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Chapter 12: Problem 68

Even Carnot engine cannot give \(100 \%\) efficiency because we cannot (A) prevent radiation. (B) find ideal sources. (C) reach absolute zero temperature. (D) eliminate friction.

Short Answer

Expert verified
The most accurate reason why even a Carnot engine cannot have 100% efficiency is because we cannot reach absolute zero temperature (option C). The efficiency of the Carnot engine depends on the difference between the high and low temperatures, and since reaching absolute zero is not possible, 100% efficiency cannot be achieved.

Step by step solution

01

Analyze the given options

Begin by analyzing the given options (A) to (D), focusing on which one is most related to the temperature aspect of the Carnot engine and will affect its efficiency the most: (A) Prevent radiation - Radiation can cause energy loss, which reduces the efficiency of the system. However, the Carnot engine is a theoretical model, assuming no energy loss. Thus, radiation is not the major factor here. (B) Find ideal sources - It is not easy to find ideal sources in reality, but in a theoretical model, idealized sources can be assumed. This reasoning does not directly relate to the efficiency limitation of the Carnot engine. (C) Reach absolute zero temperature - The efficiency of the Carnot engine depends on the difference between the high and low temperatures, where the lower temperature (T_low) can never equal absolute zero. Hence, this option seems to be a plausible candidate for our answer. (D) Eliminate friction - Friction can also cause energy loss and reduce the efficiency of the system. However, as stated before, the Carnot engine is a theoretical model, and it assumes zero friction. Thus, this aspect is not the main reason for the efficiency limitation.
02

Choose the correct option

After analyzing the given options, we can conclude that the option most related to the temperature aspect of the Carnot engine and its efficiency limitation is: (C) Reach absolute zero temperature The efficiency of the Carnot engine is given by: \( Eff = 1 - \frac{T_{low}}{T_{high}} \) If \(T_{low}\) reaches absolute zero (0 K), the efficiency would become 100%, but reaching absolute zero is not possible. Hence, even a Carnot engine cannot have 100% efficiency due to this reason.

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Key Concepts

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

Absolute Zero Temperature
Absolute zero is a theoretical temperature at which particles in a substance would have minimal motion, meaning no kinetic energy. This state is theoretically considered as 0 Kelvin, or -273.15掳C. At this temperature, molecular movement ceases, making it an interesting point in thermodynamic studies.
To achieve 100% efficiency in a Carnot engine, the cold reservoir must be at absolute zero. The formula for Carnot efficiency is given by:
  • \[ \text{Efficiency} = 1 - \frac{T_{\text{low}}}{T_{\text{high}}} \]
Since it is impossible to reach 0 Kelvin, the lower temperature \(T_{\text{low}}\) can never be absolute zero. This concept puts a limit on the maximum efficiency any heat engine can achieve. Absolute zero is a theoretical baseline for energy studies, but practical systems can't reach it, leaving efficiency less than 100%.
Thermodynamics
Thermodynamics is the branch of physics that deals with heat, work, and energy transfer. It includes several laws that dictate how energy conversion and heat interactions occur in various systems. The Carnot engine is a theoretical construct based on these principles that achieves maximum efficiency as defined by these laws.
The second law of thermodynamics particularly affects efficiency limits in practical engines. It states that energy will always spread out or dissipate over time, which means there are inevitable losses. No real engine can achieve the perfectly reversible process required for Carnot efficiency because there are always losses due to irreversibility, often referred to as entropy.
Carnot's theorem shows that no engine operating between two heat reservoirs can be more efficient than a Carnot engine. Understanding this concept helps us grasp why absolute zero plays a crucial role in defining theoretical efficiency limits.
Ideal Sources
Ideal sources are theoretical constructs where a system can perfectly take or reject energy without any losses. While they are used in physics to make calculations simpler and understand potential maximum efficiencies, in reality, such ideal conditions are non-existent.
In a perfectly ideal scenario (ideal sources), a Carnot engine assumes infinite thermal mass that can exchange heat without a change in temperature. However, real-world materials and environments always introduce inefficiencies and energy losses. This is why ideal sources are important for calculations but cannot be practically realized.
Because actual materials have imperfections and limitations, real engines will always operate below the ideal levels of efficiency predicted in theoretical models.
Friction in Engines
Friction is a force that resists motion between two surfaces or objects that interact. In engines, friction leads to wear and tear, heat generation, and inevitable energy loss.
Although a Carnot engine is a theoretical concept assuming no friction, real engines must deal with this unavoidable challenge. Friction not only reduces efficiency by converting useful energy into heat, but it also requires extra energy input to maintain movement and operations.
Reducing friction is essential for more efficient engines. Techniques include using lubricants, designing smoother surfaces, or using materials that minimize frictional effects. However, completely eliminating it is impossible, which means efficiency always takes a hit due to friction in practical scenarios.

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Most popular questions from this chapter

An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume \(V_{1}\) and contains ideal gas at pressure \(P_{1}\) and temperature \(T_{1}\). The other chamber has volume \(V_{2}\) and contains ideal gas at pressure \(P_{2}\) and temperature \(T_{2}\). If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be: (A) \(\frac{T_{1} T_{2}\left(P_{1} V_{1}+P_{2} V_{2}\right)}{P_{1} V_{1} T_{1}+P_{2} V_{2} T_{2}}\) (B) \(\frac{T_{1} T_{2}\left(P_{1} V_{1}+P_{2} V_{2}\right)}{P_{1} V_{1} T_{2}+P_{2} V_{2} T_{1}}\) (C) \(\frac{P_{1} V_{1} T_{1}+P_{2} V_{2} T_{2}}{P_{1} V_{1}+P_{2} V_{2}}\) (D) \(\frac{P_{1} V_{1} T_{2}+P_{2} V_{2} T_{1}}{P_{1} V_{1}+P_{2} V_{2}}\)

Which of the following is incorrect regarding the first law of thermodynamics? [2005] (A) It is not applicable to any cyclic process. (B) It is a restatement of the principle of conservation of energy. (C) It introduces the concept of the internal energy. (D) It introduces the concept of the entropy.

Equal amount of same gas in two similar cylinders, \(A\) and \(B\), compressed to same final volume from same initial volume one adiabatically and another isothermally, respectively, then (A) final pressure in \(A\) is more than in \(B\). (B) final pressure in \(B\) is greater than in \(A\). (C) final pressure in both equal. (D) for the gas, value of \(\gamma=\frac{C_{p}}{C_{Y}}\) is required.

An ideal heat engine working between temperatures \(T_{H}\) and \(T_{L}\) has efficiency \(\eta\). If both the temperature are raised by \(100 \mathrm{~K}\) each, the new efficiency of the heat engine will be (A) equal to \(\eta\). (B) greater than \(\eta\). (C) less than \(\eta\). (D) greater or less than \(\eta\) depending upon the nature of the working substance.

When a system is taken from state \(i\) to state \(f\) along the path iaf, it is found \(Q=50\) cal and \(W=20\) cal. Along the path ibf, \(Q=36\) cal. \(W\) along the path \(i b f\) is [2007] (A) \(6 \mathrm{cal}\) (B) \(16 \mathrm{cal}\) (C) \(66 \mathrm{cal}\) (D) \(14 \mathrm{cal}\)
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