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A heat engine operates with an efficiency of 0.5. What can the temperatures of the high-temperature and lowtemperature reservoirs be? a) TH=600 K and TL=100 K b) TH=600 K and TL=200 K c) TH=500 K and TL=200 K d) TH=500 K and TL=300 K e) TH=600 K and TL=300 K

Short Answer

Expert verified
Answer: The high-temperature and low-temperature reservoirs for a heat engine operating with an efficiency of 0.5 are TH = 600 K and TL = 200 K.

Step by step solution

01

Setup the efficiency formula

Write down the formula for efficiency and plug in the given efficiency value: 0.5=1TLTH
02

Test each option

Plug in the temperatures mentioned in each option and check if the equation is satisfied. (a) TH=600 K and TL=100 K: 0.5=1100600=116=56 Answer (a) not correct. (b) TH=600 K and TL=200 K: 0.5=1200600=113 Answer (b) is correct. (c) TH=500 K and TL=200 K: 0.5=1200500=125=35 Answer (c) not correct. (d) TH=500 K and TL=300 K: 0.5=1300500=135=25 Answer (d) not correct. (e) TH=600 K and TL=300 K: 0.5=1300600=112 Answer (e) not correct.
03

Conclusion

The heat engine operating with an efficiency of 0.5 would have the high-temperature and low-temperature reservoirs as TH=600 K and TL=200 K (Option b).

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Key Concepts

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

Heat Engine
A heat engine is a fascinating device in the realm of thermodynamics. It is designed to convert thermal energy, or heat, into mechanical work. Think of it as a transformation process where energy changes forms to power machines. Here's how it works: - **Thermal Energy Input**: The engine absorbs heat from a high-temperature reservoir, which is like a very hot source. This source provides the energy needed for the engine to do work. - **Mechanical Work Output**: Part of the absorbed heat is transformed into mechanical work. This can power anything from a car to a power plant generator. - **Heat Energy Release**: Not all the absorbed heat turns into work; some of it is expelled to a low-temperature reservoir. This is where the engine dumps the leftover heat. The operation of a heat engine is cyclical. It continuously absorbs heat, performs work, and releases residual heat. Each cycle aims to maximize work while minimizing the energy lost as waste heat. Understanding the basic cycle helps in grasping the importance of efficiency, a key measure of the engine's effectiveness.
Efficiency Formula
The efficiency of a heat engine is a crucial concept that tells us how well the engine converts heat into work. It can be defined simply as the ratio of work output to heat input. Here's the simple mathematical representation:Efficiency=WQH=1TLTHBeing a fraction of the absorbed heat turned into work, efficiency explains how effectively a heat engine operates. - **Efficiency Value**: Efficiency is a number between 0 and 1 (or 0% to 100%). A value of 0.5, for instance, implies that the engine converts 50% of the heat from the high-temperature reservoir into work.- **Ideal Conditions**: The equation also shows that efficiency increases when the temperature difference between the high-temperature (TH) and low-temperature (%TL) reservoirs increases.- **Practical Implications**: In real-world scenarios, it's rare for an engine to achieve maximum efficiency due to energy losses through processes like friction.Knowing the efficiency helps engineers and scientists enhance the design and functionality of engines, making them more environmentally friendly and cost-effective.
Temperature Reservoirs
Temperature reservoirs are integral components of a heat engine, represented by the high-temperature (TH) and low-temperature (TL) reservoirs. They essentially function as sources and sinks for heat energy within the engine system.- **High-Temperature Reservoir (TH)**: This is akin to a hot body that supplies heat to the engine. Its temperature is crucial as it determines the amount of energy available to do work. The higher the temperature, the more potential energy.- **Low-Temperature Reservoir (TL)**: This acts like a cooler that receives unwanted heat energy after work has been extracted. Maintaining a lower temperature here is necessary to maximize the engine's efficiency.- **Role in Efficiency**: The difference in temperatures (THTL) dictates the efficiency, as greater variances can potentially allow more work to be extracted from the engine.In practice, selecting appropriate temperature reservoirs is vital for optimizing performance and efficiency. Engineers carefully choose or design conditions that balance the heat flow, ensuring a smooth and effective energy conversion process. Understanding these reservoirs is key to enhancing engine performance in various applications including vehicles, industrial machinery, and power generation.

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Most popular questions from this chapter

A heat engine consists of a heat source that causes a monatomic gas to expand, pushing against a piston, thereby doing work. The gas begins at a pressure of 300.kPa, a volume of 150.cm3, and room temperature, 20.0C. On reaching a volume of 450.cm3, the piston is locked in place, and the heat source is removed. At this point, the gas cools back to room temperature. Finally, the piston is unlocked and used to isothermally compress the gas back to its initial state. a) Sketch the cycle on a pV -diagram. b) Determine the work done on the gas and the heat flow out of the gas in each part of the cycle. c) Using the results of part (b), determine the efficiency of the engine.

20.14 Imagine dividing a box into two equal parts, part A on the left and part B on the right. Four identical gas atoms, numbered 1 through 4 , are placed in the box. What are most probable and second most probable distributions (for example, 3 atoms in A,1 atom in B ) of gas atoms in the box? Calculate the entropy, S, for these two distributions. Note that the configuration with 3 atoms in A and 1 atom in B and that with 1 atom in A and three atoms in B count as different configurations.

If the Earth is treated as a spherical black body of radius 6371 km, absorbing heat from the Sun at a rate given by the solar constant (1370. W/m 2 ) and immersed in space with an approximate temperature of Tsp=50.0 K, it radiates heat back into space at an equilibrium temperature of 278.9 K. (This is a slight refinement of the model in Chapter 18.) Estimate the rate at which the Earth gains entropy in this model.

Suppose a person metabolizes 2000. kcal/day. a) With a core body temperature of 37.0C and an ambient temperature of 20.0C, what is the maximum (Carnot) efficiency with which the person can perform work? b) If the person could work with that efficiency, at what rate, in watts, would they have to shed waste heat to the surroundings? c) With a skin area of 1.50 m2, a skin temperature of 27.0C and an effective emissivity of e=0.600, at what net rate does this person radiate heat to the 20.0C surroundings? d) The rest of the waste heat must be removed by evaporating water, either as perspiration or from the lungs At body temperature, the latent heat of vaporization of water is 575cal/g. At what rate, in grams per hour, does this person lose water? e) Estimate the rate at which the person gains entropy. Assume that all the required evaporation of water takes place in the lungs, at the core body temperature of 37.0C.

What is the minimum amount of work that must be done to extract 500.0 J of heat from a massive object at a temperature of 27.0C while releasing heat to a high temperature reservoir with a temperature of 100.0C?

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