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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.

Short Answer

Expert verified
Lowering the cold reservoir temperature gives greater efficiency improvement.

Step by step solution

01

Understand Carnot Efficiency

The efficiency of a Carnot engine is determined by the temperatures of the hot and cold reservoirs. The formula for the efficiency is given by:\[\eta = 1 - \frac{T_c}{T_h}\]where \(T_h\) is the absolute temperature of the hot reservoir and \(T_c\) is the absolute temperature of the cold reservoir. The temperatures must be in Kelvin.
02

Calculate Initial Efficiency

Plug in the initial temperatures into the Carnot efficiency formula. Here, \(T_h = 650\; \text{K}\) and \(T_c = 350\; \text{K}\).\[\eta_{initial} = 1 - \frac{350}{650} = 1 - 0.5385 = 0.4615\]So, the initial efficiency of the engine is 46.15%.
03

Calculate Efficiency with Increased Hot Temperature

Increase the hot temperature by 40 K: \(T_h = 650 + 40 = 690\; \text{K}\). Calculate the new efficiency:\[\eta_{hot} = 1 - \frac{350}{690} = 1 - 0.5072 = 0.4928\]So, the efficiency with increased hot reservoir temperature is 49.28%.
04

Calculate Efficiency with Decreased Cold Temperature

Decrease the cold temperature by 40 K: \(T_c = 350 - 40 = 310\; \text{K}\). Calculate the new efficiency:\[\eta_{cold} = 1 - \frac{310}{650} = 1 - 0.4769 = 0.5231\]So, the efficiency with decreased cold reservoir temperature is 52.31%.
05

Compare the Results

Compare the efficiencies obtained in Steps 3 and 4: - With increased hot reservoir temperature: 49.28% - With decreased cold reservoir temperature: 52.31% The change that gives greater improvement is decreasing the cold reservoir temperature.

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Key Concepts

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

Carnot cycle
The Carnot cycle is a theoretical model that defines the most efficient way to convert heat into work or to refrigerate using heat. This cycle operates as an idealized thermodynamic cycle involving four key steps: two isothermal processes and two adiabatic processes. Here’s a brief look at each of these steps:
  • Isothermal Expansion: The gas within the engine expands while absorbing heat from the hot reservoir, remaining at a constant high temperature.
  • Adiabatic Expansion: The gas continues to expand, doing work on the surroundings, and reduces its temperature without any heat exchange.
  • Isothermal Compression: The gas is compressed at a low temperature, releasing heat to the cold reservoir while remaining at this constant low temperature.
  • Adiabatic Compression: The gas is compressed further, increasing its temperature back to the initial state, all without heat exchange.
The importance of the Carnot cycle lies in its definition of the upper limit of efficiency for any heat engine working between two temperatures. It's idealized because no real engine can achieve this perfect efficiency due to practical limitations like friction and heat loss.
thermodynamic efficiency
Thermodynamic efficiency in the context of a Carnot engine is defined as the measure of how effectively an engine converts heat from a high-temperature reservoir into work. The efficiency is expressed mathematically as:\[\eta = 1 - \frac{T_c}{T_h}\]where \(T_h\) is the absolute temperature of the hot reservoir, and \(T_c\) is the absolute temperature of the cold reservoir, both measured in Kelvin.
The efficiency value varies between zero and one. An efficiency of one (or 100%) would imply a perfect conversion of heat into work without any losses, as seen in the hypothetical Carnot cycle. In practice, due to various losses such as friction and heat dissipations, real engines exhibit lower efficiencies than the theoretical Carnot efficiency.
Understanding this concept helps in designing engines that optimize the conversion of heat to usable energy, though they will never quite reach Carnot's ideal.
temperature reservoirs
Temperature reservoirs in a Carnot engine are crucial components that dictate its efficiency. These reservoirs are large bodies that maintain a constant temperature even as they exchange heat with the engine.
  • Hot Reservoir: Supplies the engine with heat at a high temperature, \(T_h\). The larger this temperature, the higher the potential efficiency according to the Carnot formula.
  • Cold Reservoir: Absorbs the heat discarded by the engine at a lower temperature, \(T_c\). Reducing \(T_c\) can increase efficiency by reducing the fraction \(\frac{T_c}{T_h}\).
The study of changing these temperatures, as seen in the original problem, demonstrates different methods to enhance engine efficiency. Raising the temperature of the hot reservoir or lowering the temperature of the cold reservoir both lead to greater efficiency, but as calculated, lowering the cold side often results in a greater improvement if feasible.
Kelvin temperature scale
The Kelvin temperature scale is an absolute scale used predominantly in scientific contexts, such as physics and chemistry, and is crucial for thermodynamic calculations involving heat engines like the Carnot engine. Unlike Celsius or Fahrenheit, Kelvin does not use degrees but rather an absolute measure of thermal energy in a system.
  • The key point about Kelvin is its starting point: absolute zero (0 K), which is theoretically the lowest possible temperature where molecular motion ceases.
  • The scale progresses similarly to degrees Celsius, where a difference of one Kelvin is equivalent to a difference of one degree Celsius.
This scale is particularly important when dealing with the Carnot cycle and thermodynamic efficiency calculations because it provides a true representation of temperature free from arbitrary baselines, thus accurately reflecting thermal energy content.

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Most popular questions from this chapter

One mole of neon, a monatomic gas, starts out at conditions of standard temperature and pressure. The gas is heated at constant volume until its pressure is tripled, then further heated at constant pressure until its volume is doubled. Assume that neon behaves as an ideal gas. For the entire process, find the heat added to the gas.

The pressure of a monatomic ideal gas \(\left(\gamma=\frac{5}{3}\right)\) doubles during an adiabatic compression. What is the ratio of the final volume to the initial volume?

When a .22 -caliber rifle is fired, the expanding gas from the burning gunpowder creates a pressure behind the bullet. This pressure causes the force that pushes the bullet through the barrel. The barrel has a length of 0.61 \(\mathrm{m}\) and an opening whose radius is \(2.8 \times 10^{-3} \mathrm{m} . \mathrm{A}\) bullet (mass \(=2.6 \times 10^{-3} \mathrm{kg}\) ) has a speed of 370 \(\mathrm{m} / \mathrm{s}\) after passing through this barrel. Ignore friction and determine the average pressure of the expanding gas.

Multiple-Concept Example 6 provides a review of the concepts that play roles here. An engine has an efficiency of 64\(\%\) and produces 5500 \(\mathrm{J}\) of work. Determine \((\mathrm{a})\) the input heat and \(\quad\) (b) the rejected heat.

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.

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