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Chapter 15: Problem 90

Carnot engine A has an efficiency of \(0.60,\) and Carnot engine \(\mathrm{B}\) has an efficiency of 0.80. Both engines utilize the same hot reservoir, which has a temperature of \(650 \mathrm{K}\) and delivers \(1200 \mathrm{J}\) of heat to each engine. Find the magnitude of the work produced by each engine and the temperatures of the cold reservoirs that they use.

Short Answer

Expert verified
Engine A does 720 J of work with a 260 K cold reservoir; Engine B does 960 J of work with a 130 K cold reservoir.

Step by step solution

01

Understand Carnot Efficiency

The efficiency of a Carnot engine is given by the formula: \[\text{Efficiency} = 1 - \frac{T_C}{T_H},\]where \(T_H\) is the hot reservoir temperature and \(T_C\) is the cold reservoir temperature. We are given the efficiencies of engines A and B as 0.60 and 0.80, respectively, and \(T_H = 650\, \text{K}\). We need to solve for \(T_C\) for each engine.
02

Calculate Cold Reservoir Temperature for Engine A

For engine A, with an efficiency of 0.60:\(0.60 = 1 - \frac{T_{C_A}}{650}.\)Rearrange the equation to solve for \(T_{C_A}\):\(\\frac{T_{C_A}}{650} = 1 - 0.60 = 0.40,\)\(\T_{C_A} = 0.40 \times 650 = 260\, \text{K}.\)
03

Calculate Cold Reservoir Temperature for Engine B

For engine B, with an efficiency of 0.80:\(0.80 = 1 - \frac{T_{C_B}}{650}.\)Rearrange the equation to solve for \(T_{C_B}\):\(\frac{T_{C_B}}{650} = 1 - 0.80 = 0.20,\)\(\T_{C_B} = 0.20 \times 650 = 130\, \text{K}.\)
04

Calculate Work Produced by Engine A

The work produced by an engine is given by:\[W = Q_H \times \text{Efficiency},\]where \(Q_H = 1200 \, \text{J}\). For engine A:\[W_A = 1200 \, \text{J} \times 0.60 = 720 \, \text{J}.\]
05

Calculate Work Produced by Engine B

Similarly, for engine B:\[W_B = 1200 \, \text{J} \times 0.80 = 960 \, \text{J}.\]

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Key Concepts

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

Efficiency
The concept of efficiency in a Carnot engine is a crucial part of understanding thermodynamics. Efficiency is essentially a measure of how well an engine converts heat into work. For Carnot engines, which are ideal engines, efficiency can be calculated using a straightforward formula:
  • Efficiency = \(1 - \frac{T_C}{T_H}\)
Here, \(T_H\) is the temperature of the hot reservoir, and \(T_C\) is the temperature of the cold reservoir. The challenge is to use this formula to find out the unknowns in Carnot engine problems. In the context of engines A and B:
  • Engine A has an efficiency of 0.60, meaning 60% of the heat absorbed from the hot reservoir is converted into work.
  • Engine B has a higher efficiency of 0.80, converting 80% of the heat into work, which is more desirable as it means the engine is more effective.
The efficiency not only tells us about the engine's performance but also helps us find the temperature of the cold reservoir by rearranging and solving the efficiency equation.
Temperature Reservoirs
Temperature reservoirs are essential components of a Carnot engine. These reservoirs are large systems that provide or absorb heat without changing their temperatures.
  • The hot reservoir (\(T_H\)) provides the heat energy needed by the engine.
  • The cold reservoir (\(T_C\)) absorbs the waste heat after the engine has performed work.
In our exercise, both engines use the same hot reservoir at 650 K. To find the cold reservoirs' temperatures:
  • For engine A with 0.60 efficiency, \(260 \, \text{K}\) is the temperature of its cold reservoir.
  • For engine B with 0.80 efficiency, its cold reservoir is even colder at \(130 \, \text{K}\).
This means engine B is more efficient not only in terms of output work but also in creating a larger temperature difference between its reservoirs, a key factor in higher efficiencies.
Thermodynamics
Thermodynamics is the study of energy, entropy, and the physical laws that govern the behavior of systems exchanging heat and work. The Carnot engine is a perfect example to explain these principles.
  • Second Law of Thermodynamics: No engine can be more efficient than a Carnot engine working between the same two reservoirs.
  • Efficiency: It involves converting heat energy into usable work, where the work is calculated as \(W = Q_H \times \text{Efficiency}\).
In our example, both engines receive the same amount of heat, 1200 J, from the hot reservoir. Using the efficiency:
  • Engine A produces \(720 \, \text{J}\) of work, calculated as \(1200 \, \text{J} \times 0.60\).
  • Engine B produces more, \(960 \, \text{J}\) of work, as \(1200 \, \text{J} \times 0.80\).
This demonstrates how understanding thermodynamic laws and principles can help optimize energy use and engine design.

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Most popular questions from this chapter

You and your team are tasked with evaluating the parameters of a high- performance engine. The engine compresses the fuel-air mixture in the cylinder to make it ignite rather than igniting it with a spark plug (this is how diesel engines operate). You are given the compression ratio for the cylinders in the engine, which is the ratio of the initial volume of the cylinder (before the piston compresses the gas) to the final volume of the cylinder (after the gas is compressed): \(V_{i}: V_{\mathrm{f}}\). Assume that the fuel- gas mixture enters the cylinder at a temperature of \(22.0^{\circ} \mathrm{C}\), and that the gas behaves like an ideal gas with \(\gamma=7 / 5 .\) (a) If the compression ratio is \(15.4: 1,\) what is the final temperature of the gas if the compression is adiabatic? (b) With everything else the same as in (a), what is the final temperature of the gas if the compression ratio is increased to \(17.0: 1 ?\) (c) Everything else being the same, for which compression ratio do you think the engine runs more efficiently? Give a qualitative argument for your answer.

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 \(2.8 \times 10^{-3} \mathrm{m}\). 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.

Heat engines take input energy in the form of heat, use some of that energy to do work, and exhaust the remainder. Similarly, a person can be viewed as a heat engine that takes an input of internal energy, uses some of it to do work, and gives off the rest as heat. Suppose that a trained athlete can function as a heat engine with an efficiency of 0.11. (a) What is the magnitude of the internal energy that the athlete uses in order to do \(5.1 \times\) \(10^{4} \mathrm{J}\) of work? (b) Determine the magnitude of the heat the athlete gives off.

Argon is a monatomic gas whose atomic mass is 39.9 u. The temperature of eight grams of argon is raised by \(75 \mathrm{K}\) under conditions of constant pressure. Assuming that argon behaves as an ideal gas, how much heat is required?

Suppose a monatomic ideal gas is contained within a vertical cylinder that is fitted with a movable piston. The piston is frictionless and has a negligible mass. The area of the piston is \(3.14 \times 10^{-2} \mathrm{m}^{2},\) and the pressure outside the cylinder is \(1.01 \times 10^{5}\) Pa. Heat \((2093 \mathrm{J})\) is removed from the gas. Through what distance does the piston drop?
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