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An inventor claims that he has created a water-driven engine with an efficiency of 0.200 that operates between thermal reservoirs at \(4^{\circ} \mathrm{C}\) and \(20 .{ }^{\circ} \mathrm{C}\). Is this claim valid?

Short Answer

Expert verified
Answer: No, the inventor's claim is invalid as the engine's efficiency of 0.200 is higher than the maximum possible efficiency (Carnot efficiency) of 0.0546 for an engine operating between the given temperatures.

Step by step solution

01

Conversion of temperatures to Kelvin

Convert the given temperatures in Celsius to Kelvin. To do so, add 273.15 to the Celsius temperature value. We'll call the lower temperature T1 and the higher temperature T2. T1 = 4°C + 273.15 = 277.15 K T2 = 20°C + 273.15 = 293.15 K
02

Calculate the Carnot efficiency

The Carnot efficiency is the maximum possible efficiency of any heat engine operating between two temperatures T1 and T2. It is determined by the following formula: Carnot efficiency = 1 - (T1 / T2) Now, let's plug in the values for T1 and T2: Carnot efficiency = 1 - (277.15 / 293.15) = 1 - 0.9454 ≈ 0.0546
03

Compare the efficiencies

Now that we've calculated the Carnot efficiency (0.0546), let's compare it to the efficiency claimed by the inventor (0.200). If the inventor's engine efficiency is greater than the Carnot efficiency, his claim is invalid: Inventor's efficiency: 0.200 Carnot efficiency: 0.0546 Since 0.200 > 0.0546, the inventor's claim of an engine with an efficiency of 0.200 is invalid, as it is higher than the maximum possible efficiency (Carnot efficiency) for an engine operating between the given temperatures.

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Key Concepts

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

Thermodynamics
Thermodynamics is a branch of physics that deals with heat, work, and temperature, and their relation to energy, radiation, and physical properties of matter. The key idea of thermodynamics is the conversion of heat energy into mechanical energy, which is precisely what heat engines do.

There are several laws of thermodynamics that govern how energy is transferred and transformed. The first law, also known as the law of energy conservation, states that energy cannot be created or destroyed in an isolated system. The second law introduces the concept of entropy, indicating that thermodynamic processes are irreversible and that energy tends to disperse or spread out unless acted upon by work. This plays a fundamental role in determining the efficiency of engines, as some energy is always lost as waste heat, which is why no engine can be 100% efficient.

In examining the inventor's claim, the first step is to analyze the thermodynamic process involved, understand the energy transfers, and use relevant equations to calculate the efficiency of the claimed process—a task that involves the use of the Carnot efficiency.
Heat Engines
Heat engines are devices that convert thermal energy into mechanical work. The basic operation of a heat engine involves transferring heat from a higher temperature source (the hot reservoir) to a lower temperature sink (the cold reservoir) and, in the process, doing work. The efficiency of a heat engine is a measure of how well it can convert the input heat into useful work.

One of the most important concepts when discussing heat engines is the Carnot cycle, which is a theoretical model that defines the maximum possible efficiency a heat engine can achieve between two thermal reservoirs. Although real engines cannot reach this idealized efficiency due to various practical limitations such as friction and material imperfections, the Carnot cycle sets the benchmark for the best possible performance. The efficiency of a Carnot engine is dependent solely on the temperatures of the hot and cold reservoirs and not on the specific details of the engine itself. This ties into the inventor's claim, which can be assessed by comparing the claimed efficiency with the theoretical limit set by the Carnot cycle.
Temperature Conversion
Temperature conversion is a fundamental practice in science and engineering as different processes operate more naturally within different temperature scales. The most commonly used temperature scales are Celsius (°C) and Kelvin (K). In thermodynamics, the Kelvin scale is particularly important because it is an absolute temperature scale, starting at absolute zero, where all molecular motion ceases.

The conversion from Celsius to Kelvin is straightforward: simply add 273.15 to the Celsius temperature. This is crucial when computing the Carnot efficiency because the formula requires temperatures to be in Kelvin. It's important to perform this step accurately to ensure correct calculation of thermal efficiency.

The step-by-step solution to the exercise showed how to convert the temperatures of the thermal reservoirs from Celsius to Kelvin. Using these Kelvin temperatures, we were then able to calculate the Carnot efficiency and thereby evaluate the validity of the inventor's claim. In doing so, we illustrated not only the practical application of temperature conversion but also its critical role in thermodynamic calculations and the assessment of engine efficiencies.

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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 \(300 . \mathrm{kPa}\), a volume of \(150 . \mathrm{cm}^{3}\), and room temperature, \(20.0^{\circ} \mathrm{C}\). On reaching a volume of \(450 . \mathrm{cm}^{3}\), 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 \(p V\) -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.

A key feature of thermodynamics is the fact that the internal energy, \(E_{\text {int }}\) of a system and its entropy, \(S\), are state variables; that is, they depend only on the thermodynamic state of the system and not on the processes by which it reached that state (unlike, for example, the heat content, \(Q\) ). This means that the differentials \(d E_{\text {int }}=T d S-p d V\) and \(d S=\) \(T^{-1} d E_{\text {int }}+p T^{-1} d V,\) where \(T\) is temperature (in kelvins), \(p\) is pressure, and \(V\) is volume, are exact differentials as defined in calculus. What relationships follow from this fact?

1 .00 mole of a monatomic ideal gas at a pressure of 4.00 atm and a volume of \(30.0 \mathrm{~L}\) is isothermically expanded to a pressure of 1.00 atm and a volume of \(120.0 \mathrm{~L}\). Next, it is compressed at a constant pressure until its volume is \(30.0 \mathrm{~L}\), and then its pressure is increased at the constant volume of \(30.0 \mathrm{~L}\). What is the efficiency of this heat engine cycle?

Explain how it is possible for a heat pump like that in Example 20.2 to operate with a power of only \(6.28 \mathrm{~kW}\) and heat a house that is losing thermal energy at a rate of \(21.98 \mathrm{~kW}\).

An ideal gas is enclosed in a cylinder with a movable piston at the top. The walls of the cylinder are insulated, so no heat can enter or exit. The gas initially occupies volume \(V_{1}\) and has pressure \(p_{1}\) and temperature \(T_{1}\). The piston is then moved very rapidly to a volume of \(V_{2}=3 V_{1}\). The process happens so rapidly that the enclosed gas does not do any work. Find \(p_{2}, T_{2},\) and the change in entropy of the gas.

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