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(I) A heat engine exhausts \(7800 \mathrm{~J}\) of heat while performing \(2600 \mathrm{~J}\) of useful work. What is the efficiency of this engine?

Short Answer

Expert verified
The efficiency of the engine is 25%.

Step by step solution

01

Understanding the Given Values

We are given two key values. First, the exhaust heat which is the heat that is not converted into work: \[ Q_c = 7800 \, \text{J} \]Second, the useful work done by the engine: \[ W = 2600 \, \text{J} \]We need these values to find the efficiency of the engine.
02

Finding Total Input Heat

The first law of thermodynamics states that the total input heat energy (\(Q_h\)) into a system equals the sum of the work done (\(W\)) and the waste heat (\(Q_c\)).\[ Q_h = W + Q_c = 2600 \, \text{J} + 7800 \, \text{J} = 10400 \, \text{J} \]This is important because efficiency is calculated as the ratio of work done to total input heat.
03

Calculating Efficiency

The efficiency (\(\eta\)) of a heat engine is calculated using the formula: \[ \eta = \frac{W}{Q_h} \]Substituting the values we found:\[ \eta = \frac{2600}{10400} \]\[ \eta = 0.25 \] or 25%.This calculation shows how much of the input energy is converted into work.

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Key Concepts

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

First Law of Thermodynamics
The First Law of Thermodynamics is a fundamental principle of physics that relates to energy conservation. It states that energy cannot be created or destroyed, only transferred or changed from one form to another. In the context of a heat engine, the first law manifests itself in the way energy is accounted for in the system.

For a heat engine, the first law is expressed as:
  • Input Heat Energy (\[Q_h\]) = Work Done (\[W\]) + Waste Heat (\[Q_c\])
This equation underscores that the energy put into the system as heat (\[Q_h\]) gets converted into two outcomes: useful work done by the engine (\[W\]) and the remaining energy that is wasted or exhausted as heat (\[Q_c\]).

Understanding this law is crucial when calculating heat engine efficiency because it ensures the energy flows within the engine are properly accounted for. Recognizing how the input heat is divided helps engineers optimize engines to convert as much energy as possible into work.
Waste Heat
Waste heat is the portion of energy that is not converted into work by a heat engine and is instead transferred to the surroundings. This concept is key when discussing efficiency since waste heat can significantly impact an engine's performance. In our problem, the waste heat (\[Q_c\]) is given as 7800 J.

A large amount of waste heat indicates that a significant portion of input energy is not being utilized efficiently. This can be due to the inevitable energy losses caused by factors such as friction, thermal inefficiencies, and material limitations. Being able to manage and reduce waste heat is a critical goal for engineers.

Efficient systems aim to minimize waste heat, thus allowing more input energy to be converted into useful work. Methods such as improving insulation, designing better heat exchangers, and using high-efficiency materials can help mitigate waste heat.
Input Heat Energy
Input heat energy (\[Q_h\]) refers to the total heat energy supplied to a system, in this case, a heat engine. For analyzing and designing heat engines, it's essential to track the input heat energy, as it determines how much energy can potentially be transformed into work.

From the first law of thermodynamics, we know:
  • Total Input Heat Energy (\[Q_h\]) = \[2600 \, \text{J}\] (work done) + \[7800 \, \text{J}\] (waste heat)
  • Thus, \[Q_h = 10400 \, \text{J}\]
The efficiency of the engine depends on how effectively this input heat (\[10400 \, \text{J}\]) is converted into work (\[W\]). In our example, the efficiency is 25%, meaning a quarter of the energy is useful for performing work, while the rest is lost as waste heat.

Understanding input heat energy helps in identifying possible inefficiencies and working towards better engine designs that can utilize more of this input energy, ultimately leading to more sustainable and energy-efficient systems.

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Most popular questions from this chapter

A 35\(\%\) efficient power plant puts out 920 \(\mathrm{MW}\) of electrical power. Cooling towers are used to take away the exhaust heat. \((a)\) If the air temperature \(\left(15^{\circ} \mathrm{C}\right)\) is allowed to rise 7.0 \(\mathrm{C}^{\circ}\) , estimate what volume of air \(\left(\mathrm{km}^{3}\right)\) is heated per day. Will the local climate be heated significantly? (b) If the heated air were to form a layer 150 \(\mathrm{m}\) thick, estimate how large an area it would cover for 24 \(\mathrm{h}\) of operation. Assume the air has density 1.2 \(\mathrm{kg} / \mathrm{m}^{3}\) and that its specific heat is about 1.0 \(\mathrm{kJ} / \mathrm{kg} \cdot \mathrm{C}^{\circ}\) at constant pressure.

(II) (a) Show that the work done by a Carnot engine is equal to the area enclosed by the Carnot cycle on a \(P V\) diagram, Fig. 20-7. (See Section 19-7.) (b) Generalize this to any reversible cycle.

(II) A nuclear power plant operates at 65\(\%\) of its maximum theoretical (Carnot) efficiency between temperatures of \(660^{\circ} \mathrm{C}\) and \(330^{\circ} \mathrm{C}\) If the plant produces electric energy at the rate of \(1.2 \mathrm{GW},\) how much exhaust heat is discharged per hour?

(II) If \(0.45 \mathrm{~kg}\) of water at \(100^{\circ} \mathrm{C}\) is changed by a reversible process to steam at \(100^{\circ} \mathrm{C}\), determine the change in entropy of \((a)\) the water, \((b)\) the surroundings, and \((c)\) the universe as a whole. \((d)\) How would your answers differ if the process were irreversible?

Suppose a power plant delivers energy at \(850 \mathrm{MW}\) using steam turbines. The steam goes into the turbines superheated at \(625 \mathrm{~K}\) and deposits its unused heat in river water at \(285 \mathrm{~K}\). Assume that the turbine operates as an ideal Carnot engine. \((a)\) If the river's flow rate is \(34 \mathrm{~m}^{3} / \mathrm{s}\) estimate the average temperature increase of the river water immediately downstream from the power plant. (b) What is the entropy increase per kilogram of the downstream river water in \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K} ?\)

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