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20.39. A Carnot engine whose low-temperature reservoir is at \(-90.0^{\circ} \mathrm{C}\) has an efficiency of 40.0\(\%\) . An engineer is assigned the problem of increasing this to 45.0\(\%\) . (a) By how many Celsius degrees must the temperature of the high-temperature reservoir be increased if the temperature of the low-temperature reservoir remains constant? (b) By how many Celsius degrees must the temperature of the low-temperature reservoir be decreased if the temperature of the high-temperature reservoir remains constant?

Short Answer

Expert verified
(a) Increase high temperature by 26.84°C. (b) Decrease low temperature by 15.26°C.

Step by step solution

01

Convert temperatures to Kelvin

To use the Carnot efficiency formula, temperatures should be in Kelvin. The low-temperature T_C is initially given as \(-90.0^{\circ} \mathrm{C}\) which is equivalent to:\[T_C = -90.0 + 273.15 = 183.15 \text{ K}\]We will denote the initial high-temperature reservoir as \(T_H\).
02

Use the efficiency formula for Carnot engines

The efficiency \( \eta \) of a Carnot engine is given by:\[\eta = 1 - \frac{T_C}{T_H}\]For an efficiency of 0.40 (or 40%), the equation becomes:\[0.40 = 1 - \frac{183.15}{T_H}\]Solving for \(T_H\), we find:\[T_H = \frac{183.15}{0.60} = 305.25 \text{ K}\]
03

Calculate new high temperature for increased efficiency

To achieve an efficiency of 45% (or 0.45), set up the equation:\[0.45 = 1 - \frac{183.15}{T_{H, \text{new}}}\]Solving for \(T_{H, \text{new}}\), we get:\[T_{H, \text{new}} = \frac{183.15}{0.55} = 332.09 \text{ K}\]
04

Determine the change in the high temperature

Convert the new temperature back to Celsius:\[T_{H, \text{new}} = 332.09 - 273.15 = 58.94 ^{\circ} \mathrm{C}\]Initial high temperature was \(305.25 - 273.15 = 32.10^{\circ} \mathrm{C}\).The change in temperature is:\[\Delta T = 58.94^{\circ} - 32.10^{\circ} = 26.84^{\circ} \mathrm{C}\]
05

Calculate new low temperature for increased efficiency

If the high-temperature reservoir remains constant at \(305.25 \,\mathrm{K}\), we adjust the low temperature to achieve 45% efficiency:\[0.45 = 1 - \frac{T_{C, \text{new}}}{305.25}\]Solving for \(T_{C, \text{new}}\), we find:\[T_{C, \text{new}} = (1 - 0.45) \times 305.25 = 167.89 \text{ K}\]
06

Determine the change in the low temperature

Convert the new low temperature back to Celsius:\[T_{C, \text{new}} = 167.89 - 273.15 = -105.26 ^{\circ} \mathrm{C}\]The initial low temperature in Celsius was \(-90.0^{\circ} \mathrm{C}\).The change in temperature is:\[\Delta T = -105.26 - (-90.0) = -15.26^{\circ} \mathrm{C}\]

