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A Carnot engine is operated between two heat rescrvoirs at temperatures of \(520 \mathrm{~K}\) and \(300 \mathrm{~K}\). (a) If the engine receives \(6.45 \mathrm{~kJ}\) of heat energy Irom the reservoir at \(520 \mathrm{~K}\) in each eycle, how many joules per cycle does it discard to the reservoir at \(300 \mathrm{~K}^{\prime} ?\) (b) How much mechanical work is performed by the engine during each cycle? (c) What is the thermal efficiency of the engine?

Short Answer

Expert verified
The engine discards 3720.5 Joules of energy to the reservoir at 300 K in each cycle. It performs 2726.35 Joules of work during each cycle and its thermal efficiency is 42.3%.

Step by step solution

01

Determine Energy discarded

First, one must calculate the heat discarded to the reservoir at \(300 \mathrm{~K}\). The efficiency of a Carnot engine is given by \(1 - \frac{T_{cold}}{T_{hot}}\). Here, \(T_{cold} = 300 \mathrm{~K}\) and \(T_{hot} = 520 \mathrm{~K}\). So, the efficiency = \(1 - \frac{300}{520}\) = 0.423. Since efficiency is also equal to \(\frac{Work~done}{Heat~input}\), we can express the heat discarded (\(Q_{cold}\)) in terms of the heat input (\(Q_{hot} = 6450 \mathrm{~J}\)) and the work done (\(W\)): \(W = Q_{hot} - Q_{cold}\). Therefore, \(Q_{cold} = Q_{hot} - W = Q_{hot} - (Efficiency * Q_{hot}) = 6450(1 - 0.423) \mathrm{~J} = 3720.5 \mathrm{~J}\)
02

Determine Work performed by the engine

The work done by the Carnot engine can be calculated directly using the definition of efficiency: \(Work = Efficiency * Heat~input = 0.423 * 6450 \mathrm{~J} = 2726.35 \mathrm{~J}\)
03

Calculate the thermal efficiency of the engine

The thermal efficiency of the Carnot engine has already been calculated in Step 1 as \(0.423\) or \(42.3\%\) when expressed as a percentage. It represents the proportion of heat input that is converted to work in the engine.

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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 concerned with heat and temperature and their relation to energy and work. It forms the foundation for a variety of scientific disciplines and industrial applications. The laws of thermodynamics govern how and why energy is transferred in various systems.

At its core, thermodynamics deals with the concepts of internal energy, entropy, and the ability of a system to perform work when subjected to a temperature difference. For instance, the problem involving a Carnot engine highlights the practical application of these principles. A Carnot engine operates between two heat reservoirs, transferring heat from a high-temperature source to a lower-temperature sink, converting part of the heat into work. This process is a prime example of the second law of thermodynamics, which states that temperature differences between systems drive the spontaneous flow of heat and the potential to do work.

The efficiency of any thermodynamic cycle, especially the idealized Carnot cycle, reflects the limitations imposed by the second law. No engine can be 100% efficient, as some heat will always be lost in the process due to inevitable increases in entropy. Understanding thermodynamics is essential for grasping how thermal machines like the Carnot engine function, thus making it a critical area of study in physical sciences and engineering.
Thermal Efficiency
Thermal efficiency is a dimensionless performance measure of a heat engine that describes the fraction of heat converted into work. It's defined by the ratio of work output to the heat input from a high-temperature source. The concept of thermal efficiency is central to the study of energy systems since increasing efficiency is a key goal in the design of thermal engines, leading to energy conservation and better resource management.

In the context of our exercise, the Carnot engine displays a typical application of this concept. The thermal efficiency of the Carnot engine is calculated by the formula:
\[\begin{equation} \text{Efficiency} = 1 - \frac{T_{cold}}{T_{hot}} \end{equation}\] where \(T_{hot}\) is the temperature of the hot reservoir and \(T_{cold}\) is the temperature of the cold reservoir, both in kelvins.

For instance, with a hot reservoir at \(520 \text{K}\) and a cold reservoir at \(300 \text{K}\), the theoretical maximum efficiency is 42.3%, which means 42.3% of the heat energy taken from the hot reservoir is converted to work, while the rest is discarded as waste heat to the cold reservoir. Thermal efficiency is a key factor in environmental and energy economics, as it directly impacts fuel consumption and the cost of operating power plants and engines.
Mechanical Work
Mechanical work in the context of thermodynamics can be described as the amount of energy transferred by a force acting through a distance. In the realm of thermal machines like the Carnot engine, work is the energy converted from heat that is used to perform a physical task or produce motion.

The formula for mechanical work, when related to thermodynamics and heat engines, can be expressed as:
\[\begin{equation} \text{Work} = \text{Efficiency} \times \text{Heat Input} \end{equation}\] In our Carnot engine example, work is the useful energy output obtained after the engine operates through one cycle. The problem states that the engine has an efficiency of 42.3% and receives \(6450 \text{J}\) of heat from the hot reservoir. By using the efficiency, we can calculate the mechanical work as \(2726.35 \text{J}\) per cycle.

