature of the low-temperature reservoir \( T_L \) is increased, efficiency decreases.* Conversely, decreasing \( T_L \) increases the engine's efficiency.This equation makes it clear why efficiency cannot be 100% in real-world engines, as \( T_L \) can never be absolutely zero. It provides a useful theoretical limit for engineers to aim for in designing more efficient engines.
High-Temperature Reservoir
The high-temperature reservoir in a Carnot engine is crucial as it provides the thermal energy required for the engine to perform work.In our example:* The given high-temperature reservoir is at \(400 \mathrm{~K}\).* This temperature serves as the baseline heat source for the engine's operation.By maintaining the high-temperature reservoir at an optimal level:* We ensure that the engine functions efficiently.* Any variation in this temperature could drastically alter engine performance.The high-temperature reservoir plays a fundamental role in determining how much work output the engine can deliver. The greater the temperature difference between it and the low-temperature reservoir, the higher the potential efficiency of the engine.
Low-Temperature Reservoir
The low-temperature reservoir is another essential element of the Carnot engine, as it acts as a heat sink.In this problem:* The initial low-temperature reservoir temperature was calculated to be \(280 \mathrm{~K}\) to achieve a 30% efficiency.* To increase efficiency to 40%, the low-temperature reservoir must be decreased to \(240 \mathrm{~K}\).Reducing \(T_L\) requires careful consideration to avoid violating practical constraints. However, achieving a lower temperature in controlled environments can significantly enhance engine efficiencies. For example, cryogenic fluids and advanced insulations might be used to manage and sustain lower temperatures.

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Most popular questions from this chapter

A box contains \(N\) identical gas molecules equally divided between its two halves. For \(N=50,\) what are (a) the multiplicity \(W\) of the central configuration, (b) the total number of microstates, and (c) the percentage of the time the system spends in the central configuration? For \(N=100,\) what are (d) \(W\) of the central configuration, (e) the total number of microstates, and (f) the percentage of the time the system spends in the central configuration? For \(N=200,\) what are \((\mathrm{g}) W\) of the central configuration, \((\mathrm{h})\) the total number of microstates, and (i) the percentage of the time the system spends in the central configuration?(j) Does the time spent in the central configuration increase or decrease with an increase in \(N ?\)

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The efficiency of a particular car engine is \(25 \%\) when the engine does \(8.2 \mathrm{~kJ}\) of work per cycle. Assume the process is reversible. What are (a) the energy the engine gains per cycle as heat \(Q_{\text {gain }}\) from the fuel combustion and (b) the energy the engine loses per cycle as heat \(Q_{\text {lost }}\) ? If a tune-up increases the efficiency to \(31 \%\), what are (c) \(Q_{\text {gain }}\) and (d) \(Q_{\text {lost }}\) at the same work value?

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