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Key Concepts

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

Understanding Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat, work, temperature, and energy. It is foundational to understanding how engines, including Carnot engines, function. In a thermodynamic system, energy can be exchanged with its surroundings. Understanding thermodynamics involves comprehending a few key principles:
  • First Law of Thermodynamics: This is the law of energy conservation, which states that energy cannot be created or destroyed, only transformed from one form to another.
  • Second Law of Thermodynamics: This law introduces the concept of entropy. It explains that energy transformations are not 100% efficient and that some energy is always lost as heat.
  • Efficiency: The Carnot engine is a theoretical construct used to determine the maximum possible efficiency that any heat engine can achieve. This efficiency depends on the temperatures of the heat reservoirs.
For any engine, the efficiency is related to how much energy is converted into useful work versus how much is wasted as heat.
Converting Temperatures for Analysis
Temperature conversion is a crucial step when working with thermodynamics, particularly when applying formulas like the Carnot efficiency equation. In thermodynamics, calculations are often done using Kelvin rather than Celsius or Fahrenheit because Kelvin is an absolute temperature scale.
To convert from Celsius to Kelvin, you use the formula:\[T(K) = T(^{\circ}\mathrm{C}) + 273.15\]This conversion is essential because the Kelvin scale starts at absolute zero—the point at which all molecular motion ceases. For example, when converting a low-temperature reservoir at \(-90.0^{\circ}\mathrm{C}\) to Kelvin, it becomes \(183.15\,\mathrm{K}\). This ensures consistent and accurate application of thermodynamic principles.
Role of Heat Reservoir Temperatures
In a Carnot engine, two primary temperatures are considered: the temperature of the high-temperature reservoir (\(T_H\)) and the low-temperature reservoir (\(T_C\)). The difference between these temperatures determines the efficiency of the engine.
  • High-Temperature Reservoir \(T_H\): This is the source of heat energy. Maximizing its temperature increases the potential efficiency of the engine.
  • Low-Temperature Reservoir \(T_C\): This is where the waste heat is expelled. Decreasing its temperature can also enhance efficiency.
The Carnot efficiency formula,\[\eta = 1 - \frac{T_C}{T_H}\]shows that efficiency increases as \(T_C\) decreases or \(T_H\) increases. Adjusting these reservoir temperatures is key to optimizing an engine's performance, as seen in the exercise where increasing \(T_H\) or decreasing \(T_C\) resulted in higher efficiencies.

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Most popular questions from this chapter

20.25. A sophomore with nothing better to do adds heat to 0.350 \(\mathrm{kg}\) of ice at \(0.0^{\circ} \mathrm{C}\) until it is all melted. (a) What is the change in entropy of the water? (b) The source of heat is a very massive body at a temperature of \(25.0^{\circ} \mathrm{C}\) . What is the change in entropy of this body? (c) What is the total change in entropy of the water and the heat source?

20.34 . A box is separated by a partition into two parts of equal volume. The left side of the box contains 500 molecules of nitrogen gas; the right side contains 100 molecules of oxygen gas. The two gases are at the same temperature. The partition is punctured, and equilibrium is eventually attained. Assume that the volume of the box is large enough for each gas to undergo a free expansion and not change temperature, (a) On average, how many molecules of each type will there be in either half of the box? (b) What is the change in entropy of the system when the partition is punctured? the (c) What is the probability that the molecules will be found in the same distribution as they were before the partition was punctured - that is, 500 nitrogen molecules in the left half and 100 oxygen molecules in the right half?

20.36 . A lonely party balloon with a volume of 2.40 \(\mathrm{L}\) and containing 0.100 \(\mathrm{mol}\) of air is left behind to drift in the temporarily uninhabited and depressurized Intermational Space Station. Sunlight coming through a porthole heats and explodes the balloon, causing the air in it to undergo a free expansion into the empty station, whose total volume is 425 \(\mathrm{m}^{3}\) . Calculate the entropy change of the air during the expansion.

20.56. The maximum power that can be extracted by a wind turbine from an air stream is approximately $$ P=k d^{2} v^{3} $$ where \(d\) is the blade diameter \(v\) is the wind speed, and the constant \(k=0.5 \mathrm{W} \cdot \mathrm{s}^{3} / \mathrm{m}^{5} .\) (a) Explain the dependence of \(P\) on \(d\) and on \(v\) by considering a cylinder of air that passes over the turbine blades in time \(t(\text { Fig. } 20.31)\) . This cylinder has diameter \(d .\) length \(L=v t\) and density \(\rho .\) (b) The Mod-SB wind turbine at Kahaku on the Hawaiian island of Oahu has a blade diameter of 97 \(\mathrm{m}\) (slightly longer than a football field sits atop a \(58-\mathrm{m}\) tower. It can produce 3.2 \(\mathrm{MW}\) of electric power. Assuming 25\(\%\) efficiency, what wind speed is required to produce this amount of power? Give your answer in \(\mathrm{m} / \mathrm{s}\) and in \(\mathrm{km} / \mathrm{h}\) . (c) Commercial wind turbines are commonly located in or downwind of mountain passes. Why?

An aircraft engine takes in 9000 \(\mathrm{J}\) of heat and discards 6400 \(\mathrm{J}\) each cycle. (a) What is the mechanical work output of the engine during one cycle? (b) What is the thermal efficiency of the engine?

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