This concept is pivotal to understanding how different forms of energy are converted and utilized in various machines and engines. Mechanical work is a practical measure of the tangible output we can harness from heat engines and is central to the design and evaluation of these systems. In our everyday experience, this could translate into the work done by car engines, turbines, and even our bodies.

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Most popular questions from this chapter

If the proposed plant is built and prouluces \(10 \mathrm{MW}\) but the rate at which waste heat is exhausted to the cold water is \(165 \mathrm{MW}\), what is the plant's actual efficiency? (a) \(5.7 \% ;(\mathrm{b}) 6.1 \% ;(\mathrm{c}) 6.5 \% ;(\mathrm{d}) 16.5 \%\)

For a refrigerator or air conditioner, the coefficient of performance \(K\) (often denoted as \(\mathrm{COP}\) ) is, as in Eq. (20.9) , the ratio of cooling output \(\left|Q_{\mathrm{C}}\right|\) to the required electrical energy input \(|W|,\) both in joules. The cocfficicnt of performance is also expressed as a ratio of powers, $$ K=\frac{\left|Q_{\mathrm{c}}\right| / t}{|W| / t} $$ where \(\left|Q_{\mathrm{C}}\right| / t\) is the cooling power and \(|W| / t\) is the electrical power input to the device, both in watts. The energy efficiency ratio (FER) is the same quantity expressed in units of Btu for \(\left|Q_{\mathrm{C}}\right|\) and \(\mathrm{W} \cdot \mathrm{h}\) for \(|W|\) (a) Derive a general relationship that expresses EER in terms of \(K\). (b) For a home air conditioner, FER is generally determined for a \(95^{\circ} \mathrm{F}\) outside temperature and an \(80^{\circ} \mathrm{F}\) return air temperature. Calculate EER for a Carnot device that operates between \(95^{\circ} \mathrm{F}\) and \(80 \mathrm{~F}\). (c) You have an air conditioner with an EER of 10.9 . Your home on average requires a total cooling output of \(\left|Q_{\mathrm{C}}\right|=1.9 \times 10^{10} \mathrm{~J}\) per year. If electricity costs you 15.3 cents per \(\mathrm{kW} \cdot \mathrm{h}\), how much do you spend per year, on average. to operate your air conditioner? (Assume that the unit's EER accurately represcnts the operation of your air conditioner. A seasonal energy efficiency ratio (SFER) is often used. The SFFR is calculated over a range of outside temperatures to get a more accurate seasonal average.) (d) You are considering replacing your air conditioner with a more efficient one with an EER of \(14.6 .\) Based on the EER, how much would that save you on electricity costs in an average year?

To heat 1 cup of water \(\left(250 \mathrm{~cm}^{3}\right)\) to make coffee, you place an electric heating element in the cup. As the water lemperature increases from \(20^{\circ} \mathrm{C}\) to \(78^{\circ} \mathrm{C}\), the temperature of the heating element remains at a constant \(120^{\circ} \mathrm{C}\). Calculate the change in entropy of (a) the water, (b) the heating element; (c) the system of water and heating element. (Make the same assumption about the specific heat of water as in Example 20.10 in Section \(20.7,\) and ignore the heat that flows into the ceramic coffee cup itself.) (d) Is this process reversible or irreversible? Explain.

A Carnot engine whose high-temperature reservoir is at \(620 \mathrm{~K}\) takes in \(550 \mathrm{~J}\) of heat at this temperature in each cycle and gives up \(335 \mathrm{~J}\) to the low-temperature rescrvoir. (a) How much mechanical work does the cngine perform during cach cycle? What is (b) the temperature of the low-temperature reservoir; (c) the thermal cfficicncy of the cycle?

A cylinder contains oxygen at a pressure of \(2.00 \mathrm{~atm}\). The volume is \(4.00 \mathrm{I}_{2}\), and the temperature is \(300 \mathrm{~K}\). Assume that the oxygen may be treated as an ideal gas. The oxygen is carried through the following processes: (i) Heated at constant pressure from the initial state (state 1 ) to state 2 , which has \(T=450 \mathrm{~K}\). (ii) Cooled at constant volume to \(250 \mathrm{~K}\) (state 3 ). (iii) Compressed at constant temperature to a volume of \(4.00 \mathrm{~L}\) (state 4 ). (iv) Heated at constant volume to \(300 \mathrm{~K},\) which takes the system back to state 1 (a) Show these four processes in a \(p V\) -diagram, giving the numerical values of \(p\) and \(V\) in each of the four states. (b) Calculate \(Q\) and \(W\) for each of the four processes. (c) Calculate the net work done by the oxygen in the complete cycle. (d) What is the efficiency of this device as a heat engine? How does this compare to the efliciency of a Carnot-cycle engine operating between the same minimum and maximum temperatures of \(250 \mathrm{~K}\) and \(450 \mathrm{~K}\) ?

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